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Prove that $\mathbb R\times\mathbb R\times\mathbb R\times\mathbb R$ is not isomorphic to $M_2(\mathbb R)$.

I know that to prove that two rings are not isomorphic, you need to show that they have different properties. I know that multiplication in $\mathbb R\times\mathbb R\times\mathbb R\times\mathbb R$ is commutative but multiplication in $M_2(\mathbb R)$ is not. But how do I show the multiplication in $\mathbb R\times\mathbb R\times\mathbb R\times\mathbb R$? I'm thinking that it would be $(a,b,c,d)[(e,f,g,h)(i,j,k,l)]$ and multiply it through?

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  • $\begingroup$ One usually multiplies two elements at a time not three. $\endgroup$ – R_D Nov 16 '15 at 3:49
  • $\begingroup$ Just for the uninitiated, what operations are you using? $\endgroup$ – user223391 Nov 16 '15 at 3:49
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    $\begingroup$ How do you show that the multiplication in $R\times R\times R\times R$ what? $\endgroup$ – Mariano Suárez-Álvarez Nov 16 '15 at 3:50
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    $\begingroup$ @Bennie They are indeed vector space isomorphic, all finite-dimensional vector spaces of the same dimension are $\endgroup$ – Anthony Peter Nov 16 '15 at 3:55
  • $\begingroup$ What is $M_2(\mathbb{R})$? $\endgroup$ – mvw Nov 16 '15 at 3:55
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The usual way to construct a direct product is with componentwise addition and multiplication. That is, $$(a,b,c,d)\cdot(a',b',c',d')=(aa',bb',cc',dd')$$

If the question is about vector space isomorphism, then (as pointed out in the comments) the two are isomorphic having the same dimension. If the question is about ring isomorphism, then indeed a noncommutative example of $2\times 2$ multiplication is what you need.

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  • $\begingroup$ Thank you. I didn't realize that the two rings would be isomorphic in vector space. I am studying ring isomorphism, so this helps a lot. Thank you so much. $\endgroup$ – Bennie Joseph Vassallo Nov 16 '15 at 4:01
  • $\begingroup$ Note: vector spaces have only vector addition and scalar multiplication, so the noncommutativity of multiplication doesn't arise. As vector spaces they are just rearranged versions of each other. $\endgroup$ – vadim123 Nov 16 '15 at 4:02
  • $\begingroup$ @BennieJosephVassallo Remember that the word "isomorphism" isn't enough information unless the context is clear. You have to specify the "category" in which the isomorphism occurs $\endgroup$ – Anthony Peter Nov 16 '15 at 4:04
  • $\begingroup$ @AnthonyPeter I wasn't aware there were different "categories" of isomorphisms. I have just begun studying them and was using the basis of the question I was given. $\endgroup$ – Bennie Joseph Vassallo Nov 16 '15 at 4:06
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    $\begingroup$ @BennieJosephVassallo Not to worry. A quick google search would reveal say, "vector space isomorphisms", "group isomorphisms", "Hilbert space isomorphisms", "topological space isomorphisms (homeomorphisms)" each means something slightly different, but there is a general preservation of structure $\endgroup$ – Anthony Peter Nov 16 '15 at 4:07

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