Really just a note to remind me why the modules proof of JNF works.

Suppose $F$ is an algebraically closed field, and consider for some fixed automorphism $\alpha$ of $V$, a finite dimensional vector space over $F$, the $F[X]$-module $V$ with scalar multiplication given by $f(x) v = f(\alpha)(v)$. It’s easy to check this yields the appropriate module structure.

Now, let’s think about what happens if $M \equiv F[X]/(X-\lambda)^n$. This is telling us that $(\alpha - \lambda)$ is nilpotent with degree $n$. In this case, it is certainly true that there is a vector $v$ such that $(\alpha - \lambda)^{n-1}v$ is nonzero, and furthermore, we must have that $v_1 = v, v_2= (\alpha - \lambda)v,..., v_n=(\alpha - \lambda)^{n-1}v$ are linearly independent (if not then we can find some polynomial of degree less than $n$ which annihilates the entire module, which is clearly preposterous.

So let’s take these $n$ vectors. They are a basis, since it is clear the dimension of $M$ over $F$ is $n$. But it is obvious that if we apply $\alpha$ to an arbitrary vector $x = \sum x_i v_i$ we get

$\alpha x = \alpha (\sum x_i v_i) = \sum \alpha (v_i) x_i = \sum_{i

Writing this as a matrix,

$\alpha x = \left(\begin{array}{cccc} \lambda & 1 & & \\ & \lambda & 1 & ... \\ & & ... & \\ & & & \lambda\end{array} \right) x$.

This is a Jordan block. The JNF is now clear for an arbitrary linear map by writing (using the structure theorem for finitely generated modules) the above module as a direct sum of modules of the form above – utilising that $F$ is algebraically closed to note that all primes in $F[X]$ are linear, and hence that the primary decomposition is of the desired form, splitting the space $V$ into Jordan blocks.