Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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

1 Answer 1

up vote 3 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).

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.