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Inspired by $\operatorname{Mat}_2(\mathbb{R})$ as a field, it made me curious if, up to isomorphism, we only get two fields as subrings of the ring of two by two matrices under usual matrix addition and subtraction, $\mathbb{R}$ from $<I>$ and $\mathbb{C}$ (via the usual way): Complex number isomorphic to certain $2\times 2$ matrices?.

My instinct is yes, since we need invertible matrices that stay invertible under linear combinations, I don't see a way of doing that without forcing the patterns like we do in the above constructions, but I'm a bit shy of a proof. (This is idle curiosity)

Edit: As was pointed out in the comments, one can easily have any subfield of $\mathbb{R}$ or $\mathbb{C}$ constructed this way. To get at the heart of what I meant, can we get any fields that aren't isomorphic to a subfield of $\mathbb{C}$ this way?

Alan
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    in fact, the answer is a strong no: any subfield $F\subseteq\mathbb{R}$ can be realized as the subring $F\cdot I_2$ of $M_{2,2}(\mathbb{R})$ – Atticus Stonestrom Sep 15 '21 at 02:33
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    Maybe you want subalgebras, instead of subrings? – Noah Schweber Sep 15 '21 at 02:36
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    Perhaps a better question is this: Does every embedding $h\colon F\to M_{2,2}(\mathbb{R})$, where $F$ is a field, factor through the usual embedding $\mathbb{C}\to M_{2,2}(\mathbb{R})$? – Alex Kruckman Sep 15 '21 at 02:36
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    It's amusing to me that the three people to comment in the first 5 minutes after the question was posted are all logicians. – Alex Kruckman Sep 15 '21 at 02:37
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    @AlexKruckman: That is again trivially no since you can just pick a different embedding from the usual one (differing by a change of basis). To get something actually nontrivial you would ask whether it factors through some conjugate of the usual embedding of $\mathbb{C}$. – Eric Wofsey Sep 15 '21 at 02:38
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    @EricWofsey Make that 4 logicians (or, if you protest, people knowledgeable about logic!). Good point - I would maintain that my version is a better question, even if it's still trivial :0) – Alex Kruckman Sep 15 '21 at 02:40
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    @AtticusStonestrom Forgot about subfields of $\mathbb{R}$ and $\mathbb{C}$. Adjusting question to eliminate those more trivial cases! – Alan Sep 15 '21 at 02:55

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Your intuition regarding the revised question is correct; any subring of $M_{2,2}(\mathbb{C})$ that is also a field is isomorphic to a subfield of $\mathbb{C}$. This is actually just a special case of a more general fact; if $F$ is a field of characteristic $0$, and $|F|\leqslant 2^{\aleph_0}$, then $F$ is isomorphic to a subfield of $\mathbb{C}$. To see this, let $X$ be a transcendence basis for $F$ over $\mathbb{Q}$; this makes sense since $\operatorname{char}F=0$. We have $|X|\leqslant 2^{\aleph_0}$, and so, choosing any transcendence basis $Y$ for $\mathbb{C}$ over $\mathbb{Q}$ and any injection $i:X\to Y$, we can extend $i$ to a field morphism $\iota$ from the subfield $\mathbb{Q}(X)\subseteq F$ to $\mathbb{C}$. By definition, $F$ is algebraic over $\mathbb{Q}(X)$, and so since $\mathbb{C}$ is algebraically closed we may extend $\iota$ to a field morphism from $F$ to $\mathbb{C}$, as desired.

The desired result now follows since $|M_{2,2}(\mathbb{C})|=2^{\aleph_0}$ and $\operatorname{char}M_{2,2}(\mathbb{C})=0$, so that every subring of $M_{2,2}(\mathbb{C})$ is of characteristic $0$ and of cardinality $\leqslant 2^{\aleph_0}$.

Atticus Stonestrom
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