Let $\mathfrak{A}$ be a finite dimensional $k$-algebra for alg. closed $k$. Prove ${\rm Aut}(\mathfrak{A})$ is a closed subgroup of $GL(\mathfrak{A})$

Question

This is Exercise 7.6.3 of Humphreys', "Linear Algebraic Groups".

The Question:

Let $\mathfrak{A}$ be a finite dimensional $k$-algebra for algebraically closed field $k$. Prove that ${\rm Aut}(\mathfrak{A})$ is a closed subgroup of $GL(\mathfrak{A})$ (as a linear algebraic group).

Here "closed" is with respect to the Zariski topology.

The Details:

For the definition of a linear algebraic group I'm using, see this:

Showing the linear algebraic subgroup $\Bbb U_n$ of $\Bbb{GL}_n(k)$ is closed.

Context:

My thoughts are that this ought to fall out nicely from the definitions. After all, the exercise is early on in the theory of linear algebraic groups.

I reckon I should be able to do this myself if I were going at a leisurely pace. However, I am keen to understand things thoroughly, so I'm going back over key concepts; there's only so much time I can allocate to it.

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The goal is to show that ${\rm Aut}(\mathfrak{A})=\mathcal{V}(I)$ for some ideal $I$ of a polynomial algebra (that, to be honest, I can't identify right now - mental block - but it's something simple). This can be done indirectly.

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I've searched in all of the "Linear Algebraic Groups" books, on Google, on MSE (via Approach0), and so on. I found a paper that mentions this theorem, but that's not surprising, given the elementary nature of it.

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The type of answer I'm looking for is a basic approach to/outline of a proof, please. I think I would benefit from filling in the details myself.

Please help :)

Answer 1

I don't know if this is the best approach, but at least this is a very elementary one.

Choose a $k$-basis $a_1,\dots,a_n$ of $\mathfrak{A}$ as a $k$-vector space. Then for each $i,j\in\{1,\dots,n\}$ we have $$ a_i a_j = \sum_{\ell=1}^n c_{ij}^\ell a_\ell $$ for some scalars $c_{ij}^\ell\in k$. These scalars are called the structure constants of $\mathfrak{A}$ and determine completely its structure as a $k$-algebra.

Now recall that an element $\sigma\in\mathrm{GL}(\mathfrak{A})$ is completely determined by a non-singular $n\times n$ matrix $(A_{ij})$, where $$ \sigma(a_i) = \sum_{j=1}^n A_{ij}a_j. $$ Define $X_{ij}(\sigma) = A_{ij}$. Then $X_{ij}$ is a regular function on $\mathrm{GL}(\mathfrak{A})$, and these are the coordinate functions of $\mathrm{GL}(\mathfrak{A})$ as an algebraic variety over $k$. Then it is easy to see that $R=k[X_{ij}|i,j=1,\dots,n][\det^{-1}]$ is the ring of (global) regular functions on $\mathrm{GL}(\mathfrak{A})$, where $$ \det = \det(X_{ij}). $$ (I think you already know all of this.)

In order for $\sigma$ to belong to $\mathrm{Aut}(\mathfrak{A})$, a necessary and sufficient condition is that that $\sigma(a_ia_j)=\sigma(a_i)\sigma(a_j)$ for all $i,j=1,\dots,n$. This is equivalent to $$ \sum_{\ell=1}^n c_{ij}^\ell \sigma(a_\ell) = \sigma(a_i)\sigma(a_j), $$ which in turn is equivalent to $$ \sum_{\ell=1}^n c_{i,j}^\ell \sum_{p=1}^n A_{\ell p}a_p = \left(\sum_{\ell=1}^n A_{i\ell}a_\ell\right)\left(\sum_{m=1}^n A_{jm}a_m\right) = \sum_{\ell=1}^n\sum_{m=1}^n A_{i\ell}A_{jm}\sum_{p=1}^n c_{\ell m}^p a_p. $$ From this, you can easily see that $I$ must be the ideal in $R$ generated by the polynomials $$ \sum_{\ell=1}^n c_{ij}^\ell X_{\ell p} - \sum_{\ell=1}^n \sum_{m=1}^n c_{\ell m}^p X_{i\ell} X_{jm} $$ for $i,j,p=1,\dots,n$.