A presentation of the full matrix algebra

Lemma 1   Let $k$ be a field of characteristic $p$. Then a $k$-algebra $\mathfrak{M}$ which is generated by $\alpha_1,\alpha_2,\dots,\alpha_n,\beta_1,\beta_2,\dots,\beta_n$ with the relations

$$[\beta_j,\alpha_i]=\delta_{ji}, \beta_j^p=0 , \alpha_i^p=0 \qquad (i,j=1,2,\dots,n)$$

(where $\delta_{ij}$ is the Kronecker's delta) is isomorphic to the full matrix algebra $M_{p^n}(k)$.

Since $M_{p^n}(k)$ is isomorphic to a tensor product of $n$ copies of the matrix algebra $M_{p}(k)$, we may assume that $n=1$. We define elements ${\mu},{\nu}\in M_{p}(k)$ as follows.

$${\mu}=(\delta_{i+1,j}),\qquad {\nu}=((p-j)\delta_{i,j+1})$$ (1)

Then ${\mu},{\nu}$ satisfies the same relation as $\alpha_1,\beta_1$. In other words, we have a $k$-algebra homomorphism $\phi$ from $\mathfrak{M}$ to $M_p(k)$ with $\phi(\alpha_1)={\mu},\phi(\beta_1)={\nu}$.

On the other hand, it is easy to see that the algebra $\mathfrak{M}$ is linearly generated by $\{\alpha^i \beta^j; 0\leq i,j\leq p-1\}$ and hence that its dimension is not greater than $p^2$.

By a dimension argument we see that the algebra homomorphism $\phi$ is an isomorphism.

$\qedsymbol$

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