# $k$-subalgebra of finite extension of $k$ is a field

I am working on a proof which has

Let $\mathfrak m,\mathfrak n$ be maximal ideals and $A$ Noetherian. Given that, $A[T_1,\dots,T_n]/\mathfrak n$ is a finite extension of $A/\mathfrak m$, if $\mathfrak n_1 = \mathfrak n \cap A[T_1,\dots,T_{n-1}]$ then $A[T_1,\dots,T_{n-1}]/\mathfrak n_1$ is an $A/\mathfrak m$-subalgebra of $A[T_1, \dots, T_n]/\mathfrak n$ and therefore is a field.

Is this a special case of a more general result such as

Let $K/k$ be a finite extension. If $L$ is a $k$-subalgebra of $K$, then $L$ is a field.

If the above result is true, how can one justify it?

If the above result is false, how should one justify the original proof?

A $k$-subalgebra $A$ of a finite extension $K/k$ is (a subring of a field, hence) an integral domain, so multiplication by any nonzero $a \in A$ is injective on $A$. But since $A$ is (a $k$-subalgebra of a finite-dimensional algebra, hence) finite-dimensional, this means multiplication is also surjective. So $A$ is a field.