An unusual application of the Skolem-Noether theorem

last edited around 2020-08-02

\(\DeclareMathOperator{\End}{End} \DeclareMathOperator{im}{Im} \DeclareMathOperator{diag}{diag}\)

CSAs and Skolem-Noether

Definition 1 A \(k\)-Algebra is a ring \(A\) with a designated inclusion \(\iota\colon k\hookrightarrow Z(A)\), and a homomorphism of \(k\)-Algebras is a ring homomorphism commuting with this inclusion. Out of brevity, we will omit the inclusion and identify \(k\) with \(\iota(k)\) where unambiguous. \(A\) is centrally simple (“CSA”) if it has no two-sided ideals and \(Z(A) = k\).

It is quite easy to prove:

Lemma 2If \(V\) is an \(n\)-dimensional \(k\)-Vector space, then \(\End_k(V)\) is a CSA.

Proof

Choosing a basis gives us an isomorphism to \(M_n(k)\). Fix any matrix \(M\). Since two-sided ideals are closed under addition as well as multiplication of any matrix from both sides, we can bring \(M\) into a form where the first column contains one \(1\) and otherwise zeroes. Multiplying from the right with \(\diag(1, 0, \ldots)\), we get a matrix \(E_{i,j}\) whose only non-zero enrty is a single \(1\) at \((i, j)\). Appropriate conjugation can move \((i,j)\) around arbitrarily, so the generated two-sided ideal \((M)\) must contain \(\{E_{i,j}\mid 1\leq i,j\leq n\}\), which of course is enough to build all matrices given addition and scaling. Thus \((M) = M_n(k)\).

To identify the center, note that a matrix \(M\) commutes with \(\diag(0,\ldots,1,\ldots,0)=E_{ii}\) if and only if projection on the \(i\)-th column and row produce the same result. This can only be the case if the \(i\)-th row and column only contain a nonzero entry in \((i,i)\). Thus, the only way to commute with all matrices – in particular, all \(E_{ii}\) – is if it is diagonal. On the other hand, if \(P_{ij}\) is the matrix corresponding to the elementary row operation of adding the \(j\)-th row from the \(i\)-th when multiplied from the left, doing so from the right results in the column operation of adding the \(i\)-th column to the \(j\)-th. These things can only amount to the same if \(E_{ii}=E_{jj}\). This works for arbitrary \(1\leq i,j\leq n\), effectively restricting us to a multiple of our identity. Therefore, \(Z(M_n(k))\subseteq \{\lambda\mathbb 1\mid \lambda \in k\} = \iota(k)\), hence equality.

Skolem-Noether now says:

Theorem 3Let \(f, g\colon A\to B\) be two morphisms of \(k\)-Algebras. If \(B\) is a finite-dimensional CSA, then \(f\) and \(g\) are conjugated by an element \(b\in B\), i.e. \(f(a) = bg(a)b^{-1}=: {}^b g(a)\) for all \(a\in A\).

As a corollary, all automorphisms of a finite-dimensional CSA must be inner. Powerful stuff!

Application: Linear operators

Now for the core idea: If \(S, T\in \End_k(V)\) are two operators with the same minimal polynomial \(\mu\), can't we somehow use that structural similarity to find a conjugating element? Because then we would already be finished: For any \(\mu(x)\) that splits into \(\prod_i{(x-\lambda_i)}^{\alpha_i}\) (in particular if \(k\) is algebraically closed) and any given basis \(\mathscr B\), the operator corresponding to the Jordan matrix \(J\) with eigenvalues \(\lambda_i\) and multiplicities \(\alpha_i\) has \(\mu\) as its minimal polynomial. In that case, any \(T\) with minimal polynomial \(\mu\) would be conjugated to \(J\) via, say, \(S = {}^B J\), Implying that \(\{Bb_i\mid b_i\in \mathscr B\}\) is a basis bringing \(S\) into jordan normal form.

You might have guessed how to formally prove that \(S\) and \(T\) are indeed conjugated:

Theorem 4Let \(S, T\in \End_k(V)\) where \(V\) is finite dimensional. If \(S\) and \(T\) have the same minimal polynomial \(\mu\in k[x]\), they are conjugated.

Proof

Let \(f, g\colon k[x]/(\mu)\to \End_k(V)\) be given by the evaluations \(f( p + (\mu)) := p(S), g(p+(\mu)):=p(T)\). Observing that \(f,g\) are \(k\)-Algebra homomorphisms and that \(\End_k(V)\) is a CSA, apply Skolem-Noether to \(f\) for the polynomial \(x+(\mu)\).

As a closing remark: Please don't ever teach linear algebra that way.