Here are two examples of results which can be deduced from commutative algebra:

  • Any $n\times n$ complex matrix is conjugate to a Jordan canonical matrix (can be proven using the structure theorem for modules over a PID, in this case $\mathbf{C}[T]$ - see for example these course notes).
  • Commuting matrices have a common eigenvector (this can be seen as a consequence of Hilbert's Nullstellensatz, according to Wikipedia).

My question is, does anyone know of other examples of results in linear algebra which can be deduced (non-trivially*) from results in commutative algebra?

More precisely: Which results about modules over fields can be optained via modules over more general commutative rings? (Thanks to Martin's comment below for suggesting this precision of the question).

.* by "non-trivially," I mean you have to go deeper than simply applying module theory to modules over a field.

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    $\begingroup$ I like your question, but I feel like your asterisk rules out all possible answers to your question :( You want to apply commutative algebra results to linear algebra, but you won't allow the application of module results to vector spaces? $\endgroup$
    – rschwieb
    Jul 4, 2012 at 18:25
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    $\begingroup$ @rschwieb Note that the first example I give uses a result in module theory. E.g. "linear transformations correspond to matrices, because this is true for free modules" is a trivial deduction which isn't worth mentioning. I only meant to avoid getting responses like "but every $k$-vector space is a $k$-module, so everything comes from commutative algebra." $\endgroup$
    – vgty6h7uij
    Jul 4, 2012 at 18:41
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    $\begingroup$ Often it is the other way round: Algebraic geometry comes down to commutative algebra via charts, and in the residue fields this comes down to linear algebra. A friend of mine, who teached algebraic geometry, once said: "It's all just Linear Algebra + Nakayama." $\endgroup$ Jul 4, 2012 at 20:07
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    $\begingroup$ Somehow I like your question, but on the other hand it is hard to make it precise. After all, all parts in pure mathematics are connected with each other, and linear algebra is just a part of commutative algebra (namely, over a field). But perhaps the following makes the question more precise: Which results about modules and their homomorphisms over fields can be optained via modules over more general commutative rings? $\endgroup$ Jul 4, 2012 at 20:13
  • $\begingroup$ @MartinBrandenburg Thanks for clarifying the question; it seems I had some trouble stating the question precisely. Maybe I should add it to the original post. $\endgroup$
    – vgty6h7uij
    Jul 4, 2012 at 20:16

3 Answers 3


I think this is an interesting question, and I have several responses to it.

  1. You say (correctly, of course) that Jordan form is proved using the structure theory of finitely generated modules over a PID. But to me this latter theory is part of linear algebra rather than commutative algebra. It may be "graduate level" linear algebra, but it is generally found in general-purpose algebra texts rather than commutative algebra texts and often in the context of linear algebra.

1$'$) To go deeper, how "linear algebraic" you feel this structure theorem is probably depends upon which proof you give. If you follow the route which first establishes Smith normal form for matrices over a PID, this is strongly reminiscent of undergraduate linear algebra. It seems though that the fashion in many contemporary texts is to give a less matrix-oriented approach. Very recently I realized that I had never really absorbed (and maybe was never taught) the linear algebra approach to structure theory and also that I needed it in some work I am doing in lattices and the geometry of numbers. Maybe a serious undergraduate algebra sequence should include treatment of the Hermite and Smith normal forms and applications to module theory rather than more abstract stuff which will surely be covered later on.

1$''$) One of the few commutative algebra texts I know that gives a proof of the structure theorem for finitely generated modules over a PID is mine. The proof I give is (I think) not one of the standard ones, so to me this is an example of a linear algebra result proved by commutative algebraic methods. At the moment the argument is scattered through various sections:

$\bullet$ In $\S 3.9.2$ it is shown that a finitely generated torsionfree module over a PID is free. Since free modules are projective, it follows that every finitely generated module over a PID is the direct sum of a free module and a torsion module, so it remains to classify finitely generated torsion modules.

$\bullet$ In $\S 17.5.3$ I give the structure theorem for finitely generated torsion modules over a discrete valuation ring, which takes advantage of the fact that its quotients are "self-injective rings".

$\bullet$ In $\S 20.6$ I explain how an easy localization argument reduces the case of finitely generated torsion modules over an arbitrary PID -- in fact over an arbitrary Dedekind domain -- to the already proved case of a DVR.

  1. Although it may be interesting to know that the Nullstellensatz can be used to prove that commuting square matrices $A,B$ over an algebraically closed field have a common eigenvector, it seems to me that this is significant overkill. What I imagine to be the standard proof is simple and elementary:

Let $v$ be an eigenvector for $A$ -- say $Av = \alpha v$ -- and choose $k \in \mathbb{Z}^+$ minimal so that the set $\{v,Bv,B^2v,\ldots,B^k v\}$ is linearly dependent. Then $W = \operatorname{span}(v,Bv,\ldots,B^{k-1} v)$ is $B$-stable, so it has an eigenvector for $B$: say $Bw = \beta w$. On the other hand, since for all $0 \leq j \leq k-1$, $AB^j v = B^j A v = B^j \alpha v = \alpha B^j v$, every vector in $W$ is an eigenvector for $A$, and thus $w$ is an eigenvector for both $A$ and $B$.

I am tempted to say that it often works like this: you can sometimes bring in commutative algebraic methods to prove linear algebraic results in nontrivial and interesting ways, but in most cases I know the standard linear algebraic methods are simpler and more efficient. One exception is that in proving polynomial identities (like the Cayley-Hamilton theorem) in linear algebra it is often useful to note that it is sufficient to show that the identity holds "generically", or even on a Zariski-dense subset of the appropriate vector space.

  • $\begingroup$ Thanks for sharing these interesting and detailed pedagogical remarks. $\endgroup$
    – vgty6h7uij
    Jul 4, 2012 at 19:52

This answer doesn't contain anything new and it was already alluded via the JNF in the question, but let me elaborate this point of view:

Given a vector space $V$ over a field $K$, then an endomorphism $f : V \to V$ is the same as giving $V$ the structure of a left $K[T]$-module (which restricts to the given $K$-module structure). The multiplication with $T$ on the left is just $f$. This is just the most natural gadget when you think about polynomials $f^n + a_{n-1} f^{n-1} + \dotsc + a_0$ and their action on $V$. Many notions of linear algebra can be understood in more concise terms with the help of this module (and actually this can be used as a motivation for an introduction to modules in a linear algebra class):

  • A $f$-invariant subspace of $V$ is just a $K[T]$-submodule of $V$
  • That $V$ is $f$-cyclic just means that $V$ is cyclic as a $K[T]$-module, i.e. generated by one element, i.e. isomorphic to $K[T]/(p)$ for some $p$. We have that $V$ is finite-dimensional iff $p \neq 0$.
  • It immediately follows that $f$-invariant subspaces of $f$-cyclic spaces are also $f$-cyclic. Try to prove this without the language of modules.
  • That $V$ is $f$-indecomposable just means that it is indecomposable as a $K[T]$-module
  • More general and concisely: The arrow category of $\mathsf{Vect}_K$ is isomorphic to the category of $K[T]$-modules.
  • The minimal polynomial $p$ of $f$ is just the unique normed generator of the kernel of the ring homomorphism $K[T] \to \mathsf{End}_K(V)$ given by the module.
  • When you think about the endomorphism of $K[T]/(p)$ which multiplies with $T$ in terms of the canonical basis, you will arrive automatically at the companion matrix of $p$ and the relation with $f$-cyclic vector spaces. You don't have to "learn" this, it is just there.
  • The structure theorem for finitely generated modules over principal ideal domains implies the general normal form (which holds over every field $K$): Every endomorphism of a finite-dimensional vector space is similar to a direct sum of cyclic endomorphisms. In coordinate language: Every matrix is similar to a block matrix of companion matrices.
  • If $K$ is algebraically closed, you may reduce to polynomials of the form $(T-\lambda)^n$, but $K[T]/(T-\lambda)^n \cong K[T]/T^n$ has a very simple matrix, namely the Jordan block. Hence, the general normal form implies the Jordan normal form: Every matrix is similar to a block matrix of Jordan blocks.

Criteria for unitarily similarity between two matrices is given by Specht in 1940 in a paper titled “Zur Theorie der Matrizen” (“Toward Matrix Theory”).

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    $\begingroup$ Zur Theorie der Matrizen is On Matrix Theory in English (in this context zur does not mean towards). $\endgroup$
    – Lennart
    Jul 4, 2012 at 18:51

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