Embedding a field into a matrix ring

Question

I'm interested in subrings of matrix rings over a field and would like to understand what kinds of fields can be realised as subrings of $M_n(L).$ If $K$ is a finite extension of $L$ or $L$ is an extension of $K$, we easily see that there exists $n\ge 1$ such that $K\subset M_n(L)$. But in other situations, it seems a little more difficult. The following is what I have figured out so far.

1. If $K$ and $L$ can be embedded into an algebraically closed field so that $n=[K:K\cap L]$ is finite, then $K$ is isomorphic to some subring of $M_n(K\cap L)$ and thus of $M_n(L).$ For example, when $K=\mathbb{C}$ and $L=\mathbb{R}(X)$, they can be embedded into the algebraic closure of $\mathbb{R}(X)$ so that $[\mathbb{C}:\mathbb{C}\cap\mathbb{R}(X)]=[\mathbb{C}:\mathbb{R}]=2$ and you can regard $\mathbb{C}\subset M_2(\mathbb{R})\subset M_2(\mathbb{R}(X))$ by expressing the elements of $\mathbb{C}$ as a $\mathbb{R}$-linear transformation on $\mathbb{C}$.
2. However, there exist embeddings that cannot be constructed this way. For example, we have $\mathbb{C}\cong \mathbb{R}\left[\left(\begin{array}{cc}0&i\\i&0\end{array}\right)\right]\subset M_2(\mathbb{C}).$

In the example given in statement 2, we take $K=L=\mathbb{C}$, yet these fields still satisfy the assumption in statement 1. Is the assumption necessary for a pair of fields $(K, L)$ to admit some $n\ge1$ such that $K$ is isomorphic to a subring of $M_n(L)$?

Edit:
I will clarify my question. I ultimately want to find the answer to the following question.

Let $L$ and $K$ be fields and $n\ge 1$ an integer. What is the necessary and sufficient condition for the existence of an embedding of $K$ into $M_n(L)$?

I found a sufficient condition mentioned in claim 1 above and I believe that this is not only sufficient, but also necessary. Could someone provide a proof or a counterexample to this belief?

Answer 1

Let $\sigma : K \to M_n(L)$ be an embedding and consider the subring $R$ of $M_n(L)$ generated by $\sigma(K)$ and $L$ (embedded as scalar multiples of the identity), which is commutative. $R$ is a finite-dimensional $L$-algebra into which $K$ embeds; conversely given such an embedding of $K$ into a finite-dimensional $L$-algebra we get an embedding into $M_n(L)$. So

$K$ embeds into $M_n(L) \Leftrightarrow K$ embeds into a finite-dimensional commutative $L$-algebra.

The virtue of passing to $R$ is that we can now take quotients. If $m$ is any maximal ideal, then $R/m$ is a finite extension of $L$ into which $K$ continues to embed. So

$K$ embeds into $M_n(L) \Leftrightarrow K$ embeds into a finite extension of $L$.

So, this condition says "$K$ is a subfield of a finite extension of $L$," while your condition says "$K$ is a finite extension of a subfield of $L$." Personally I can't tell whether these conditions are equivalent but this one provably works and I find it easier to think about.

It seems to me difficult to control the size of the intersection $K \cap L$ due to examples like the following. Take $L = \mathbb{R}$ and $K = \mathbb{Q}(\alpha + \beta i) \subset L' = \mathbb{C}$ where $\alpha, \beta \in \mathbb{R}$ are independent transcendentals, so by construction $K$ is a subfield of a finite extension of $L$. But $K \cap L = \mathbb{Q}$, since $K$ contains no nontrivial real elements. So $[K : K \cap L]$ can be infinite while $[KL : L]$ is finite. This is not a counterexample because of course $K \cong \mathbb{Q}(t)$ abstractly which admits a different embedding into $L$ itself. But overall I don't know what to think either way about whether a counterexample exists.

Edit: Here's a proof that $K \cap L = \mathbb{Q}$. By hypothesis $\alpha, \beta$ are algebraically independent, which means the same is true for $z = \alpha + \beta i, \bar{z} = \alpha - \beta i$ (in $\mathbb{C}$, over $\mathbb{Q}$). Suppose $q(t) \in \mathbb{Q}(t)$ is such that $q(z) \in K \cap L \subset \mathbb{R}$, so in terms of complex numbers $q(z)$ is real. Then

$$q(z) = \overline{q(z)} = q(\overline{z})$$

is an algebraic dependence between $z$ and $\bar{z}$, which is impossible, unless $q \in \mathbb{Q}$ is constant.

This means that, contrary to what you might expect, although it is true that $z$ is algebraic over $L = \mathbb{R}$, with minimal polynomial $(t - z)(t - \bar{z}) = t^2 - 2 \alpha t + \alpha^2 + \beta^2$, neither of the coefficients of its minimal polynomial are in $K$.

Answer 2

I've decided to add another answer, with an explicit counterexample to the question clarified after my first answer. Thanks to Qiaochu Yuan's comments I streamlined some arguments.

Any field $K \subseteq \mathbb C$ embeds into $M_2(L)$ for $L=\mathbb R$. We are wondering if such $K$ always has at least some embedding (as per clarified answer: possibly different from the "original" inclusion) $i: K\hookrightarrow \mathbb C$ that makes $i(K)$ finite over $i(K)\cap \mathbb R$. Since a purely transcendental extension of $\mathbb Q$, like in Qiaochu's answer, can just be embedded into $\mathbb R$ and thus fails as a counterexample to the clarified question, the task was to find an infinite algebraic extension of $\mathbb Q$ (= a tower of number fields) which "stays away from $\mathbb R$" as much as possible. My first idea had been to adjoin square roots of more and more negative square-frees -- say, $\mathbb Q (\{\sqrt{-p} : p \text{ prime} \})$ -- but OP quickly pointed out this has big real subfields. (Duh: Any Galois extension of the rationals has a real subfield of index $\le 2$.) So my next idea was to build a very non-real and non-Galois tower by instead fixing one negative square-free, and then taking layers of square roots of square roots of square roots ... of that. And I think it works:

---

Fix any odd prime $p$. For $n \ge 1$, let $K_n := \mathbb Q(\omega_n)$ where $\omega_n := \sqrt[2^n]{p} \cdot \exp(\frac{2\pi i}{2^{n+1}})$ is the "principal" (i.e. having smallest positive argument) $2^n$-th complex root of the negative number $-p$. (Here, $\sqrt[2^n]{p}\in \mathbb R$ denotes the usual (positive, real) $2^n$-th root of $p$, but note that this number is not contained in $K_n$.)

Since $\omega_{n+1}^2=\omega_n$, $K_n \subset K_{n+1}$ for all $n$, so $K:=\bigcup_{n\ge 1} K_n$ is a field.

I claim this $K$, and $L=\mathbb R$, gives a counterexample to your "necessity" conjecture. That is, while obviously there are embeddings $K \hookrightarrow \mathbb C \hookrightarrow M_2(L)$, I claim that for any algebraically closed field $F$ and embeddings $i:K\hookrightarrow F, j: L\hookrightarrow F$, we have that $[i(K) : i(K) \cap j(L)] =\infty$.

First off, $i(K).j(L)$ (the compositum inside $F$) is an algebraic extension of $j(L) \simeq \mathbb R$, hence is isomorphic to $\mathbb C$, with $j(L)$ being a subfield of degree $2$. So w.l.o.g. we can replace $F$ by this field, and further assume that $F=\mathbb C$, $L=\mathbb R$ with its standard embedding, fixed by complex conjugation.

What is left to show is that $[i(K) : i(K) \cap \mathbb R] =\infty$ for any embedding $i:K\hookrightarrow \mathbb C$. In fact, I make the seemingly stronger claim that $i(K_n) \cap \mathbb R = \mathbb Q$ for all $n$. The rest of this answer will justify this claim, with some arguments more intricate than expected. If somebody sees a shorter line of proof, I'd be happy to know it.

---

Abstractly, $K_n$ could be written as $\mathbb Q(\sqrt[2^n]{-p})$ (the above choices of "principal" roots just make sure we have a tower of fields with consistent embeddings), or $K_n \simeq \mathbb Q[X]/(X^{2^n}+p)$. For $n\ge 2$ this is not Galois over $\mathbb Q$, i.e. the above choice of $\omega_n$ is not the only embedding of such $K_n$ into $\mathbb C$. However, the splitting field of the polynomial $X^{2^n}+p$ has a unique embedding into the complex numbers, it is the Galois closure

$$E_n := K_n(\zeta_{2^n}) = \mathbb Q(\zeta_{2^n},\omega_n), $$

where $\zeta_{2^n}$ denotes any (hereupon fixed) primitive $2^n$-th root of unity. Using that $X^{2^n}+p$ is still irreducible over the cyclotomic field $\mathbb Q(\zeta_{2^n})$ (not trivial but true: $p\neq 2$ does not ramify in $\mathbb Q(\zeta_{2^n})$), Galois theory shows

$$G_n := \mathrm{Gal}(E_n|\mathbb Q) \simeq (\mathbb Z/2^n)^* \ltimes (\mathbb Z/2^n, +)$$

(with the natural action in the semidirect product): Each $g\in G_n$ is fully described by $g(\zeta_{2^n}) = \zeta_{2^n}^x$ and $g(\omega_n) = \zeta_{2^n}^y \cdot \omega_n$, where $(x,y) \in (\mathbb Z/2^n)^* \ltimes (\mathbb Z/2^n, +)$. In particular, complex conjugation restricts to the element

$$\sigma := (-1, -1)\in G_n$$

and the original inclusion $K_n \subset \mathbb C$ (via $\omega_n$) identifies $K_n$ with the fixed field $E^{H_n}$ of the (non-normal) subgroup

$$H_n := \{ (x,0): x\in (\mathbb Z/2^n)^* \}.$$

But $\sigma$ and $H_n$ together generate the full group $G_n$. So by Galois theory, for the original inclusion $K_n \subset \mathbb C$, we have $K_n \cap \mathbb R$ = the fixed field of the full group = $\mathbb Q$.

Now finally, what about other embeddings $i: K_n \hookrightarrow \mathbb C$? Well, by Galois theory, such $i$ restricts to an element $g_i \in G_n$; then, $i(K_n)$ is the fixed field of the conjugate subgroup $g_i H_n (g_i)^{-1}$; and $i(K_n) \cap \mathbb R$ is the fixed field of

$$S_{i} = \text{ the subgroup of } G_n \text{ generated by } \sigma \text{ and } g_i H_n g_i^{-1}.$$

To see that this is still the full group $G_n$ without tediously writing out all conjugates of $H_n$, one can e.g. note that

$$g_i^{-1}S_i g_i = \text{ the subgroup of } G_n \text{ generated by } g_i^{-1}\sigma g_i \text{ and } H_n, $$

and for any $g= \in G_n$, the conjugate $g^{-1}\sigma g$ is of the form $(-1, k)$ with $k$ a generator of the normal subgroup $\{1\} \times (\mathbb Z/2^n, +)$. Multiplying with $(-1,0)\in H_n$ from the left shows that our generated group contains that "second factor" subgroup, and also the "first factor" $H_n$, so it is all of $G_n$. A fortiori, so is $S_i$, and $E^{S_i} = \mathbb Q$ q.e.d.

Answer 3

Looking at your first sentence where you talk only of a single field $L$ here is my attempt at an answer. As $M_n(L)$ is a finite dimensional (non-commutative) algebra over $L$ any field (or commutative subalgebra) extension of $L$ contained in $M_n(L)$ has to be necessarily of dimension at most $n^2$. Assuming $n$ is not fixed, one answer to your question is: every field $K$ which is a finite extension of $L$ can be realised as an $L$-subalgebra of $M_n(L)$ for some $n$.

The $n$ can be chosen as the degree of the extension $K$ over $L$. For any $x$ in $K$ consider the map $T_x\colon K\to K$ defined as $ T_x(y) = xy$ (here $xy$ means multiplication of two elements of the field $K$).

Clearly $T_x$ is a linear transformation of $K$ regarded as an $L$-vector space.

This $x\mapsto T_x$ is a map $L\to \mathrm{End}_L(V)$ is easily verified to be an $L$-algebra homomorphism. Of course $\mathrm{End}_L(V)$ is, after a choice of basis for $K$ over $L$, isomporphic to $M_n(L)$.

Answer 4

Here's a problem: Let $\overline{\mathbb C(X)}$ denote any algebraic closure of $\mathbb C(X)$. Then one of the notorious consequences of the axiom of choice is that there exist field isomorphisms $s: \overline{\mathbb C(X)} \simeq \mathbb C $. Cf. 1, 2, 3.

Now take your example, just the other way around: $K= \mathbb R(X), L= \mathbb C$. If we use the standard embeddings of both into $\overline{\mathbb C(X)}$, we have $K\cap L = \mathbb R$ and hence $[K: K\cap L]$ is infinite.

However, composing that standard embedding $\mathbb R(X) \hookrightarrow \overline{\mathbb C(X)}$ with an aforementioned isomorphism $s:\overline{\mathbb C(X)} \simeq \mathbb C$ produces an embedding $K \hookrightarrow L = M_1(L)$ and of course this also gives embeddings into any $M_n(L)$.

So this is a counterexample to the question as you asked it: "Is the criterion $[K: K\cap L] < \infty$ necessary for such embeddings $K \hookrightarrow M_n(L)$ to exist?"

Arguably, such "strange" isomorphisms should be excluded via making the question more precise. The trick here is that my embedding does not preserve the original embedding of $K\cap L$ into $L$. So maybe one should demand that, and ask: Are there such embeddings which (after first embeddings into the common field extension of $K$ and $L$) fix the subfield $K\cap L$? I would not know the answer to that question, but it would invalidate this counterexample.