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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.

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It's a "subfield" in the sense that it is a subring that happens to be a field. – Arturo Magidin Feb 7 '12 at 16:38
@ArturoMagidin Ah, re-reading the question that seems more like the issue. Maybe the terminology isn't good. – Dylan Moreland Feb 7 '12 at 16:41
The point is that $K$ is isomorphic to a subfield of the $F$-algebra $\mathcal L_{F-lin}(K,K)$, end of story. Of course if one wants to uglify this nice result, one can always find a non-natural isomorphism of said algebra with a matrix ring by artificially choosing a basis of $K$ over $F$. – Georges Elencwajg Feb 7 '12 at 16:59
@GeorgesElencwajg Do you mean that the "subfield" in the problem statement is not referring to a subset of $M_{n}(F)$ which has the structure of a field, but that $K$ is isomorphic to a subfield of the collection $\mathcal{L}(K,K)$? I'm sorry, I haven't seen that notation before: $\mathcal{L}(K,K)$ refers to the collection of $F$-linear transformations from $K$ to $K$ correct? – David K. Feb 7 '12 at 18:04
@GeorgesElencwajg, «The introduction of coordinates is an act of violence,» quoth (more or less) Weyl :) – Mariano Suárez-Alvarez Feb 7 '12 at 18:39
up vote 4 down vote accepted

"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).

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Thanks for the replies. I think I found a very simple answer to my original question. Since $K$ is a field, $\alpha$ and $\beta$ commute, so $\psi(\alpha\beta)=\psi(\beta\alpha)$ and since $\psi(\alpha\beta)=\psi(\alpha)\psi(\beta)$, and $\psi(\beta\alpha)=\psi(\beta)\psi(\alpha)$ it follows that $\mathcal{M}_\mathcal{B}(T_\alpha)\mathcal{M}_\mathcal{B}(T_\beta) =\mathcal{M}_\mathcal{B}(T_\beta)\mathcal{M}_\mathcal{B}(T_\alpha)$. Hence, the matrices in $\text{im}\,(\psi)$ form a commutative subring of $M_n(F)$. – David K. Feb 11 '12 at 19:34

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