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.
Best Answer
"Noncommutative" doesn't mean nothing commutes, it just means that things don't necessarily commute. Any noncommutative ring admits commutative subrings, for example the subrings generated by a single element. To give a more explicit example, $\mathbb{C}$ is isomorphic to a subring of $M_2(\mathbb{R})$ via the map $$a + bi \mapsto \left[ \begin{array}{cc} a & -b \\ b & a \end{array} \right]$$
(and this is in fact a special case of the result quoted above).