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Hi I am trying to find if there is a more simple way to prove the following theoreom without going through the two page proof in the Linear Algebra textbook I have been using.

Let $V$ be a finite-dimensional vector space over an infinite field $F$ and let $T:V \rightarrow V$ be a linear operator. Give to $V$ the structure of a module over the polynomial ring $F[x]$ by definiting $x \alpha = T(\alpha)$

Question: How do we sketch a proof that $V$ is a direct sum of cyclic $F[x]$-modules?

The proof I know for a vector space $V$ comes from Hoffman and Kunze's "Linear Algebra" and has four steps but is in the languae of cyclic subspaces $Z(\alpha,T)$. In particular I could reproduce the proof of the Cyclic decompostion theorem on p.233 which states if $W_0$ is a $T$-admissible subspace of $V$ there exists non-zero vectors $\alpha_1, \ldots, \alpha_r \in V$ with respective $T$-annihalators $p_1, \ldots, p_r$ such that $V = W_0\oplus Z(\alpha_1;T) \oplus \ldots \oplus Z(\alpha_r;T)$. I feel like there are some shortcuts to be made here.

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My favorite reference for this is Algebraic Theory of Numbers by Pierre Samuel…). – Pierre-Yves Gaillard Aug 4 '11 at 10:32
This is not the primary decomposition theorem at all. That theorem does not ask for cyclic summands, while this question does not involve anything about primes. – Marc van Leeuwen Mar 23 '15 at 11:19
up vote 3 down vote accepted

I don't know if this is what you are looking for, but perhaps? Here is a sketch of the proof in Artin, Ch. 12.

1) Prove a matrix presenting a finitely-generated module over a Euclidean domain can be reduced to a diagonal matrix presenting the same module.

The idea is that invertible elementary row and column operations on the matrix don't actually change the module. Artin gives a specific algorithm for getting the matrix in diagonal form using these operations. Sketch of the algorithm: permute rows and columns to get the smallest element (w.r.t. the Euclidean norm) in the upper left corner; divide other elements in the first row and column by this one; since it is a Euclidean domain, either we have cleared out the first row and column except for the upper-left entry, or else something smaller has shown up. If the latter, permute rows or columns to get it in the upper left corner and repeat until the first row and column are cleared out; otherwise, ignore the first row and column and begin the whole procedure on the submatrix that doesn't include them.

2) Interpret this result in the present case!

$V$ is finite-dimensional so it is finitely generated as an $F[x]$-module; thus the above result applies, and it is presented by some diagonal matrix! This makes it a direct sum of cyclic submodules!

What I love about this proof is that regarding an abelian group as a $\mathbb{Z}$-module and replacing $F[x]$ with $\mathbb{Z}$, the same exact argument gives us the structure theorem for abelian groups.

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