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<n} (v_{i+1}+\lambda v_i)x_i + \lambda v_n.

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.

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