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I'm struggling a fair amount with this exercise:

Find a subring of $M(2,\mathbb{Q})$ which is isomorphic to a) $\mathbb{Q}$ x $ \mathbb{Q}$ b) $\mathbb{Q}$ c) $\mathbb{Q}[x]$/$x^2$

Now I know a subring must be a subgroup, must contain the elements $0,1$ and must be closed under multiplication. As we are looking for isomorphisms then they must also be ring homomorphisms and must be bijective.

I tried to come up with a random subring and attempt to prove it is an isomorphism. E.g $\left\{\left.\begin{matrix} a & b \\ c & 0 \end{matrix}\right|a,b,c \in \mathbb{Q}\right\}$ is a subgroup of $M(2,\mathbb{Q})$ and then try and show it's an isomorphism. However I'm stuck here as I don't actually know how to go about finding these specific isomorphisms and I completely lose what I'm doing. Any help would be great.

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    $\begingroup$ "got very confused with the form of answer the question needs" - what exactly is your confusion? Terminology, notation, etc? $\endgroup$ – lisyarus Feb 26 '19 at 12:54
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    $\begingroup$ Welcome to MSE. Please show your attempts and efforts to solve this question. $\endgroup$ – Alan Muniz Feb 26 '19 at 12:56
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    $\begingroup$ See math.stackexchange.com/a/3128210/589 $\endgroup$ – lhf Feb 27 '19 at 13:53
  • $\begingroup$ Why do you ask the same question again and again? $\endgroup$ – Dietrich Burde Feb 27 '19 at 15:17
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Yes, we're talking about a ring homomorphism.

If we're looking for something isomorphic to $\mathbb{Q}$, note that we can get the integers by adding up copies of $1$. Then adding $n$ copies of $\frac1n$ gets us $1$; we're going to need all the scalar multiples of whatever we're mapping to $1$.

So, the subring we want to map to $\mathbb{Q}$ will consist of the scalar multiples of some $A$, and that $A$ will be mapped to $1$. Now, it's time to look at multiplication. What can we say about multiplication by $A$?

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  • $\begingroup$ ah okay that makes a lot more sense thank you, would it be very similar then when checking if its isomorphic to another ring, like $\mathbb{Q}$ x $\mathbb{Q}$ $\endgroup$ – L G Feb 26 '19 at 13:01
  • $\begingroup$ Similar, yes. Any ring homomorphism will also be a $\mathbb{Q}$-linear map. $\endgroup$ – jmerry Feb 26 '19 at 13:09
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Hints: Start with part b), and take the scalar multiples of the identity matrix.
For a), consider the diagonal matrices.
Finally for c), map $1$ to the identity matrix and map $x$ to a nontrivial matrix $M$ that satisfies $M^2=0$.

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  • $\begingroup$ Thanks a lot for the hints, but if i wasn't given those, how would one normally come about thinking that up? $\endgroup$ – L G Feb 27 '19 at 13:49
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Hint: Write the three rings in the form $\{ a + b u : a,b \in \mathbb Q \}$ with $u$ satisfying a quadratic equation, $u^2=cu+d$. Then identity the ring with a matrix subring using the matrix representation: $$ a+bu \leftrightarrow \pmatrix{ a & bd \\ b & a+bc} $$ This representation comes from the map $z \mapsto (a+bu)z$ in the basis $1,u$.

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