Let $K$ be a field, let $L/K$ be a field extension, and let $A$ be a unital associative finite dimensional $K$-algebra.
In what follows, "morphism" stand for "$K$-algebra morphism".
If $A$ is a field, it is known that $\lvert \operatorname{Hom}(A, \bar{K})\rvert \leq [A:K]$, and it is not difficult to get convinced that it is true if we replace $\bar{K}$ by any field extension $L$ (since $A/K$ is finite, hence algebraic, the image of any morphism $A\to L$ is contained in $\bar{K}$).
For fun sake, I just wondered if it was true for any $A$. I think I found a proof, but I would be happy if someone confirms everything is correct.
If $Hom(A, L)$ is empty, there is nothing to do, so we may assume that there exists a morphism $f:A\to L$. Since $A$ is finite dimension over $K$, $A$ is artinian, and the Jacobson radical $J(A)$ is a nilpotent ideal. If $k\geq 1$ satisfies $J(A)^k=0$, then for all $x\in J(A)$, we have $f(x)^k=f(x^k)=f(0)=0$, and since $L$ is a field, $f(x)=0$.
Hence $f$ induces a morphism $A/J(A)\to L$ , and $Hom(A,L)$ and $Hom(A/J(A),L)$ has the same number of elements.
Since $A$ is artinian, so is $A/J(A)$. Since $A/J(A)$ is artinian with zero Jacobson radical, $A/J(A)$ is semi-simple. Since $\dim_K(A/J(A))\leq \dim_K(A)$, one may assume, replacing $A$ by $A/J(A)$, that $A$ is a semi-simple $K$-algebra.
Hence $A\simeq M_{n_1}(D_1)\times\cdots\times M_{n_r}(D_r)$, where $D_i$ is a finite dimensional division $K$-algebra.
Let $f:M_{n_1}(D_1)\times \cdots\times M_{n_r}(D_r)\to L$ be a morphism,and let $e_i=(0,\ldots,0,1,0,\ldots)$. Since $e_i$ is an idempotent, $f(e_i)$ is an idempotent of $L$, ie $0$ or $1$. Since $f((1,\ldots,1))=1=f(e_1)+\cdots+f(e_r)$, there is a $j$ such that $f(e_j)=1$. For all $i\neq j$, we have $f(e_ie_j)=f(0)=0=f(e_i)f(e_j)=f(e_i)$. Hence there is exactly one $j$ for which $f(e_j)=1$, and the other $f(e_i)'s$ are zero.
Since $(x_1,\ldots,x_{j-1},0,x_{j+1},\ldots,x_r)=\displaystyle\sum_{i\neq j}(0,\ldots,x_i,0,\ldots,0)=\displaystyle\sum_{i\neq j}(0,\ldots,x_i,0,\ldots,0)e_i$, applying $f$ shows that $I=\displaystyle\prod_{i\neq j}M_{n_i}(D_i)$ satisfies$ f(I)=0$. Hence we have an induced morphism $M_{n_j}(D_j)\to L$. Since $M_{n_j}(K)$ is simple, this morphism is injective. This forces $M_{n_j}(D_j)$ to be commutative, so $n_j=1$ and $D_j$ is a field extension of $K$ of finite degree.
Hence $A\simeq D_1\times\cdots\times D_r$, where $D_i/K$ has finite degree, so $\dim_K(A)=[D_1:K]+\cdots+[D_r:K]$. The commutative case shows that there is at most $[D_j:K]$ choices for the induced morphism , and the previous analysis shows that there is at most $[D_1:K]+\cdots+[D_r:K]=\dim_K(A)$ morphisms $f:D_1\times \cdots\times D_r\to L$.
Question 1. Is everything correct?
Question 2. Is there a simpler proof ?
Question 3. I'm pretty sure all of this is well-known. Does anyone could provide me a reference ?
Thanks !