If $[E:F] = n$, $m < n$ show that there cannot be an injective F-algebra homomorphism $L:E \rightarrow M(m, F)$.

Question

My initial thoughts on this question is a sort of Rank-Nullity argument, which I don't know if it's on the right path or not. Note that $E/F$ is a finite field extension.

Let $n = [E:F]$. We know in this case that $L: E \rightarrow M(n, F)$ is an injective $F$-Algebra homomorphism (by prior problem). In this case, we have that the $Ker(L) = \emptyset$. However, we now consider $L: E \rightarrow M(m, F)$, $m < n$. We can note that $dim(M(m,F)) < dim(M(n,F))$. By Rank-Nullity, treating these as vector spaces, we have $dim(Ker(L)) + dim(Im(L)) = dim(E/F) = n$. Note that $dim(Im(L))$ cannot be greater than $m$, since this is the dimension of the codomain. This then implies $dim(Ker(L)) \neq 0$, or that we do not have a trivial Kernel. Hence, we cannot have an injective mapping.

I could have completely misunderstood this problem, so if this is entirely off I apologize. I would greatly appreciate any hints on what the right track would be.

Answer 1

Let $a_1,\ldots, a_n$ be a basis of $E$. If $L$ exists, then the first rows of $L(a_1),\ldots, L(a_n)\in F^m$ are linearly dependent, hence we find coefficients (not all zero) such that $$L(\underbrace{\sum c_ia_i}_{=:\alpha})=\sum c_iL(a_i)$$ has a zero first row, hence is not invertible even though $\alpha$ is.

(This works even if you do not require that $L(1)=1$, by noticing that $L(1)$ must project onto a subspace of $F^m$ and we can restrict $L$ to the endomorphism ring of that).

Answer 2

Here is a partial answer:

When $E=F(a)$ this is easy because $L(a)$ would have a minimal polynomial of degree $n$ but all elements in $M(m,F)$ have minimal polynomial of degree at most $m$.

If $E/F$ is separable, then this argument suffices because of the primitive element theorem.