Representing an finite-dimensional algebra over a field using a quotient ring

Question

My professor was talking about different ways to think about an algebra $A$ over a field $k$. One that she mentioned briefly but didn't go into much detail on was that of a quotient ring. Roughly the idea was to take vectors $v_1,...,v_n$ as a basis for $A$ and construct the algebra as $A = k[v_1,...,v_n]/I$, where $I$ was an ideal for which we didn't get a specific definition in lecture.

Is there a way we can pick $I$ such that $A$ will be an algebra that’s finite-dimensional as s vector space over $k$, and in particular that $k[v_1,...,v_n]/I$ will be isomorphic to $\mathrm{span}_k(v_1,...,v_n)$?

(P.S. we were specifically talking about algebras that are also integral domains, but AFAIK that's super relevant)

Answer 1

I think the following is what you're talking about.

Let $A$ be a commutative, finite-dimensional $k$-algebra. Let $v_1,\ldots,v_n$ be a basis for $A$ over $k$. For each $i$ and $j$, let $v_iv_j = \sum_kc_{ijk}v_k$. Then $A\cong k[x_1,\ldots,x_n]/I$, where $I$ is the ideal generated by $$\left\{x_ix_j-\sum_k c_{ijk}x_k : 1\le i,j\le n \right\}.$$

Proof:

Let $R=k[x_1,\ldots,x_n]$ for notational convenience.

Let $\phi : k[x_1,\ldots,x_n]\to A$ be the map sending $x_i$ to $v_i$. By construction, $I\subseteq \ker\phi$, so $\phi$ induces a map $\psi : k[x_1,\ldots,x_n]/I\to A$.

We want to show that $\psi$ is an isomorphism. We will proceed by constructing an inverse. Let $\tau : A \to R$ be the $k$-linear map defined by $v_i\mapsto \bar{x_i}$, where $\bar{x_i}$ represents the image of $x_i$ in the quotient. Then $\psi(\tau(v_i))=v_i$, so $\psi\tau = 1_A$, and $\tau(\psi(\bar{x_i}))=\bar{x_i}$. We have two choices for how to proceed here. If we prove that the $\bar{x_i}$ span $R$ as a $k$-vector space, then we will have $\tau\psi = 1_R$. However, we can also prove that $\tau$ is a ring homomorphism, which will also give us $\tau\psi =1_R$, since while we don't know yet that $\bar{x_i}$ span $R$, we do know that they generate $R$ as a ring. It seems to me that proving $\tau$ is a ring homomorphism is the easier route, so we'll go that way.

Let $a=\sum_i a_iv_i$, $b=\sum_j b_jv_j$ be elements of $A$ expressed in terms of the basis elements.

Then $$ab = \sum_i\sum_j a_ib_jv_iv_j = \sum_i\sum_j\sum_k a_ib_jc_{ijk}v_k.$$ Hence $$\tau(ab) = \sum_i\sum_j\sum_k a_i b_j c_{ijk}\bar{x_k} = \sum_i\sum_j a_ib_j\bar{x_i}\bar{x_j} = \sum_i a_i\bar{x_i}\sum_j b_j\bar{x_j} = \tau(a)\tau(b),$$ so $\tau$ is in fact a ring homomorphism.