# Jordan-Chevalley decomposition of $T$ acting on $k[T]/(\pi(T)^e)$

Given an algebraically closed field $K$, a f.d. vector space $V$ over $K$ and $A\in{\rm GL}(V)$, we can view the space $V$ as a $K[T]$-module, where $T$ acts by $A$. Using the fundamental theorem of modules over principal ideal domains we can decompose $V$ as a direct sum of quotients $K[T]/(T-\lambda)^e$ for some scalars $\lambda\in K$ and naturals $e\ge1$. This is because all irreducible elements of $K[T]$ are linear factors, since $K$ is algebraically closed. If we collect all these quotients for fixed $\lambda$, we get the generalized eigenspace of $T$ associated to the eigenvalue $\lambda$. If we explicitly write the matrix for how multiplication-by-$T$ acts on $K[T]/(T-\lambda)^e$, we get the $e\times e$ Jordan block again associated with the eigenvalue $\lambda$. If we're dealing with a single summand of $V$, writing $T=\lambda+(T-\lambda)$ gives us the Jordan-Chevalley decomposition of $T$ (writing $T$ as a sum of a semisimple, i.e. diagonalizable, part and a nilpotent part). Do this over all summands to get the JC decomposition over $V$.

I find this very pleasant - a lot of stuff that could be done using matrices and coordinate vectors, which can be a headache, are neatly organized with polynomials and "purer" algebra.

But if our scalar field $k$ is not algebraically closed then our irreducibles in $k[T]$ needn't be linear so we can't use the $T=\lambda+(T-\lambda)$ trick. However, according to Wikipedia, the semisimple and nilpotent parts of $T$ are polynomials in $T$. This means (for convenience, focusing on a single block), the semisimple and nilpotent parts of $T$ acting on $k[T]/(\pi(T)^e)$ are multiplication-by-$s(T)$ and multiplication-by-$n(T)$ respectively for some polynomials $s,n$ that depend on $\pi(T)$ and $e$. Can we write these polynomials down explicitly?

## 1 Answer

Yes, you can find these polynomials by using the Chinese Remainder Theorem to solve a system of congruences as given in section 4.2 of Humphreys's Introduction to Lie Algebras and Representation Theory (pp. 17-18). However, I think you still need the eigenvalues of the transformation to be contained in the field, since the eigenvalues are used in the coefficients of these polynomials. Here is the relevant passage (note that Humphreys calls the linear transformation $x$ and the polynomial ring $F[T]$):

Let $a_1, \ldots, a_k$ (with multiplicities $m_1, \ldots, m_k$) be the distinct eigenvalues of $x$, so the characteristic polynomial is $\prod_i (T - a_i)^{m_i}$...Now apply the Chinese Remainder Theorem (for the ring $F[T]$) to locate a polynomial $p(T)$ satisfying the congruences, with pairwise relatively prime moduli: \begin{align*} p(T) &\equiv a_i \pmod{(T - a_i)^{m_i}}\\ p(T) &\equiv 0 \pmod{T} \end{align*} (Notice that the last congruence is superfluous if $0$ is an eigenvalue of $x$, while otherwise $T$ is relatively prime to the other moduli.

Humphreys shows that $x_s := p(x)$ and $x_n := x - p(x)$ are semisimple and nilpotent, respectively, and satisfy the other criteria, and are unique.

• There is a unique solution $p(T)$ of minimal degree (its leading coefficient can't be altered). If $\sigma$ is any field automorphism then $\sigma p(T)$ also satisfies the congruence and has the same degree, so $\sigma p(T)=p(T)$. Assuming the eigenvalues are separable, by Galois theory this implies $p(T)$'s coefficients are in the original scalar field. Not sure about what happens if any scalars are inseparable. It's interesting that the actual polynomial expressions that give us the semisimple/nilpotent parts depend only on the characteristic polynomial, and not more refined data. – whacka Aug 14 '14 at 21:47