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Define the set $S$ of matrices by

$$S = \left\{A = (a_{ij}) \in M_2(R) : a_{11} = a_{22}, a_{12}=-a_{21}\right\}$$

Assume that $M_2(R)$ has already been proven to be a ring. If you wanted to prove that $S$ was a ring, which axioms would follow immediately from the fact that $M_2(R)$ is a ring, and for which axioms would there be something else to check?

This is what I'm thinking. Since $S$ is a subset of $M_2(R)$, most of the axioms follow. These include the associative, identity, inverse, and commutative additive laws and the associative multiplicative law. (Also both of the mixed laws)

For the axioms which I believe need something else to be checked, are the additive and multiplicative closure laws because they both require $a+b$ and $ab$ to be in $S$, and this means the condition has to be satisfied (An extra step).

Am I correct in thinking this way?

Thanks in advance.

rschwieb
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Bill Cogn
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    I fixed the pseudo-math into proper LaTeX form. May you check if it's accurate? – Accelerator Apr 13 '23 at 23:44
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    Distributivity, associativity, commutativity follow automatically. You should check $S$ includes the zero matrix and identity matrix (in general, a subset may not), and is closed under various operations, with additive inverses in particular. BTW, $S$ here is the $2\times2$ matrix representation of the complex numbers. – anon Apr 13 '23 at 23:58
  • Why not simply apply the subring test? – Bill Dubuque Apr 14 '23 at 20:05

1 Answers1

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All axioms of the form "for all x,y,z, ... (something that involves only $+$ and $\times$)" are automatic.

The ones that state "there exists (something)" are not automatic, as you need to check that the thing that needs to exist, exists in the subring.

The ones that need to be checked are the existence of 0, the existence of additive inverses, the existence of 1, and of course, that $S$ is closed under $+$ and $\times$

David Lui
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  • One thing, don't the closure laws fall under your statement "for all x,y,z, ... (something that involves only + and ×)"? – Bill Cogn Apr 14 '23 at 07:23
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    @BillCogn No, because it also involves $S$: The closure axiom states $\forall x,y : x+y \in S$ and $\forall x,y : x*y \in S$, which has $S$ in the condition. – David Lui Apr 14 '23 at 07:53