Comparing Free Modules via Homomorphisms

We discuss some well-known results that compare finitely-generated free modules via homomorphisms. Let \(R\) be a nonzero commutative ring with \(1\).

Proposition 1 ([1, exercise 3.15]) Any \(R^{\oplus n} \to R^{\oplus n}\) epimorphism \(\varphi\) must be an isomorphism.

Obtain a split monomorphism (a right inverse of \(\varphi\)) from the fact that \(R^{\oplus n}\) is projective. Use the determinant to conclude.

Remark

Remark. There’s actually a more general result: for any finitely-generated \(R\)-module \(M\), every epimorphism \(M \to M\) is an isomorphism. To show this, view \(M\) as an \(R[x]\)-module via the action \(x \cdot m = \varphi(m)\), then \((x) \cdot M = x \cdot M = M\) by the surjectivity of \(\varphi\). By Nakayama’s lemma, there exists \(sx = r \in (x) \subseteq R[x]\) such that \((1-sx) \cdot M = (1-r) \cdot M = 0\), i.e. \(sx \cdot m = m\) for all \(m \in M\), that is, \(m \mapsto s \cdot m\) is an inverse of \(\varphi\). Hence \(\varphi\) is an isomorphism.

This result is due to [2]. We also referred to Math StackExchange.

Proposition 2 An \(R^{\oplus n} \to R^{\oplus n}\) monomorphism \(\varphi\) may not be an isomorphism in general.

Consider \(\varphi: \mathbb Z \to \mathbb Z\), \(z \mapsto 2z\).

[TODO] Is it true when \(R\) is a local ring?

Proposition 3 ([1, exercise 2.11]) An \(R^{\oplus n} \to R^{\oplus m}\) epimorphism \(\varphi\) implies \(n \geq m\).

Tensor with any residue field of \(R\). Right side is still exact, hence a vector space epimorphism.

Proposition 4 An \(R^{\oplus n} \to R^{\oplus m}\) isomorphism \(\varphi\) implies \(n = m\). [1, exercise 2.11]

Do the same as (3) again with the inverse map.

Proposition 5 ([1, exercise 2.11]) An \(R^{\oplus n} \to R^{\oplus m}\) monomorphism \(\varphi\) implies \(n \leq m\).

\(R\) is said to satisfy the strong rank condition if the above statement always holds.

It’s easy to see that this is true for \(R\) an integral domain: tensoring with the field of fractions of \(R\) does not destroy the injectivity.

For the general case of arbitrary nonzero commutative rings, we provide some references:

• [3]. This paper proves a general result that for any finitely-generated \(R\)-module \(M\) and its submodule \(N\), every epimorphism \(N \to M\) is an isomorphism.

  The strong rank condition follows immediately from this result: view \(R^n\) as a submodule of \(R^m\) via the monomorphism \(\varphi\), assume otherwise that \(n > m\), then we have an epimorphism \(R^m \to R^n\) by projection, hence an isomorphism \(R^m \to R^n\) by the result of [3]. This contradicts (4).

  Note that [2]’s result stated in Proposition 1 is a special case of this result by taking \(N = M\).

• Strong rank condition for commutative rings - Mathlib 4 Docs

• Search Keyword: McCoy’s Rank Theorem

• Another proof seems to make use of the exterior algebra.

References

[1]

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

[2]

W. V. Vasconcelos, “On finitely generated flat modules,” Trans. Amer. Math. Soc., vol. 138, pp. 505–512, 1969, doi: 10.1090/S0002-9947-1969-0238839-5.

[3]

M. Orzech, “Onto endomorphisms are isomorphisms,” The American Mathematical Monthly, vol. 78, no. 4, pp. 357–362, Apr. 1971, doi: 10.1080/00029890.1971.11992759.