Selected solutions to Atiyah-Macdonald’s exercises

Abstract

A non-comprehensive collection of solutions to Atiyah-Macdonald’s exercises, with some additional comments and discussions. There may be errors, for reference only.

1 Chapter 2

Exercise 1 ([1]-exr-2.1) \(\mathbb Z / m \mathbb Z \otimes_{\mathbb Z} \mathbb Z / n \mathbb Z = 0\) when \(m, n\) are coprime.

Proof

Proof. By the Chinese Remainder Theorem, there exists integers \(a, b\) such that \[ a m + b n = 1 \] Then for any \(x \otimes y \in \mathbb Z / m \mathbb Z \otimes_{\mathbb Z} \mathbb Z / n \mathbb Z\), \[ x \otimes y = 1 \cdot (x \otimes y) = (a m + b n)(x \otimes y) = a m (x \otimes y) + b n (x \otimes y) = 0 + 0 = 0. \] Thus \(\mathbb Z / m \mathbb Z \otimes_{\mathbb Z} \mathbb Z / n \mathbb Z = 0\).

Exercise 2 ([1]-exr-2.3) For \(A\) a local ring, \(M, N\) finitely generated \(A\)-modules. If \(M \otimes_A N = 0\), then \(M=0\) or \(N=0\).

Proof

Proof. Tensoring with the residue field \(k := A / \mathfrak m\) (where \(\mathfrak m\) is the maximal ideal of \(A\)), we have \[ (M \otimes_A k) \otimes_k (N \otimes_A k) = 0 \] Since its a tensor product of vector spaces over field \(k\), we must have \(M \otimes_A k = 0\) or \(N \otimes_A k = 0\), by dimension counting. Without loss of generality, say \(M \otimes_A k = 0\). By Nakayama’s lemma, this implies \(M=0\).

Exercise 3 ([1]-exr-2.11) Let \(A\) be a nonzero ring. Show that \(A^m \cong A^n\) iff \(m=n\).

Additionally, show that

• if \(\varphi: A^m \to A^n\) is surjective, then \(m \geq n\).
• If \(\varphi\) is injective, is it true that \(m \leq n\)?

Proof

Proof. By taking tensor product with \(k := A / \mathfrak m\) (where \(\mathfrak m\) is any maximal ideal of \(A\), existence by non-zeroness) and counting dimensions as vector spaces. The first additional question by the right-exactness of tensor product. The second question seems rather hard [TODO].

Exercise 4 ([1]-exr-2.14: Direct limits) Definition of direct limit of a direct system of \(A\)-modules \((M_i, \mu_{i j})_{i, j \in I}\) indexed by a directed set \(I\). \(M := C / D\) where \(C\) is the direct sum and \(D\) is the relations.

Proof

Proof. There’s nothing to prove.

Exercise 5 ([1]-exr-2.15) Every element of \(M := \varinjlim_i M_i\) is of the form \(\mu_i (x_i)\) for some \(i \in I\) and \(x_i \in M_i\).

If \(\mu_i (x_i) = 0\) in \(M\), then \(\exists j \geq i\) s.t. \(\mu_{i j} (x_i) = 0\) in \(M_j\).

Proof

Proof. Say \(M \ni x = \sum_i \mu_i (x_i)\), where \(i\) runs over a finite subset of \(I\) and \(x_i \in M_i\). By directedness of \(I\), \(\exists k \in I\) s.t. \(k \geq i\) for all \(i\) in the finite subset. Then \(\mu_k \left( \sum_i \mu_{i k} (x_i) \right) = x\).

If \(\mu_i (x_i) = 0\) in \(M\), then \(x_i \in M_i \cap D\), that is, a finite sum in \(C\) \[ x_i = \sum_{j<k} y_{jk} - \mu_{jk}(y_{jk}) = \sum_{k} \left( \sum_{k<l} y_{kl} - \sum_{j<k} \mu_{jk} (y_{jk}) \right) \] Comparing the summation componentwise in \(C\) gives \[ \begin{aligned} x_i &= \sum_{i<l} y_{il} - \sum_{j<i} \mu_{ji} (y_{ji}) \\ 0 &= \sum_{k<l} y_{kl} - \sum_{j<k} \mu_{jk} (y_{jk}) \quad \text{ for all } k \neq i \end{aligned} \] Let \(p\) be an upper bound of all indices appearing in the above equations. Apply \(\mu_{i p}\) to the first equation and \(\mu_{k p}\) to the second equation for all \(k \neq i\). We get \[ \begin{aligned} \mu_{i p} (x_i) &= \sum_{i<l} \mu_{i p} (y_{il}) - \sum_{j<i} \mu_{j p} (y_{ji}) \\ 0 &= \sum_{k<l} \mu_{k p} (y_{kl}) - \sum_{j<k} \mu_{j p} (y_{jk}) \quad \text{ for all } k \neq i \end{aligned} \] Summing up all these equations gives \(\mu_{i p} (x_i) = 0\).

Exercise 6 ([1]-exr-2.16) The universal property of direct limits.

Proof

Proof. “Up to isomorphism” part by abstract nonsense of category theory.

Say we have \(\alpha_i : M_i \to N\) s.t. \(\alpha_j \circ \mu_{i j} = \alpha_i\) for all \(i \leq j\). Define \(\varphi : C \to N\) by \(\varphi \circ \mu_i := \alpha_i\). By the universal property of direct sum, such \(\varphi\) is well-defined. Also note that this is the only possible definition of \(\varphi\) satisfying \(\varphi \circ \mu_i = \alpha_i\). This uniqueness also passes down to the quotient \(M = C/D\). Say the quotient map is \(\bar \varphi : M = C / D \to N\), it remains to show that \(\bar \varphi\) is well-defined. For any generator of \(D\), say \(x_j - \mu_{j k} (x_j)\) for some \(j \leq k\) and \(x_j \in M_j\), we have \(\varphi(x_j - \mu_{j k} (x_j)) = \alpha_j (x_j) - \alpha_k (\mu_{j k} (x_j)) = 0\).

Exercise 7 ([1]-exr-2.17) Any \(A\)-module is a direct limit of its finitely generated \(A\)-submodules.

Proof

Proof. It’s clear that \(\sum_i M_i = \bigcup_i M_i\) when the direct system is directed by inclusion. It’s also easy to verify that \(\bigcup_i M_i\) satisfies the universal property of direct limits.

Exercise 8 ([1]-exr-2.18) \(\mathbf M := (M_i, \mu_{i j})\), \(\mathbf N := (N_i, \nu_{i j})\) direct systems of \(A\)-modules indexed by the same directed set \(I\). Let \(M\), \(N\) be their direct limits and \(\mu_i\), \(\nu_i\) be the canonical maps, respectively.

A homomorphism \(\Phi : \mathbf M \to \mathbf N\) is defined to be a family of homomorphisms \(\varphi_i : M_i \to N_i\) satisfying \(\nu_{i j} \circ \varphi_i = \varphi_j \circ \mu_{i j}\) for all \(i \leq j\). Show that \(\Phi\) defines a unique homomorphism \(\varphi = \varinjlim \Phi : M \to N\) such that \(\varphi \circ \mu_i = \nu_i \circ \varphi_i\) for all \(i\).

That is, taking direct limits is a functor from the category of direct systems of \(A\)-modules to the category of \(A\)-modules.

Proof

Proof. Let \(\bar \varphi_i\) be the composition \(\nu_i \circ \varphi_i : M_i \to N_i \to N\). Define our map \(\varphi : M \to N\) by these maps, the required compatibility condition \(\varphi \circ \mu_i = \nu_i \circ \varphi_i = \bar \varphi_i\) is guaranteed by hypothesis.

Exercise 9 ([1]-exr-2.19) Say \(\mathbf M := (M_i)_{i \in I}\), \(\mathbf N := (N_i)_{i \in I}\), \(\mathbf P := (P_i)_{i \in I}\) are direct systems of \(A\)-modules indexed by the same directed set \(I\). If the sequence of direct systems \[ \mathbf M \longrightarrow \mathbf N \longrightarrow \mathbf P \] is exact, then the sequence of direct limits \[ M \longrightarrow N \longrightarrow P \] is also exact. That is, taking direct limits is an exact functor from the category of direct systems of \(A\)-modules to the category of \(A\)-modules.

Proof

Proof. Say the maps in the direct systems are \(\varphi_i : M_i \to N_i\) and \(\psi_i : N_i \to P_i\). The maps in the direct limits are \(\varphi : M \to N\) and \(\psi : N \to P\) induced by \(\varphi_i\) and \(\psi_i\) respectively. The canonical maps are \(\mu_i : M_i \to M\), \(\nu_i : N_i \to N\) and \(\pi_i : P_i \to P\). These maps make the following diagram commute:

Figure 1

We do diagram chasing. First, We have \((\psi \circ \varphi) \circ \mu_i = \psi \circ \nu_i \circ \varphi_i = (\pi_i \circ \psi_i) \circ \varphi_i = 0\). By the universal property of direct limits, \(\psi \circ \varphi = 0\). Second, for any \(y \in N\) with \(\psi(y) = 0\), say \(y = \nu_i (y_i)\) for some \(i \in I\) and \(y_i \in N_i\) (by Exercise 5). Then by the exactness at \(N_i\), there exists \(x_i \in M_i\) s.t. \(\varphi_i (x_i) = y_i\). Let \(x := \mu_i (x_i) \in M\), then \(\varphi(x) = \varphi \circ \mu_i (x_i) = \nu_i \circ \varphi_i (x_i) = \nu_i (y_i) = y\), hence the exactness at \(N\) follows.

Exercise 10 ([1]-exr-2.20) show that \[ \varinjlim_{i}\left(M_i \otimes N\right) \cong \left(\varinjlim_{i} M_i\right) \otimes N \]

i.e. tensor product commutes with direct limits.

Proof

Proof. It is known from category theory that a right exact functor commutes with colimits [TODO: prove this], but we write some concrete nonsense here anyway.

Suffices to show the universal property of direct limits holds for \(\left(\varinjlim_{i} M_i\right) \otimes N\) with respect to the direct system \(\left(M_i \otimes N, \mu_{ij} \otimes \mathrm{id}_N\right)_{i,j \in I}\).

First, determine the canonical inclusions. Define \(M_i \otimes N \to \left( \varinjlim_{i} M_i \right) \otimes N\) by \(\mu_i \otimes \mathrm{id}_N\), i.e. sending \(m_i \otimes n \mapsto \mu_i(m_i) \otimes n\) for \(m_i \in M_i\), \(n \in N\).

Next, verify the universal property. Say we have maps \(\alpha_i : M_i \otimes N \to P\) satisfying the compatibility condition \(\alpha_i = \alpha_j \circ (\mu_{i j} \otimes \mathrm{id}_N)\) for all \(i \leq j\). To define a map \(\bar \alpha_i : \left(\varinjlim_{i} M_i\right) \otimes N \to P\), it suffices to give a map \(\varinjlim_{i} M_i \to \operatorname{Hom}_A(N, P)\). This in turn require us to define maps \(M_i \to \operatorname{Hom}_A(N, P)\), and we define it as \(m_i \mapsto (n \mapsto \alpha_i(m_i \otimes n))\). One verifies they commutes with transition maps \(\mu_{i j}\) thanks to the compatibility condition \(\alpha_i = \alpha_j \circ (\mu_{i j} \otimes \mathrm{id}_N)\), thus the map \(\varinjlim_{i} M_i \to \operatorname{Hom}_A(N, P)\) is well-defined, an in turn \(\bar \alpha_i : \left(\varinjlim_{i} M_i\right) \otimes N \to P\) is defined.

Finally, note that this choice of definition satisfies, and is unique to satisfy the universal property condition \(\bar \alpha_i \circ (\mu_i \otimes \mathrm{id}_N) = \alpha_i\). Hence we have verified the universal property of \(\left(\varinjlim_{i} M_i\right) \otimes N\) (with the defined canonical inclusions) w.r.t. the direct system \(\left(M_i \otimes N, \mu_{ij} \otimes \mathrm{id}_N\right)_{i,j \in I}\).

Exercise 11 ([1]-exr-2.21) \((A_i)_{i \in I}\) a family of rings indexed by a directed set \(I\). For each \(i \leq j\), let \(\alpha_{i j} : A_i \to A_j\) be a ring homomorphism such that \(\alpha_{i i} = \mathrm{id}_{A_i}\) and \(\alpha_{j k} \circ \alpha_{i j} = \alpha_{i k}\) for all \(i \leq j \leq k\). As \(\mathbb Z\)-module we may view them as a direct system and obtain their direct limit \(A := \varinjlim_i A_i\). Show that \(A\) has a natural ring structure such that the canonical maps \(\alpha_i : A_i \to A\) are ring homomorphisms. This defines the direct limit in the category of rings.

In addition, show that \(A=0\) implies \(A_i = 0\) for some \(i\).

Proof

Proof. We define the multiplication on \(A\) as follows. For any \(x, y \in A\), say \(x = \alpha_i (x_i)\) and \(y = \alpha_j (y_j)\) for some \(i, j \in I\) and \(x_i \in A_i\), \(y_j \in A_j\) (by Exercise 5). Let \(k \in I\) be an upper bound of \(i, j\). Define \(x \cdot y := \alpha_k (\alpha_{i k} (x_i) \cdot \alpha_{j k} (y_j))\). We verify that this definition is independent of the choice of representatives \(x_i, y_j\) and the choice of upper bound \(k\).

If we have another choice of representatives \(x_{i'} \in A_{i'}\), \(y_{j'} \in A_{j'}\) for some \(i', j' \in I\) with \(x = \alpha_{i'} (x_{i'})\) and \(y = \alpha_{j'} (y_{j'})\). Let \(k' \in I\) be an upper bound of \(i', j'\). First we claim that there exist an \(l \in I\) s.t. \(\alpha_{i l} (x_i) = \alpha_{i l} (x_{i'})\). Indeed, since \(\alpha_k (\alpha_{i k} (x_i)) = x = \alpha_k (\alpha_{i'k} (x_{i'}))\), by Exercise 5 there exists an upper bound \(l\) of \(k, i'\) s.t. \(\alpha_{i' l} (x_{i'}) = \alpha_{i l} (x_i)\). Similarly, we can find an upper bound s.t. \(x_j\) and \(x_{j'}\) agree after applying the transition map to that upper bound. WLOG, still use the letter \(l\) be an upper bound of these two upper bounds. Then \[ \begin{aligned} \alpha_k (\alpha_{i k} (x_i) \cdot \alpha_{j k} (y_j)) &= \alpha_l (\alpha_{k l} (\alpha_{i k} (x_i) \cdot \alpha_{j k} (y_j))) \\ &= \alpha_l (\alpha_{i l} (x_i) \cdot \alpha_{j l} (y_j)) \\ &= \alpha_l (\alpha_{i' l} (x_{i'}) \cdot \alpha_{j' l} (y_{j'})) \\ &= \alpha_{k'} (\alpha_{i' k'} (x_{i'}) \cdot \alpha_{j' k'} (y_{j'})) \end{aligned} \] Thus the definition is independent of the choice of representatives and upper bounds.

\(\alpha_i\) is a ring homomorphism is straightforward to verify.

if \(A=0\), then for any \(i \in I\), \(\alpha_i (1_{A_i}) = 0\) in \(A\). By Exercise 5, there exists an upper bound \(j \in I\) of \(i\) s.t. \(\alpha_{i j} (1_{A_i}) = 0\) in \(A_j\). But the ring homomorphism \(\alpha_{i j}\) must preserve identity, thus \(A_j = 0\).

Exercise 12 ([1]-exr-2.22) \((A_i, \alpha_{ij})\) direct system of rings. Let \(\mathfrak R_i\) be the nilradical of \(A_i\). Show that \(\varinjlim_i \mathfrak R_i\) is the nilradical of \(\varinjlim_i A_i\).

As a corollary, if each \(A_i\) is an integral domain, so is \(\varinjlim A_i\).

Proof

Proof. For \(i \in I\), \(a_i \in A_i\), TFAE:

• \(\alpha_i(a_i) \in \mathfrak R(\varinjlim_j A_j)\)
• Exists some \(n\) s.t. \(\alpha_i(a_i^n) = 0\)
• Exists some \(n\), \(j \geq i\) s.t. \(\alpha_{i j} (a_i^n) = 0\) (by Exercise 5)
• Exists some \(j \geq i\) s.t. \(\alpha_{i j} (a_i) \in \mathfrak R_j\)
• \(\alpha_i (a_i) \in \mathfrak \varinjlim_j \mathfrak R_j\)

Thus \(\mathfrak R(\varinjlim_j A_j) = \varinjlim_j \mathfrak R_j\) by Exercise 5.

Exercise 13 ([1]-exr-2.23) \((B_\lambda)_{\lambda \in \Lambda}\) a family of \(A\)-algebras. For each finite subset of \(\Lambda\), let \(B_J\) be the tensor product of the \(B_\lambda\) for \(\lambda \in J\). By inclusion of \(J\), all \(B_J\) form a direct system. Let \(B\) denote its direct limit. Then \(B\) has a natrual \(A\)-algebra structure with \(B_J \to B\) are \(A\)-algebra homomorphisms. This \(B\) is called the tensor product of the family \((B_\lambda)_{\lambda \in \Lambda}\) over \(A\).

Proof

Proof. There is nothing to prove.

Exercise 14 ([1]-exr-2.24) \(M\) is flat iff \(\operatorname{Tor}_i(M, N)=0\) for all \(N\) iff \(\operatorname{Tor}_1(M,N)=0\) for all \(N\).

Proof (i \(\implies\) ii:)

Proof (i \(\implies\) ii:). Say \[ 0 \rightarrow K \rightarrow P \rightarrow N \] where \(P\) is projective. Write the long exact sequence: \[ \begin{matrix} \operatorname{Tor}_2(K, M) & \longrightarrow & \operatorname{Tor}_2 (P, M) &\longrightarrow & \operatorname{Tor}_2(N, M) & \longrightarrow \\ \operatorname{Tor}_1(K, M) & \longrightarrow & \operatorname{Tor}_1(P, M) &\longrightarrow & \operatorname{Tor}_1(N, M) & \longrightarrow \\ K \otimes M & \longrightarrow & P \otimes M &\longrightarrow & N \otimes M & \longrightarrow & 0 \end{matrix} \]

Since \(-\otimes M\) is an exact functor. \(K \otimes M \rightarrow P \otimes M\) is injective, i.e. \(\operatorname{Tor}_1(N, M) \xrightarrow{0} K \otimes M\). Meanwhile, \(\operatorname{Tor}_1(P,N)=0\) since \(P\) is projective. This forces \(\operatorname{Tor}_1(N, M)=0\). Thus for all \(N\), \(\operatorname{Tor}^1(N, M)=0\). In particular, \(\operatorname{Tor}^1(K, M)=0\). With \(\operatorname{Tor}_2(P, M)=0\), this fonces \(\operatorname{Tor}^2(N, M)=0\). Then inductively work upward gives \(\operatorname{Tor}_i(N, M)=0\) for all \(i \geq 1\).

Proof (ii \(\implies\) iii)

Proof (ii \(\implies\) iii). Trivial.

Proof (iii \(\implies\) i)

Proof (iii \(\implies\) i). The long exact sequence.

Exercise 15 ([1]-exr-2.25) \[ 0 \longrightarrow N_1 \longrightarrow N_2 \longrightarrow N_3 \longrightarrow 0 \] is exact. \(N_3\) is flat. Then \(N_1\) is flat iff \(N_2\) is flat.

Proof

Proof. Write the long exact sequence: \[ \begin{matrix} \operatorname{Tor}_2(N_1, M) &\longrightarrow& \operatorname{Tor}_2 (N_2, M) &\longrightarrow& \operatorname{Tor}_2(N_3, M) &\longrightarrow& \\ \operatorname{Tor}_1(N_1, M) &\longrightarrow& \operatorname{Tor}_1(N_2, M) &\longrightarrow& \operatorname{Tor}_1(N_3, M) &\longrightarrow& \\ N_1 \otimes M &\longrightarrow& N_2 \otimes M &\longrightarrow& N_3 \otimes M &\longrightarrow& 0 \end{matrix} \]

\(\operatorname{Tor}_i(N_3, M)=0\) by \(N_3\) is flat.

If \(N_1\) is flat. i.e. \(\operatorname{Tor}_i(N_1, M)=0\), then this force \(\operatorname{Tor}_i(N_2, M)=0\). i.e. \(N_2\) is flat.

Similarly for the case that \(N_2\) is flat.

Exercise 16 ([1]-exr-2.26: Characterization of flatness via finitely generated ideals) \(N\) is an \(A\)-module. Then \(N\) is flat iff \(\operatorname{Tor}_1(A/\mathfrak a, N) = 0\) for all finitely generated ideals \(\mathfrak a\) in \(A\).

Proof (\(\implies\))

Proof (\(\implies\)). Trivial (by Exercise 14).

Proof (\(\impliedby\))

Proof (\(\impliedby\)). The hypothesis is that for all finitely generated ideals \(\mathfrak a\) of \(A\), the sequence \[ 0 \longrightarrow \mathfrak a \otimes N \longrightarrow A \otimes N \longrightarrow A / \mathfrak a \otimes N \longrightarrow 0 \] is exact.

We first prove that for any ideal \(\mathfrak a\) we have \(\operatorname{Tor}_1(A/\mathfrak a, N)=0\). Consider the directed system of all finitely generated subideals of \(\mathfrak a\), say \((\mathfrak a_i)_{i \in I}\). It’s direct limit is \(\mathfrak a\). Then we have an exact sequences of direct systems: \[ 0 \longrightarrow (\mathfrak a_i)_{i \in I} \otimes N \longrightarrow \mathbf A \otimes N \longrightarrow (A / \mathfrak a_i)_{i \in I} \otimes N \longrightarrow 0 \] Taking direct limits, we get an exact sequence \[ 0 \longrightarrow \mathfrak a \otimes N \longrightarrow A \otimes N \longrightarrow A / \mathfrak a \otimes N \longrightarrow 0 \] (by Exercise 9 exactness of taking direct limits, Exercise 10 commutativity of direct limit and tensor product). This shows that \(\operatorname{Tor}_1(A/\mathfrak a, N)=0\) for any ideal \(\mathfrak a\). i.e. \(\operatorname{Tor}_1(A/\mathfrak a, N)=0\).

Note any cyclic \(A\)-modules are naturally isomorphic to \(A/\mathfrak a\) for some ideal \(\mathfrak a\) of \(A\), thus \(\operatorname{Tor}_1(M, N)=0\) for any cyclic \(A\)-module \(M\).

Next we show that for any finitely generated \(A\)-module \(M\), \(\operatorname{Tor}_1(M, N)=0\). Consider a filtration of \(M\): \[ 0 = M_0 \subset M_1 \subset \cdots \subset M_m=M \] such that each successive quotient \(M_{i} / M_{i-1}\) is cyclic. Such a filtration exists by taking generators of \(M\) step by step. Consider the short exact sequence \[ 0 \longrightarrow M_{i-1} \longrightarrow M_i \longrightarrow M_i / M_{i-1} \longrightarrow 0 \] for each \(1 \leq i \leq m\). Since \(M_i / M_{i-1}\) is cyclic, we have \(\operatorname{Tor}_1(M_i / M_{i-1}, N)=0\). By the proof of Exercise 15, we have \(\operatorname{Tor}_1(M_i, N) \cong \operatorname{Tor}_1(M_{i-1}, N)\). Thus \[ \operatorname{Tor}_1(M, N) \cong \operatorname{Tor}_1(M_{m-1}, N) \cong \cdots \cong \operatorname{Tor}_1(M_1, N) \cong \operatorname{Tor}_1(M_0, N)=0. \]

Finally, for any \(A\)-module \(M\), we can write it as a direct limit of its finitely generated submodules, say \((M_i)_{i \in I}\). Then by a similar argument as above using Exercise 9 and Exercise 10 applied to the exact sequences \[ 0 \longrightarrow \operatorname{Tor}_1(M_i, N) \longrightarrow K_i \otimes N \longrightarrow P_i \otimes N \longrightarrow M_i \otimes N \longrightarrow 0 \] where \(P_i\) is a free module, we know that for all \(A\)-module \(M\), \(\operatorname{Tor}_1(M, N)=0\) and hence \(N\) is flat.

Remark

Remark. The last step essentially shows that the \(\operatorname{Tor}\) functor commutes with direct limits, i.e. for a directed system of \(A\)-modules \((M_i)_{i \in I}\), \[ \operatorname{Tor}_n\left(\varinjlim_i M_i, N\right) \cong \varinjlim_i \operatorname{Tor}_n\left(M_i, N\right) \] for all \(n \geq 0\). In particular, \(n=0\) is exactly the result of Exercise 10. [TODO] The proof above is just a vague sketch for now.

1.1 Extra explorations on Tor and flatness

The etymology of Tor: Quotient perspective

Say \(A\) is an integral domain and \(\mathfrak a\) is a principal ideal generated by \(a \in A\). Then from the short exact sequence \[ 0 \longrightarrow A \xrightarrow{a \cdot -} A \longrightarrow A / \mathfrak a \longrightarrow 0 \] tensoring \(N\) we obtain the long exact sequence (leftmost item is zero since \(A\) is projective) \[ 0 \longrightarrow \operatorname{Tor}_1(A / \mathfrak a, N) \longrightarrow N \xrightarrow{a \cdot -} N \longrightarrow N / \mathfrak a N \longrightarrow 0 \] We see that \(\operatorname{Tor}_1(A / \mathfrak a, N) \cong \ker (a \cdot -) =: N[a]\), the \(a\)-torsion submodule of \(N\). This explains the etymology of \(\operatorname{Tor}\).

The etymology of Tor: Localization perspective

Wikipedia if you want a quick reference for this part.

Another (somehow dual) way of understanding \(\operatorname{Tor}\) is via localization. Say \(A_S\) is any localization of \(A\) with respect to a multiplicative set \(S\). Consider the short exact sequence \[ 0 \longrightarrow A \longrightarrow A_S \longrightarrow A_S / A \longrightarrow 0 \] Tensoring \(N\) to obtain the long exact sequence (leftmost item is zero since \(A_S\) as a localization of \(A\) is flat) \[ 0 \longrightarrow \operatorname{Tor}_1(A_S / A, N) \longrightarrow N \longrightarrow S^{-1} N \longrightarrow (A_S / A) \otimes N \longrightarrow 0 \] We see that \(\operatorname{Tor}_1(A_S / A, N) \cong \ker (N \to S^{-1} N)\). Thus \(\operatorname{Tor}_1(A_S / A, N)\) measures those elements in \(N\) that collapse to zero upon localization \(N \to S^{-1} N\). What are these elements? When \(A\) is an integral domain, they are precisely those elements \(x \in N\) such that there exists \(s \in S\) with \(s \cdot x = 0\), i.e. the \(S\)-torsion submodule of \(N\).

In particular, if we take the localization to be the field of fractions \(k\), then we have realize the torsion submodule \(T(M)\) of \(M\) as \(\operatorname{Tor}_1(k / A, M)\), or the kernel of the localization map \(M \to k \otimes M\). This gives another (probably better) explanation of the etymology of \(\operatorname{Tor}\).

Another example is to consider the localization at a prime ideal \(\mathfrak p\), i.e. \(S = A \setminus \mathfrak p\). In this case the terminology goes a little subtle… [TODO] continue this thought on PID.

It worths noting that the kernel of \(k \otimes N \to (k / A) \otimes N\) captures the torsion-free part of \(N\): \[ \ker (k \otimes N \to (k / A) \otimes N) \cong \operatorname{im}(N \to k \otimes N) \cong N / \ker (N \to k \otimes N) \cong N / T(N) \] [TODO] continue this thought.

Characterizing flatness

We continue our discussion in Section 1.1.1. Recall Exercise 16 suggests that, to characterize the flatness of \(A\)-module \(N\), it suffices to understand \(\operatorname{Tor}_1(A / \mathfrak a, N)\) for all finitely generated ideals \(\mathfrak a\) of \(A\). So we ask: If \(A\) is not in general an integral domain, and \(\mathfrak a\) is not in general principal, what is the meaning of \(\operatorname{Tor}_1(A / \mathfrak a, N)\)?

In this case, we can still use the short exact sequence \[ 0 \longrightarrow \mathfrak a \longrightarrow A \longrightarrow A / \mathfrak a \longrightarrow 0 \] to obtain the long exact sequence \[ 0 \longrightarrow \operatorname{Tor}_1(A / \mathfrak a, N) \longrightarrow \mathfrak a \otimes N \xrightarrow{- \cdot -} N \longrightarrow A / \mathfrak a N \longrightarrow 0 \] From this we see that \(\operatorname{Tor}_1(A / \mathfrak a, N) \cong \ker(\mathfrak a \otimes N \xrightarrow{- \cdot -} N)\), which can be viewed as the module of relations among the elements of \(\mathfrak a\) when they act on \(N\) via the \(A\)-scalar multiplication. This can be interpreted as somehow a generalization of the torsion elements in the principal ideal case.

Combine this with Exercise 16, one may say that the flatness of \(A\)-module \(N\) is precisely characterized by the fact that any finite subset \(\{ a_1, \dots, a_n \} \subseteq A\) acts on \(N\) without introducing any “\(\mathfrak a\)-torsion” or “\(\mathfrak a\)-relations”, where \(\mathfrak a\) is the ideal generated by \(\{ a_1, \dots, a_n \}\). This intuition can be extremely useful, as the examples below will show.

Example 1 (Pure ideals)  

• For \(n \geq 2\), \(\mathbb Z / n \mathbb Z\) is not a flat \(\mathbb Z\)-module because it is not torsion-free.

  Recall that a finitely generated module over a PID is free iff it is torsion-free, and in particular flat.

• \(2 \mathbb Z / 6 \mathbb Z\) is a flat \(\mathbb Z / 6 \mathbb Z\)-module, not only by that it is projective (a direct summand of \(\mathbb Z / 6 \mathbb Z\)), but also because

  • \(2\) won’t annihilate anything in \(2 \mathbb Z / 6 \mathbb Z\).
  • Dispite that \(3\) annihilates \(2 \mathbb Z / 6 \mathbb Z\), we have \(3 \mathbb Z / 6 \mathbb Z \otimes_{\mathbb Z / 6 \mathbb Z} 2 \mathbb Z / 6 \mathbb Z = 0\). For example, \(3 \otimes 2 = 3 \otimes 8 = 6 \otimes 4 = 0\). Thus \(3\)-torsion won’t give any nontrivial relation in this case.

  Blame \(\mathbb Z / 6 \mathbb Z\) for this subtlety! Recall that over an integral domain, any flat module is torsion-free. But \(\mathbb Z / 6 \mathbb Z\) is not an integral domain. In fact, any direct product of nontrivial rings ruins this property.

• Say \(k\) is a field, \(A := k[x,y]\), \(\mathfrak a\) is any nontrivial ideal of \(A\). Then \(A / \mathfrak a\) is not a flat \(A\)-module.

  In fact, pick any nonzero \(a \in \mathfrak a\), then \(a \otimes 1 \in \mathfrak a \otimes_A A / \mathfrak a\) is a nonzero element in the kernel of the multiplication map \(\mathfrak a \otimes_A A / \mathfrak a \to A / \mathfrak a\).

  Note that \(a \otimes 1\) being nonzero is guaranteed by the fact that \(A\) is an integral domain. Looks like arguments above can be generalized to any integral domain.

  Geometrically, this is an embedding of varieties \(\operatorname{Spec}A / \mathfrak a \hookrightarrow \operatorname{Spec}A\).

Refer to the Stacks projects 04PQ for general discussions on when a quotient ring \(R/J\) is flat over \(R\). Refer to 04PU, 04PV and 04PW for an algebraic geometry perspective.

Example 2 (Localizations) Any localization \(S^{-1}A\) of a ring \(A\) is always a flat \(A\)-module, because \(S^{-1}A \otimes_A \mathfrak a\) is natrually isomorphic to \(S^{-1}\mathfrak a\).

[TODO] Geometric examples.

Example 3 (A non-flat projection) Let \(k\) be a field and \(A = k[x,y] / (xy)\), \(B = k[t]\). Define an \(B\)-module structure on \(A\) via the homomorphism \(\varphi: B \to A\), \(t \mapsto x\) (that is, identifying \(x\) in \(A\) with \(t\)). Then \(A\) is not a flat \(B\)-module.

This is because, if we pick \(\mathfrak t := (t) = tB\) an ideal of \(B\), then \(y\) is a \(t\)-torsion element in \(A\) (by \(t \cdot y = \varphi(t) y = xy = 0\)). That is, \(t \otimes_{\mathfrak t} y\) is a nonzero element in the kernel of the multiplication map \(\mathfrak t \otimes_B A \to A\).

Geometrically, this corresponds to projecting the union of two coordinate axes \(\operatorname{Spec}A\) onto the \(x\)-axis, i.e. \(\operatorname{Spec}\varphi : (x,y) \mapsto x\). Picking \(\mathfrak t = (t)\) corresponds to looking at the fibers around \(t=0\). Note that the fiber over the origin \((0,0)\) is the entire \(y\)-axis, while fibers over other points are singletons. This “jumping” of fiber dimension is reflected algebraically by the non-flatness of \(B\) as an \(A\)-module.

Figure 2: \(\operatorname{Spec}\varphi\) and its fibers

Example 4 (A non-flat parametrization) Let \(k\) be a field and \(A = k[x,y] / (y^2-x^3)\), \(B=k[t]\). Define an \(A\)-module structure on \(B\) via the homomorphism \(\varphi: A \to B\), \((x,y) \mapsto (t^2, t^3)\). Then \(B\) is not a flat \(A\)-module.

This is because, if we pick \(\mathfrak a := (x,y)\) an ideal of \(A\), then \(x \cdot t = y \cdot 1_B\) becomes a nontrivial \(\mathfrak a\)-relation in \(B\), i.e. \(x \otimes_A t - y \otimes_A 1_B\) is a nonzero element in the kernel of the multiplication map \(\mathfrak a \otimes_A B \to B\).

Geometrically, this corresponds to the fact that the parametrized curve \(\operatorname{Spec}\varphi : t \mapsto (t^2, t^3)\) has a cusp at the origin. Picking \(\mathfrak a = (x,y)\) corresponds to looking at the fibers of this map around the origin. [TODO] Understand this thoroughly.

Figure 3: \(\operatorname{Spec}\varphi\)

Exercise 17 ([1]-exr-2.27: Absolute flatness) A ring \(A\) is absolutely flat (or von Neumann regular) if every \(A\)-module is flat. Show that TFAE:

• 1. \(A\) is absolutely flat;
• 2. every principal ideal of \(A\) is idempotent;
• 3. every finitely generated ideal of \(A\) is a direct summand of \(A\).

Proof

Proof. i \(\implies\) ii: Consider the commutative diagram

As inclusion, the left vertical arrow is injective, hence so do the right vertical arrow. As projection, the bottom horizontal arrow is surjective, hence so do the top horizontal arrow. Note that \((x) \to A \to A / (x)\) is zero, this forces the top horizontal arrow to be zero. But it is also surjective, hence \((x) \otimes A / (x) = 0\). Now note that \((x) \otimes A / (x) \cong (x) / (x^2)\), hence \((x) = (x^2)\).

ii \(\implies\) iii: First we show that every principal ideal \((x)\) is generated by an idempotent element \(e\). Since \((x) = (x^2)\), we have \(x = a x^2\), then \(e := ax\) is idempotent and \((x) = (e)\). Next, for any two principal ideals \((e)\) and \((f)\) generated by idempotent elements \(e\) and \(f\), we have \((e,f) = (e+f - ef)\) which is also generated by an idempotent element. Then inductively we know that any finitely generated ideal is generated by an idempotent element, hence is a direct summand of \(A\) since \(A \cong A / (e) \oplus A / (1-e) \cong (1-e) \oplus (e)\).

iii \(\implies\) i: By Exercise 16, to verify any \(A\)-module \(M\) is flat, it suffices to verify the flatness for for any finitely generated ideal \(\mathfrak a\). Say \(\mathfrak a \oplus \mathfrak b = A\), then \[ 0 \longrightarrow \mathfrak a \longrightarrow A \] tensoring \(M\) gives the sequence \[ 0 \longrightarrow \mathfrak a \otimes M \longrightarrow A \otimes M = \mathfrak a \otimes M \oplus \mathfrak b \otimes M \] hence is exact.

Exercise 18 ([1]-exr-2.28) A Boolean ring is absolutely flat.

Proof

Proof. Because in a Boolean ring, every element is idempotent.

2 Chapter 3

Exercise 19 ([1]-exr-3.2) Let \(\mathfrak a\) be an ideal of a ring \(A\), and let \(S = 1 + \mathfrak a\). Show that \(S^{-1} \mathfrak a\) is contained in the Jacobson radical of \(S^{-1} A\).

Proof

Proof. Easily verify that \(S\) is an multiplicative set. It suffices to prove for all \(\frac{a_1}{1+a_2} \in S^{-1} a\), \(1+\frac{a_1}{1+a_2} \cdot \frac{x}{1+a_3}\) is a unit in \(S^{-1} A\) for all \(\frac{x}{1+a_3} \in S^{-1} A\). The inverse \(\frac{\left(1+a_2\right)\left(1+a_3\right)}{\left(1+a_2\right)\left(1+a_3\right)+a_1 x}\) would do.

Proving [1]-cor-2.5: \(M\) f.g., \(\mathfrak a\) is an ideal of \(A\) s.t. \(\mathfrak a M=M\). Taking \(S^{-1}\) gives \[ \left(S^{-1} \mathfrak a \right)\left(S^{-1} M\right)=\left(S^{-1} \mathfrak a \right) M=S^{-1} M \] By Nakayama’s lemma, \(S^{-1} M=0\). This shows that \(\exists s \in S\) s.t. \(s M=0\) (by [1]-exr-3.1). Say \(s=1+a\). Done.

Exercise 20 ([1]-exr-3.12: Torsion and torsion-free modules) Let \(A\) be an integral domain, \(M\) be a finitely generated \(A\)-module, and \(T(M)\) the torsion submodule of \(M\). Show that

1. \(T(M)\) is a indeed a submodule of \(M\).
2. \(M/T(M)\) is torsion-free.
3. Module homomorphism \(f: M \to N\) maps \(T(M)\) into \(T(N)\).
4. Taking torsion is an left exact functor.
5. Let \(K\) be the field of fractions of \(A\). Then \(T(M)\) is exactly the kernel of the localization map \(M \to K \otimes_A M\). Equivalently, there is a natural isomorphism \(T(M) \cong \operatorname{Tor}_1^A (K / A, M)\).

Proof

Proof.

1. For any \(x, y \in T(M)\), \(\exists a, b \in A \setminus \{0\}\) s.t. \(a x = 0\), \(b y = 0\). Then \(ab(x+y) = a b x + a b y = 0 + 0 = 0\). Thus \(x+y \in T(M)\). Similarly for scalar multiplication.

2. For any \(x \in M\), if \(\bar x \in M/T(M)\) is torsion, then \(\exists a \in A \setminus \{0\}\) s.t. \(a \cdot \bar x = \bar 0\). This means \(ax \in T(M)\), hence \(x \in T(M)\) and \(\bar x = \bar 0\). Thus \(M/T(M)\) is torsion-free.

3. For any \(x \in T(M)\), \(\exists a \in A \setminus \{0\}\) s.t. \(a x = 0\). Then \(a f(x) = f(a x) = f(0) = 0\). Thus \(f(x) \in T(N)\).

4. Say \(0 \to M_1 \xrightarrow{f} M_2 \xrightarrow{g} M_3\) is exact. We need to show that \(0 \to T(M_1) \xrightarrow{f} T(M_2) \xrightarrow{g} T(M_3)\) is exact.

First, \(g \circ f = 0\) passes to the torsion submodules.

Secondly, \(f\) is injective on \(T(M_1)\) by the injectivity of \(f\) on \(M_1\).

Finally, for any \(y \in T(M_2)\) with \(g(y) = 0\), by the exactness of the original sequence, there exists \(x \in M_1\) s.t. \(f(x) = y\). Since \(y\) is torsion, \(\exists b \in A \setminus \{0\}\) s.t. \(b \cdot y = 0\). Then \(0 = b \cdot y = b \cdot f(x) = f(b \cdot x)\), which implies \(b \cdot x = 0\) by the injectivity of \(f\). Thus \(x \in T(M_1)\) and \(f(x) = y\). This shows the exactness at \(T(M_2)\).

5. See Section 1.1.2, where we systematically treat the etymology of \(\operatorname{Tor}\).

Exercise 21 ([1]-exr-3.13) Taking torsion commutes with localization, i.e. for any multiplicative set \(S\) of \(A\), \(T(S^{-1} M) \cong S^{-1} T(M)\).

As a consequence, show that torsion-free is a local property.

Proof

Proof. Let \(k\) be the field of fractions of \(A\). By Exercise 20, consider the exact sequence \[ 0 \longrightarrow T(M) \longrightarrow M \longrightarrow k \otimes M \] Taking localization \(S^{-1}\) gives the exact sequence \[ 0 \longrightarrow S^{-1} T(M) \longrightarrow S^{-1} M \longrightarrow k \otimes M \] Consider another exact sequence \[ 0 \longrightarrow T(S^{-1} M) \longrightarrow S^{-1} M \longrightarrow k \otimes S^{-1} M \cong k \otimes M \]

Note that both \(S^{-1} T(M)\) and \(T(S^{-1} M)\) are realized as the kernels of the same map \(S^{-1} M \to k \otimes M\). Thus they are naturally isomorphic.

For the consequence, say for all maximal ideals \(\mathfrak m\) of \(A\), \(M_{\mathfrak m}\) is torsion-free. Then by the isomorphism above, \(T(M)_{\mathfrak m} \cong T(M_{\mathfrak m}) = 0\). Being a zero module is a local property, thus \(M\) is torsion-free.

Example 5 ([1]-exm-3.19: Support of a module) Let \(A\) be a ring, \(M\) an \(A\)-module. Define the support of \(M\) to be the set \(\operatorname{Supp} M\) of prime ideals \(\mathfrak p\) of \(A\) such that \(M_{\mathfrak p} \neq 0\).

Show that

1. \(M = 0\) iff \(\operatorname{Supp} M = \varnothing\).

2. \(V(\mathfrak a) = \operatorname{Supp} (A/\mathfrak a)\).

3. If \(0 \to M_1 \to M_2 \to M_3 \to 0\) is exact, then \(\operatorname{Supp} M_2 = \operatorname{Supp} M_1 \cup \operatorname{Supp} M_3\).

4. If \(M = \sum_i M_i\), then \(\operatorname{Supp} M = \bigcup_i \operatorname{Supp} M_i\).

5. If \(M\) is finitely generated, then \(\operatorname{Supp} M = V(\operatorname{Ann}(M))\) (and hence is a closed subset of \(\operatorname{Spec}A\)).

6. If \(M\) and \(N\) are finitely generated, then \(\operatorname{Supp} (M \otimes_A N) = \operatorname{Supp} M \cap \operatorname{Supp} N\).

7. If \(M\) is finitely generated and \(\mathfrak a\) is an ideal of \(A\), then \(\operatorname{Supp} (M/\mathfrak a M) = V(\mathfrak a + \operatorname{Ann}(M))\).

8. If \(f: A \to B\) is a ring homomorphism and \(M\) is a finitely generated \(A\)-module, then \(\operatorname{Supp} (B \otimes_A M) = (\operatorname{Spec}f)^{-1}(\operatorname{Supp} M)\).

Proof

Proof.

1. If \(M=0\), then for all prime ideal \(\mathfrak p\), \(M_{\mathfrak p} = 0\). Conversely, if \(\operatorname{Supp} M = \varnothing\), then for all prime ideal \(\mathfrak p\), \(M_{\mathfrak p} = 0\). This implies that for all \(x \in M\), \(\exists s \in A \setminus \mathfrak p\) s.t. \(s x = 0\). Thus the annihilator ideal \(\operatorname{Ann}(x)\) is not contained in any prime ideal, which forces \(\operatorname{Ann}(x) = A\) and hence \(x=0\). Thus \(M=0\).

2. For any prime ideal \(\mathfrak p \in V(\mathfrak a)\), \(\mathfrak a \subseteq \mathfrak p\). Thus the annihilated port of \(A_\mathfrak p\) never touches \(\mathfrak a\) and hence \((A/\mathfrak a)_{\mathfrak p} \neq 0\). Conversely, for any prime ideal \(\mathfrak p \notin V(\mathfrak a)\), \(\exists a \in \mathfrak a \setminus \mathfrak p\). Then \(a/1\) is a unit in \(A_{\mathfrak p}\), which forces \((A/\mathfrak a)_{\mathfrak p} = 0\).

3. By the exactness of localization, we have the exact sequence \[ 0 \longrightarrow (M_1)_{\mathfrak p} \longrightarrow (M_2)_{\mathfrak p} \longrightarrow (M_3)_{\mathfrak p} \longrightarrow 0 \] for all prime ideal \(\mathfrak p\). Thus \((M_2)_{\mathfrak p} \neq 0\) iff either \((M_1)_{\mathfrak p} \neq 0\) or \((M_3)_{\mathfrak p} \neq 0\).

4. Localization, as a special tensor functor, commutes with taking sums (the later is a colimit, or more concretely, a direct limit) [TODO: Detail this nonsense] (A concrete proof can be found in Exercise 10). Thus for all prime ideal \(\mathfrak p\), we have \[ M_{\mathfrak p} = \left(\sum_i M_i\right)_{\mathfrak p} = \sum_i (M_i)_{\mathfrak p} \] Hence the conclusion follows.

5. Say \(M = \sum_i M_i\) where each \(M_i\) is cyclic, i.e. \(M_i \cong A / \operatorname{Ann}(m_i)\) for some \(m_i \in M_i\). Then by 4, we have \[ \operatorname{Supp} M = \bigcup_i \operatorname{Supp} M_i = \bigcup_i V(\operatorname{Ann}(m_i)) = V\left(\bigcap_i \operatorname{Ann}(m_i)\right) = V(\operatorname{Ann}(M)) \] Note that the second last equality holds by the finiteness of the generating set \(\{ m_i \}\).

6. Recall that localization commutes with tensor product, i.e. for all prime ideal \(\mathfrak p\), \[ (M \otimes_A N)_{\mathfrak p} \cong M_{\mathfrak p} \otimes_{A_{\mathfrak p}} N_{\mathfrak p} \] Thus by Exercise 2, \((M \otimes_A N)_{\mathfrak p} = 0\) iff both \(M_{\mathfrak p} = 0\) or \(N_{\mathfrak p} = 0\). The conclusion follows.

7. We have \[ \begin{aligned} \operatorname{Supp}(M/\mathfrak a M) &= \operatorname{Supp}(A/\mathfrak a \otimes_A M) = \operatorname{Supp}(A / \mathfrak a) \cap \operatorname{Supp}(M) \\ &= V(\mathfrak a) \cap V(\operatorname{Ann}(M)) = V(\mathfrak a + \operatorname{Ann}(M)) \end{aligned} \]

8. Say \(\mathfrak q \in \operatorname{Spec}B\), and denote \(\mathfrak p := f^{-1}(\mathfrak q)\), we are to show that \((B \otimes_A M)_{\mathfrak q} = 0\) iff \(M_{\mathfrak p} = 0\). It’s clear that if \(M_{\mathfrak p} = 0\), then \((B \otimes_A M)_{\mathfrak q} = 0\) by the isomorphism \[ (B \otimes_A M)_{\mathfrak q} \cong B_{\mathfrak q} \otimes_{A_{\mathfrak p}} M_{\mathfrak p} \] Conversely, if \((B \otimes_A M)_{\mathfrak q} = 0\), by the isomorphism above we have \(B_{\mathfrak q} \otimes_{A_{\mathfrak p}} M_{\mathfrak p} = 0\). Note that we can’t use Exercise 2 since \(B_{\mathfrak p}\) may not be finitely generated over \(A_{\mathfrak p}\). However, tensoring with the residue field \(A_{\mathfrak p} / \mathfrak p\), we still have \(B_{\mathfrak q} / \mathfrak q B_{\mathfrak q} \otimes_{A_{\mathfrak p}} \left( M_{\mathfrak p} \otimes_{A_{\mathfrak p}} A_{\mathfrak p} / \mathfrak p A_{\mathfrak p} \right) = 0\), where \(M_{\mathfrak p} \otimes_{A_{\mathfrak p}} A_{\mathfrak p} / \mathfrak p A_{\mathfrak p}\) is a finite-dimensional vector space over the field \(A_{\mathfrak p} / \mathfrak p A_{\mathfrak p}\), and the whole LHS is a finite-dimensional vector space over the field \(B_{\mathfrak q} / \mathfrak q B_{\mathfrak q}\). Thus it must be of zero dimension. In turn \(M_{\mathfrak p} \otimes_{A_{\mathfrak p}} A_{\mathfrak p} / \mathfrak p A_{\mathfrak p} = 0\). By Nakayama’s lemma, \(M_{\mathfrak p} = 0\).

Remark

Remark. [TODO] Best appreciated in a geometric view. If \(B\) is an \(A\)-algebra defined by the ring map \(\varphi : A \to B\), then \(\operatorname{Supp}_A B\) is exactly the image of \(\operatorname{Spec}\varphi : \operatorname{Spec} B \to \operatorname{Spec} A\), since for any prime ideal \(\mathfrak p\) of \(A\), the fiber over \(\mathfrak p\) is \(\operatorname{Spec} (B \otimes_A k(\mathfrak p))\), which is nonempty iff \(B \otimes_A k(\mathfrak p) \neq 0\) iff \(B_{\mathfrak p} \neq 0\).

Exercise 22 ([1]-exr-3.20) \(f : A \to B\) ring homomorphism. Show that

1. Every prime ideal of \(A\) is a contracted ideal iff \(\operatorname{Spec}f\) is surjective.
2. Every prime ideal of \(B\) is an extended ideal implies that \(\operatorname{Spec}f\) is injective.
3. The converse of (2) is not true in general.

Proof

Proof.

1. By definition, \(\mathfrak p \in \operatorname{Spec}A\) is a contracted ideal iff \(\mathfrak p = f^{-1}(\mathfrak q)\) for some prime ideal \(\mathfrak q\) of \(B\), exactly the surjectivity of \(\operatorname{Spec}f\).

2. Denote \(\mathfrak a^\mathrm{e}\) for ideal extension of \(\mathfrak p\) alongside \(f\), and \(\mathfrak b^\mathrm{c}\) for contraction. For any \(\mathfrak q \in \operatorname{Spec}B\), we have \(\mathfrak q = \mathfrak a^\mathrm{e}\) for some ideal \(\mathfrak a\) of \(A\). This makes \(\mathfrak q \mapsto \mathfrak q^\mathrm{ce}\) an identity map on \(\operatorname{Spec}B\) by the fact that \(\mathfrak q^\mathrm{ce} = \mathfrak a^\mathrm{ece} = \mathfrak a^\mathrm e = \mathfrak q\). Hence \(\operatorname{Spec}f: \mathfrak q \mapsto \mathfrak q^\mathrm c\) is injective.

3. Consider \(f: \mathbb Z \to \mathbb Q\). \(\operatorname{Spec}f\) is injective since \(\mathbb Q\) is a field. However, the prime ideal \((0)\) of \(\mathbb Q\) cannot be extended from any ideal of \(\mathbb Z\).

Remark

Remark. If \(f\) is an integral extension, then the converse of (2) holds true. (By the going-up / lying-over / incomparability theorems.)

Exercise 23 ([1]-exr-3.21)  

1. Let \(A\) be a ring, \(S\) a multiplicative set of \(A\), \(\varphi : A \to S^{-1} A\) the localization homomorphism. Show that the map \(\operatorname{Spec}\varphi : \operatorname{Spec}S^{-1} A \to \operatorname{Spec}A\) is a homeomorphism onto its image in \(X = \operatorname{Spec}A\). Denote this image by \(S^{-1} X\).

   In particular, if \(f \in A\), then the map \(\operatorname{Spec}A_f \to \operatorname{Spec}A\) is a homeomorphism onto the basic open subset \(D_A (f)\) of \(X\).

2. Let \(f : A \to B\) be a ring homomorphism. Let \(X = \operatorname{Spec}A\) and \(Y = \operatorname{Spec}B\), \(\operatorname{Spec}f : \operatorname{Spec}B \to \operatorname{Spec}A\). Identify \(S^{-1} X \subseteq X\) and \(S^{-1} Y \subseteq Y\). Show that \(\operatorname{Spec}(S^{-1} f) : \operatorname{Spec}S^{-1} B \to \operatorname{Spec}S^{-1} A\) is the restriction of \(\operatorname{Spec}f\) to \(S^{-1} Y\), and that \(S^{-1} Y = (\operatorname{Spec}f)^{-1} (S^{-1} X)\).

3. Let \(\mathfrak a\) be an ideal of \(A\) and \(\mathfrak b = \mathfrak a^\mathrm e\) its extension alongside \(f : A \to B\). Let \(\bar f : A / \mathfrak a \to B / \mathfrak b\) be the induced homomorphism. Identify \(\operatorname{Spec}(A / \mathfrak a) = V(\mathfrak a) \subseteq X\) and \(\operatorname{Spec}(B / \mathfrak b) = V(\mathfrak b) \subseteq Y\), show that \(\operatorname{Spec}(\bar f)\) is the restriction of \(\operatorname{Spec}f\) to \(V(\mathfrak b)\).

4. Let \(\mathfrak p \in \operatorname{Spec}A\). Take \(S = A \setminus \mathfrak p\) in (2) and reduce mod \(S^{-1} \mathfrak p\) as in (3). Deduce that the subspace \((\operatorname{Spec}f)^{-1} (\mathfrak p)\) of \(\operatorname{Spec}B\) is homeomorphic to \(\operatorname{Spec}(B \otimes_A k(\mathfrak p))\), the fiber of \(\operatorname{Spec}f\) over \(\mathfrak p\).

Proof

Proof.

1. Continuity and onto-ness are clear. Injectivity follows from Exercise 23 and the fact that every prime ideal of \(S^{-1} A\) is an extended ideal from \(A\). It suffices to show that \(\operatorname{Spec}\varphi\) is an open map onto its image.

   Fix some \(f/s \in S^{-1} A\). Consider any contracted ideal \(\mathfrak p \in S^{-1} X\). Then it can be contracted from \(S^{-1} \mathfrak p\). Suppose \(S^{-1} \mathfrak p \in D_{S^{-1} A}(f/s)\), i.e. \(f/s \notin S^{-1} \mathfrak p\) or equivalently any \(t \in S\) won’t make \(f \in (\mathfrak p : t)\). Note that \(\mathfrak p\) is contracted, hence \((\mathfrak p : t) = \mathfrak p\). Thus \(f \notin \mathfrak p\), i.e. \(\mathfrak p \in D_A(f)\). Above arguments can be reversed, hence we have shown that \[ \operatorname{Spec}\varphi \left( D_{S^{-1} A}(f/s) \right) = D_A(f) \cap S^{-1} X \]

2. Apply the \(\operatorname{Spec}\) functor to the commutative diagram gives Figure 4.

   The equation \(S^{-1} Y = (\operatorname{Spec}f)^{-1} (S^{-1} X)\) is given by the fact that \[ \begin{aligned} &\phantom{\iff} \mathfrak q \in S^{-1} Y\\ &\iff f(S) \cap \mathfrak q = \varnothing \\ &\iff \forall s \in S,\, f(s) \notin \mathfrak q \\ &\iff \forall s \in S,\, s \notin f^{-1} (\mathfrak q) \\ &\iff S \cap f^{-1} (\mathfrak q) = \varnothing \\ &\iff f^{-1} (\mathfrak q) \in S^{-1} X \\ &\iff \mathfrak q \in (\operatorname{Spec}f)^{-1} (S^{-1} X) \end{aligned} \]

Figure 4

3. Similar to (2), see Figure 5.

Figure 5

4. Analyzing Figure 6 gives the desired homeomorphism.

Figure 6

3 Chapter 4

Exercise 24 ([1]-exr-4.4) In the polynomial ring \(\mathfrak Z[t]\), the ideal \(\mathfrak m = (2,t)\) is maximal and the ideal \(\mathfrak q = (4,t)\) is \(\mathfrak m\)-primary, but is not a power of \(\mathfrak m\).

Proof

Proof. \(\mathfrak m\) being maximal is clear. \(\sqrt{\mathfrak q} = (2,t)\) is also clear. For any \(ab \in \mathfrak q\), if \(a \notin \mathfrak m\), then \(a\) is a odd integer. Hence \(4 \mid b\) or \(t \mid b\), i.e. \(b \in \mathfrak q\), showing that \(\mathfrak q\) is \(\mathfrak m\)-primary. Finally, note that \((2,t)^2 = (4,2t,t^2) \neq (4,t) = \mathfrak q\).

Exercise 25 ([1]-exr-4.5) Consider the polynomial ring \(K[x,y,z]\) where \(K\) is a field. Let \(\mathfrak p_1 = (x,y)\), \(\mathfrak p_2 = (x,z)\), \(\mathfrak m = (x,y,z)\). Note that \(\mathfrak p_1\) and \(\mathfrak p_2\) are prime ideals, and \(\mathfrak m\) is a maximal ideal. Let \(\mathfrak a = \mathfrak p_1 \mathfrak p_2 = (x^2, x y, x z, y z)\). Show that \(\mathfrak a := \mathfrak p_1 \cap \mathfrak p_2 \cap \mathfrak m^2\) is a reduced primary decomposition of \(\mathfrak a\). Identify the associated primes of \(\mathfrak a\).

Proof

Proof. The mutual incomparability of \(\mathfrak p_1\), \(\mathfrak p_2\) and \(\mathfrak m\) is clear. \(\mathfrak p_1\) and \(\mathfrak p_2\) are minimal and \(\mathfrak m\) is embedded.

Exercise 26 ([1]-exr-4.6) Let \(X\) be an infinite compact Hausdorff topological space, \(C(X)\) be the ring of real-valued continuous functions on X (cf. [1]-exr-1.26). Is the zero ideal decomposable in this ring?

Proof

Proof. It can’t be. If it is, recall that by the remark below [1, proposition 4.7], the set of zero-divisors of \(C(X)\) is exactly the union of the associated primes. Consider the maximal ideals in these associated primes, they correspond to the points in \(X\) by [1, exr–1.26], say \(x_1\), \(x_2\), \(\cdots\), \(x_n\). Then every zero-divisor \(f\) must vanish at one of these points, i.e. \(f(x_i) = 0\) for some \(1 \leq i \leq n\). However, since \(X\) is infinite, we can always find some point \(x \in X \setminus \{ x_1, x_2, \cdots, x_n \}\). To reach a contradiction, we shall construct a continuous function \(g\) on \(X\) such that \(g(x_i) \neq 0\) for all \(1 \leq i \leq n\), but still a zero-divisor.

By the Hausdorff property, for each \(1 \leq i \leq n\), we can find some open neighborhood \(U_i\) of \(x_i\) and some open neighborhood \(V_i\) of \(x\) such that \(U_i \cap V_i = \varnothing\). Let \(U := \bigcup_{i=1}^n U_i\), \(V := \bigcap_{i=1}^n V_i\), then \(U\) and \(V\) are still open neighborhoods of \(\{ x_1, x_2, \cdots, x_n \}\) and \(x\) respectively, and \(U \cap V = \varnothing\). By the Urysohn lemma, there exists some continuous function \(g : X \to [0,1]\) such that \(g(x_i) = 1\) for all \(1 \leq i \leq n\) and \(g \vert_{X \setminus U} = 0\). It suffices to show that \(g\) is a zero-divisor. It’s counterpart can be constructed in a symmetric way: By Urysohn lemma again, there exists a continuous function \(h : X \to [0,1]\) such that \(h(x) = 1\) and \(h \vert_{X \setminus V} = 0\). Then \(g h = 0\), showing that \(g\) is a zero-divisor.

Exercise 27 ([1]-exr-4.7) \(A\) is a ring.

1. \(\mathfrak a[x]\) is the extension of \(\mathfrak a\) alongside the inclusion \(A \to A[x]\).
2. If \(\mathfrak p\) is a prime ideal in \(A\), then \(\mathfrak p[x]\) is a prime ideal in \(A[x]\).
3. If \(\mathfrak q\) is \(\mathfrak p\)-primary in \(A\), then \(\mathfrak q[x]\) is \(\mathfrak p[x]\)-primary in \(A[x]\).
4. If \(\mathfrak a = \bigcap_{i=1}^n \mathfrak q_i\) is a primary decomposition of \(\mathfrak a\) in \(A\), then \(\mathfrak a[x] = \bigcap_{i=1}^n \mathfrak q_i[x]\) is a primary decomposition of \(\mathfrak a[x]\) in \(A[x]\).
5. If \(\mathfrak p\) is a minimal prime ideal of \(\mathfrak a\), then \(\mathfrak p[x]\) is a minimal prime ideal of \(\mathfrak a[x]\).

Proof

Proof.

1. Trivial.
2. \(A[x] / p[x] \cong (A / p)[x]\) is an integral domain since \(A / p\) is so.
3. \(A[x] / \mathfrak q[x] \cong (A / \mathfrak q)[x]\). Any zero-divisor in \((A / \mathfrak q)[x]\) must have all its coefficients being zero-divisors in \(A / \mathfrak q\) by [1]-exr-1.2-iii, hence nilpotent since \(\mathfrak q\) is \(\mathfrak p\)-primary. Thus this zero-divisor is nilpotent by [1]-exr-1.2-ii, and \(\mathfrak q[x]\) is \(\mathfrak p[x]\)-primary. It’s radical is at least \(\mathfrak p[x]\) because \(\mathfrak q\) can reach \(\mathfrak p\) by taking radicals. Regarding that \(\mathfrak p[x]\) is radical, we have \(r(\mathfrak q[x]) = \mathfrak p[x]\).
4. There is nothing to prove.
5. There is nothing to prove.

Exercise 28 ([1]-exr-4.20: Primary decomposition of modules) Let \(A\) be a ring, \(M\) be a fixed \(A\)-module, \(N\) a submodule of \(M\). The radical of \(N\) in M is defined to be \[ r_M(N) = \{ x \in A : x^q M \subseteq N \text{ for some } q > 0 \} \] Show that \(r_M(N) = r(N : M) = r(\operatorname{Ann}(M/N))\).

Proof

Proof. Recall that \((N : M) = \{ a \in A : a M \subseteq N \}\), hence the first equality is straightforward. The second equality follows from the fact that \(\operatorname{Ann}(M/N) = (N : M)\).

Remark

Remark. Analogous to [1]-exr-1.13, we have

• \(r_M(N) \supseteq (N : M)\)
• \(r(r_M(N)) = r_M(N)\)
• \(r_M(N \cap P) = r_M (N) \cap r_M(P)\)
• \(r_M(N) = A\) iff \(M=N\)
• \(r_M(N+P) \supseteq r(r_M(N) + r_M(P))\)

Note that the last identity may not be an equality in general. For example, let \(M = A \oplus A\), \(N = A \oplus 0\), \(P = 0 \oplus A\). Then \(r_M(N) = r_M(P) = 0\), while \(r_M(N+P) = r_M(M) = A\).

Exercise 29 ([1]-exr-4.21) For any \(x \in A\), \(\varphi_x\) denotes the multiplication endomorphism \(M \to M\), \(m \mapsto x m\). Then \(x\) is said to be a zero-divisor on \(M\) iff \(\varphi_x\) is not injective, is a nilpotent on \(M\) iff \(\varphi_x\) is nilpotent. A submodule \(Q\) of \(M\) is primary in \(M\) if \(Q \neq M\) and every zero-divisor in \(M/Q\) is nilpotent.

Show that if \(Q\) is primary in \(M\), then \((Q:M)\) is a primary ideal and hence \(r_M(Q)\) is a prime ideal \(\mathfrak p\). We say that \(Q\) is \(\mathfrak p\)-primary in \(M\).

Proof

Proof. Recall that \((Q:M) = \operatorname{Ann}(M/Q)\). Note that \(Q \neq M\) by definition of primary submodule. Say \(ab \in (Q:M)\) for some \(a,b \in A\).

In case that \(a\) is a zero-divisor on \(M/Q\), then by the primary-ness of \(Q\), \(a\) is nilpotent on \(M/Q\) and hence \(a^n \in (Q:M)\) for some \(n>0\). If \(a\) is strong enough to kill \(M/Q\), then \(a \in (Q:M)\) and we are done. Otherwise \(b\) kills \(a M / Q \neq 0\). Thus \(b\) is a zero-divisor on \(M/Q\), hence nilpotent on \(M/Q\), i.e. \(b \in r(Q:M)\).

In case that \(a\) is not a zero-divisor on \(M/Q\), then \(a M/Q \neq 0\). Since \(ab\) kills \(M/Q\), \(b\) must kill \(a M / Q \neq 0\). Thus \(b\) is a zero-divisor on \(M/Q\), hence nilpotent on \(M/Q\), i.e. \(b \in r(Q:M)\).

This shows that \((Q:M)\) is a primary ideal.

Exercise 30 ([1]-exr-4.22) A primary decomposition of \(N\) in \(M\) is a representation of \(N\) as an intersection \[ N = Q_1 \cap Q_2 \cap \cdots \cap Q_n \] of primary submodules of \(M\); it is a minimal primary decomposition if the ideals \(\mathfrak p_i = r_M(Q_i)\) are all distinct and if none of the components \(Q_i\) can be omitted from the intersection.

Prove that the prime ideals \(\mathfrak p_i\) depend only on \(N\) (and \(M\)), not on the particular minimal primary decomposition of \(N\) in \(M\).

Proof

Proof. Note that \((N : M) = \bigcap_{i=1}^n (Q_i : M)\). By Exercise 29, each \((Q_i : M)\) is a \(\mathfrak p_i\)-primary ideal. Thus by the uniqueness of primary decomposition of ideals, the set \(\{ \mathfrak p_i \}\) is uniquely determined by \((N : M)\) and hence by \(N\) and \(M\).

4 Chapter 5

Exercise 31 ([1]-exr-5.1) Let \(f: A \to B\) be an integral homomorphism of rings. Show that \(\operatorname{Spec}f : \operatorname{Spec}B \to \operatorname{Spec}A\) is a closed map, i.e. the image of a closed set is closed.

Proof

Proof. Let \(V(\mathfrak b) \subseteq \operatorname{Spec}B\) be a closed set for some ideal \(\mathfrak b\) of \(B\). We are to show that \(f^{-1} (V(\mathfrak b))\) is closed in \(\operatorname{Spec}A\). Note that \[ f^{-1} (V(\mathfrak b)) = \{ \mathfrak p \in \operatorname{Spec}A : \exists \mathfrak q \in V(\mathfrak b),\, f^{-1} (\mathfrak q) = \mathfrak p \} \] while \[ V(f^{-1}(\mathfrak b)) = \{ \mathfrak p \in \operatorname{Spec}A : f^{-1}(\mathfrak b) \subseteq \mathfrak p \} \] Thus it suffices to show the two sets above are equal.

Write \(f : A \to f(A) \subseteq B\), the former is surjective and the later is an integral extension. We have the ideal one-to-one correspondence between all ideals containing \(\ker f\) of \(A\), and all ideals of \(f(A)\). By [1], theorem 5.10, every prime ideals of \(f(A)\) can be contracted from some prime ideal of \(B\). To summarize, every prime ideal of \(A\) containing \(\ker f\) can be contracted from some prime ideal of \(B\). We call this the lying-over theorem.

“\(\subseteq\)”: For any \(\mathfrak p \in f^{-1} (V(\mathfrak b))\), there exists some \(\mathfrak q \in V(\mathfrak b)\) such that \(f^{-1} (\mathfrak q) = \mathfrak p\). Thus \(f^{-1}(\mathfrak b) \subseteq f^{-1}(\mathfrak q) = \mathfrak p\), i.e. \(\mathfrak p \in V(f^{-1}(\mathfrak b))\).

“\(\supseteq\)”: For any \(\mathfrak p \in V(f^{-1}(\mathfrak b))\), we have \(\ker f \subseteq f^{-1}(\mathfrak b) \subseteq \mathfrak p\). By the lying-over theorem, there exists some prime ideal \(\mathfrak q\) of \(B\) such that \(f^{-1}(\mathfrak q) = \mathfrak p\). Note that \(\mathfrak b \subseteq \mathfrak q\) since \(f^{-1}(\mathfrak b) \subseteq \mathfrak p = f^{-1}(\mathfrak q)\). Thus \(\mathfrak q \in V(\mathfrak b)\) and hence \(\mathfrak p \in f^{-1} (V(\mathfrak b))\).

Exercise 32 ([1]-exr-5.2) Let \(A \subseteq B\) is an integral extension of rings. Let \(f \to \Omega\) be a homomorphism of \(A\) into an algebraically closed field \(\Omega\). Show that there exists an extension of \(f\) to a homomorphism \(g : B \to \Omega\).

Proof

Proof. By Zorn’s lemma it suffices to extend \(f\) to any intermediate ring \(A[x]\) where \(x \in B\). Note that \(x\) is integral over \(A\), hence there exists some monic polynomial \(p(t) \in A[t]\) such that \(p(x) = 0\). Applying \(f\) to the coefficients of \(p(t)\) gives a polynomial \(f(p)(t) \in \Omega[t]\). Since \(\Omega\) is algebraically closed, there exists some \(a \in \Omega\) such that \(f(p)(a) = 0\). We extend \(f\) to \(g : A[x] \to \Omega\) by sending \(x \mapsto a\). It’s straightforward to verify that \(g\) is a well-defined ring homomorphism.

Exercise 33 ([1]-exr-5.3) The \(A\)-algebra functor \(- \otimes_A C\) preserves integrality of \(A\)-algebra homomorphisms, i.e. if \(f : B_1 \to B_2\) is an integral homomorphism of \(A\)-algebras, then so is \(f \otimes \operatorname{id}_C : B_1 \otimes_A C \to B_2 \otimes_A C\) for any \(A\)-algebra \(C\).

Proof

Proof. We are to show that any finite sum \(\sum_i b_i \otimes c_i \in B_2 \otimes_A C\) for some \(b_i \in B_2\) and \(c_i \in C\) is integral over \(B_1 \otimes_A C\). Since the integral closure is closed under addition, it suffices to show any \(b \otimes c \in B_2 \otimes_A C\) is integral over \(B_1 \otimes_A C\). By the integrality of \(f\), there exists some monic polynomial \(t^n + \sum_{i=0}^{n-1} a_i t^i \in B_1[t]\) such that \(p(b) = 0\). Now \[ (b \otimes c)^n + \sum_{i=0}^{n-1} (a_i \otimes 1) (b \otimes c)^i = \left( b^n + \sum_{i=0}^{n-1} a_i b^i \right) \otimes c^n = 0 \]

Exercise 34 ([1]-exr-5.6) Let \(B_1,\dots,B_n\) be integral \(A\)-algebras. Show that \(\prod_{i=1}^n B_i\) is an integral \(A\)-algebra.

Proof

Proof. By that integral closure is closed under finite sums, it suffices to show that any element of the form \((0,\dots,0,b,0,\dots,0) \in \prod_{i=1}^n B_i\) is integral over \(A\). Say \(b\) vanishes in some monic polynomial \(f \in B_i[t]\). Then \((0,\dots,0,b,0,\dots,0)\) vanishes in the same polynomial, where each coefficient is embedded via the canonical homomorhism \(A \to \prod_{i=1}^n B_i\).

5 Chapter 7

Exercise 35 TFAE for a Noetherian local ring \((A, \mathfrak m, k)\) and an \(A\)-module \(M\):

1. \(M\) is free.
2. \(M\) is flat.
3. \(\mathfrak m \otimes M \to A \otimes M\) is injective.
4. \(\operatorname{Tor}_1^A (k, M) = 0\).

Proof

Proof.

1. \(\implies\) (2) is by that free modules are projective (by pulling back basis) and projective modules are flat (by the currification property of tensor functor).

2. \(\iff\) (3) is by the definition of flatness.

3. \(\implies\) (4) is by the long exact sequence of \(\operatorname{Tor}\) alongside the short exact sequence \[ 0 \longrightarrow \mathfrak m \longrightarrow A \longrightarrow k \longrightarrow 0 \]

It remains to show that (4) \(\implies\) (1). Cosider the exact sequence \[ 0 \longrightarrow Q \longrightarrow A^m \longrightarrow M \longrightarrow 0 \] where \(m\) is the minimal number of generators of \(M\). By Nakayama’s lemma, such number is also the dimension of \(k \otimes M\) as a \(k\)-vector space. Applying \(k \otimes -\) gives the long exact sequence \[ \operatorname{Tor}_1^A (k, M) = 0 \longrightarrow k \otimes Q \longrightarrow k^m \longrightarrow k^m \longrightarrow 0 \] Hence \(k \otimes Q = 0\). By Nakayama’s lemma again, we have \(Q = 0\) and \(M \cong A^m\) is free.

6 Chapter 9

Exercise 36 ([1]-exr-9.9: Chinese Remainder Theorem over Dedekind domains) Let \(\mathfrak a_1, \dots, \mathfrak a_n\) be ideals and let \(x_1, \dots, x_n\) be elements in a Dedekind domain \(A\). Then the system of congruences \(x \equiv x_i \pmod{\mathfrak a_i}\) for \(i=1,\dots,n\) has a solution \(x \in A\) iff for all \(i,j\), \(x_i \equiv x_j \pmod{\mathfrak a_i + \mathfrak a_j}\).

Proof

Proof. Note that this is equivalent of the exactness of the following sequence: \[ A \longrightarrow \bigoplus_{i=1}^n A / \mathfrak a_i \longrightarrow \bigoplus_{i<j} A / (\mathfrak a_i + \mathfrak a_j) \] where the map on the left is the natural projection, and the map on the right sends \((\bar x_1, \dots, \bar x_n)\) to \((\bar x_i - \bar x_j)_{i<j}\). The exactness is a local property, hence it suffices to assume \(A\) is a DVR since Dedekind domains are locally DVRs. In this case, each ideal \(\mathfrak a_i\) is of the form \((m^{k_i})\) for some \(k_i \geq 0\), where \((m)\) is the unique maximal ideal of \(A\). WLOG we assume \(k_1 \leq k_2 \leq \cdots \leq k_n\). Then for any \(i<j\), we have \(\mathfrak a_i + \mathfrak a_j = (m^{k_i})\). Thus the exactness follows from the exactness of \[ A \longrightarrow \bigoplus_{i=1}^n A / (m^{k_i}) \longrightarrow \bigoplus_{i<j} A / (m^{k_i}) \] Say \((\bar x_1, \dots, \bar x_n)\) lies in the kernel of the right map. Then for any \(i<j\), \(m^{k_i} \mid x_i - x_j\). Let \(\delta_{i,j}\) be its divisor. We construct the preimage \(x\) as follows: \[ x = x_1 + \delta_{1,2} m^{k_1} + \delta_{2,3} m^{k_2} + \dots + \delta_{n-1,n} m^{k_n} \] It’s strightforward to verify that it satisfy the congurence relation \(x \equiv x_i \pmod{m^{k_i}}\).

7 Chapter 11

Exercise 37 ([1]-exr-11.1) Let \(f \in k[x_1, \dots, x_r]\) be an irreducible polynomial over an algebraically closed field \(k\). A point \(P\) on the variety \(f(x)=0\) is non-singular iff not all the partial derivatives \(\frac{\partial f}{\partial x_i}\) vanish at \(P\). Let \(A=k[x_1,\dots,x_n] / (f)\), and let \(\mathfrak m\) be the maximal ideal of \(A\) corresponding to \(P\). Show that \(P\) is non-singular iff the localization \(A_{\mathfrak m}\) is a regular local ring.

Proof

Proof. WLOG we assume \(P\) is the origin. Write \[ f(x_1,\dots,x_r) = \sum_{i=1}^r \left.\frac{\partial f}{\partial x_i}\right\vert_P x_i + \text{higher order terms} \] Note that via pulling pack from \(A\) to \(k[x_1,\dots,x_r]\), we have \[ \mathfrak m / \mathfrak m^2 \cong \left( (x_1, \dots, x_r) + (f) \right) / \left( (x_1, \dots, x_r)^2 + (f) \right) = (x_1, \dots, x_r) / \left( (x_1, \dots, x_r)^2 + (\sum_{i=1}^r \left.\frac{\partial f}{\partial x_i}\right\vert_P x_i) \right) \] In case that all partial derivatives vanish at \(P\), then the dimension of the RHS is \(r\) since \(x_1, \dots, x_r\) form a basis. Otherwise, a nontrivial linear combination of \(x_1, \dots, x_r\) lies in the denominator, hence the dimension of the RHS is \(r-1\).

Exercise 38 Reformulate [1], theorem 11.1 in terms of the Grothendieck group \(K(A_0)\) (cf. [1]-exr-7.25)

Solution

Solution. Given a Noetherian graded ring \(A = \bigoplus_{n=0}^{+\infty} A_n\), an additive function \(\lambda\) over all finitely-generated \(A_0\)-modules. By the universal property of Grothendieck group, for any additive group \(G\), \(\lambda\) induces a group homomorphism \(\tilde \lambda : K(A_0) \to G\). Then \[ P(M, t) = \sum_{n=0}^{+\infty} \lambda(M_n) t^n = \sum_{n=0}^{+\infty} \tilde \lambda([M_n]) t^n \] is a rational function of the form \(\frac{f(t)}{\prod_{i=1}^s (1-t^{k_i})}\), where \(f(t) \in \mathbb Z [t]\) and \(k_i\) are positive integers.

[TODO] What’s the significance of this reformulation?

Exercise 39 ([1]-exr-11.6) Let \(A\) be a ring, not necessarily Noetherian. Show that \[ 1 + \dim A \leq \dim A[x] \leq 1 + 2 \dim A \]

Proof

Proof. Note that for any maximal ideal \(\mathfrak m\) of \(A\), \(\mathfrak m[x] + (x)\) is a maximal ideal of \(A[x]\) because it is the kernel of \(A[x] \to A \to A / \mathfrak m\). Now say we have an maximal chain of prime ideals in \(A\) of length \(\dim A\). The top prime ideal must be an maximal ideal of \(A\), say \(\mathfrak m\). Lifting it to \(A[x]\), we have a chain of prime ideals in \(A[x]\) of length \(\dim A\), with top prime ideal \(\mathfrak m[x]\). Extend this chain by adding one more prime ideal \(\mathfrak m + (x)\) on the top, giving a chain of length \(1 + \dim A\). Thus \(\dim A[x] \geq 1 + \dim A\).

Consider any chain of prime ideals in \(A[x]\). By contracting to \(A\), we get a chain of prime ideals in \(A\). Between any two consecutive prime ideals in \(A\), there can be at most two prime ideals in \(A[x]\): The fiber over \(\mathfrak p \in \operatorname{Spec}A\) is homeomorphic to the spectrum of \(k(\mathfrak p) \otimes A[x] = k(\mathfrak p)[x]\) by Exercise 23, which has dimension \(1\). Thus the length of the chain in \(A[x]\) is at most \(1 + 2 \dim A\).

Exercise 40 ([1]-exr-11.7) Let \(A\) be a Noetherian ring, then \(\dim A[x] = 1 + \dim A\).

Proof

Proof. Say \(\mathfrak p\) is a prime ideal of \(A\) of height \(m\). Localizing at \(\mathfrak p\) gives a Noetherian local ring \(A_{\mathfrak p}\) of dimension \(m\). By the characterization of dimension [1], theorem 11.14, there exists an ideal \(\mathfrak a = (a_1, \dots, a_m)\) with \(\mathfrak p\) as one of its minimal prime ideals.

Now consider the height of \(\mathfrak p[x]\). By Exercise 27, it is still a minimal associated prime ideal of \(\mathfrak a[x]\), the latter is still generated by \(m\) elements. Thus by [1], corollary 11.16, the height of \(\mathfrak p[x]\) is at most \(m\).

We also have that the height of \(\mathfrak p[x]\) is at least \(m\), since every chain of prime ideals in \(A\) can be lifted to a chain of prime ideals in \(A[x]\) via the \(\mathfrak p_i \mapsto \mathfrak p_i[x]\). So the height of \(\mathfrak p[x]\) is exactly \(m\).

Now, for any prime ideal \(\mathfrak q\) of \(A[x]\) whose contraction to \(A\) is \(\mathfrak p\), since \(\mathfrak p[x]\) is of height \(m\) and also contracts to \(\mathfrak p\), by the same fiber argument in Exercise 39, the height of \(\mathfrak q\) is at most \(m+1\). Thus the dimension of \(A[x]\) is at most \(1 + \dim A\). Combining with the other direction of inequality in Exercise 39, we have \(\dim A[x] = 1 + \dim A\).

8 Extra

Exercise 41 Let \(R:=k[x_1, \dots, x_n]\), \(I := (x_1, \dots, x_n)\) is an \(R\)-ideal. Show that \[ R(I) \cong k[x_1, \dots, x_n, y_1, \dots, y_n] / (x_i y_j - x_j y_i) \] where \(R(I)\) is the Rees algebra of \(I\), that is, \[ R(I) := \bigoplus_{d=0}^{+\infty} I^d t^d \subseteq R[t] \]

Proof

Proof. Define a homomorphism \[ \varphi : k[x_1, \dots, x_n, y_1, \dots, y_n] \to R(I), \quad x_i \mapsto x_i,\, y_i \mapsto x_i t \] It’s clearly surjective. Note that \[ \varphi (x_i y_j - x_j y_i) = x_i x_j t - x_j x_i t = 0 \] Hence the ideal \((x_i y_j - x_j y_i)\) lies in the kernel of \(\varphi\). It remains to show that the kernel is exactly this ideal.

If some polynomial \(f(x_1, \dots, x_n, y_1, \dots, y_n)\) lies in the kernel, then \(f(x_1, \dots, x_n, x_1 t, \dots, x_n t) = 0\) in \(R(I)\). Say \(f\) \[ f(x_1, \dots, x_n, y_1, \dots, y_n) = \sum_{i_1,\dots,i_n, j_1,\dots,j_n} a_{i_1,\dots,i_n,j_1,\dots,j_n} x_1^{i_1} \dots x_n^{i_n} y_1^{j_1} \dots y^{j_n}_n \]

\(f(x_1, \dots, x_n, x_1 t, \dots, x_n t) = 0\) shows that \[ \sum_{i_1,\dots,i_n; \sum j_{\bullet} = d} a_{i_1,\dots,i_n,j_1,\dots,j_n} x_1^{i_1+j_1} \dots x_n^{i_n+j_n} = 0 \tag{1}\] and in \(k[x_1, \dots, x_n, y_1, \dots, y_n] / (x_i y_j - x_j y_i)\), note that \[ x_1^{i_1} \dots x_n^{i_n} y_1^{j_1} \dots y_n^{j_n} = x_1^{i_1'} \dots x_n^{i_n'} y_1^{j_1'} \dots y_n^{j_n'} \] when \(i_k + j_k = i_k' + j_k'\) for all \(1 \leq k \leq n\). Thus collecting terms with the same \(i_k + j_k\), and apply Equation 1, \(f = 0\) in \(k[x_1, \dots, x_n, y_1, \dots, y_n] / (x_i y_j - x_j y_i)\).

References

[1]

Atiyah, M. F. and Macdonald, I. G., Introduction to commutative algebra. Addison-Wesley Publishing Company, Inc., 1969.