I'm working on a problem from Dummit & Foote's Abstract Algebra and I'm a bit confused about one part of the problem. The problem reads:
Let $K$ be an extension of $F$ of degree $n$.
$\bf\text{(a)}$ For any $\alpha\in K$ prove that $\alpha$ acting by left multiplication on $K$ is an $F$-linear transformation of $K$.
$\bf\text{(b)}$ Prove that $K$ is isomorphic to a $\bf\underline{subfield}$ of the ring of $n\times n$ matrices over $F$, so the ring of $n\times n$ matrices over $F$ contains an isomorphic copy of every extension of degree $\leq n$.
I've already worked out $\bf\text{(a)}$, it's the part in $\bf\text{(b)}$ about a "$\bf\underline{subfield}$ of the ring of $n\times n$ matrices" that I'm a bit confused on.
What I have done so far is defined a map $\psi:K\to M_{n}(F)$ from $K$ to the ring of $n\times n$ matrices over $F$ given by $\psi(\iota)=\mathcal{M}_{\mathcal{B}}(T_{\iota})$ where $\iota$ is any element of $K$, $\mathcal{M}_{\mathcal{B}}(T_{\iota})$ is the matrix that represents the $F$-linear transformation $T_{\iota}:K\to K$, with respect to a basis $\mathcal{B}$ of the vector space $K$.
I can readily establish that $\psi$ is an injective homomorphism that is surjective to its image in $M_{n}(F)$ and since $K$ is a field, every nonzero $\alpha\in K$ has an inverse $\alpha^{-1}\in K$ so that $\psi(\alpha^{-1})=\psi(\alpha)^{-1}\in M_{n}(F)$.
So this establishes an isomorphism between $K$ and the image of $K$ under $\psi$.
This is the part that I'm confused about: the elements in $\text{im}\,(\psi)$ are elements of the noncommutative ring of $n\times n$ matrices over $F$. How is it that $\text{im}\,(\psi)$ is a subfield of $M_{n}(F)$ if a subfield is commutative and $M_{n}(F)$ is a noncommutative ring?
I should mention that I'm an undergraduate student in a graduate Galois Theory course and my linear algebra is a bit weak. So if I've left out some important or illuminating details I'd greatly appreciate having them pointed out.