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I need to prove the following:

Let $D$ be a square free integer. Show that $ \lbrace\begin{pmatrix} a & bD \\ b & a \end{pmatrix} \mid a,b\in\mathbb{Q}\rbrace $ is a field isomorphic to $\mathbb{Q}[X]/(X^2-D)$.

I am not even really sure where to begin with this proof. My first thought was to try and use something like the Universality Theorem and try to construct a map between $\mathbb{Q}[X]$ and $\mathbb{Q}^n$ since there is already a morphism between $\mathbb{Q}[X]$ and $\mathbb{Q}[X]/(X^2-D)$. I think that if I could do that all would have to do is show that the right functions in the relation are injective or surjective as needed so that the composition of the two functions would be a bijective morphism between the field of matrices and $\mathbb{Q}[X]/(X^2-D)$ meaning that they are isomorphic.

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  • $\begingroup$ What exactly do you mean by "Universality Theorem"? Your link goes to "universal property" of which there are many such universal properties. Try defining a surjective ring homomorphism from $\mathbb{Q}[X]$ into the proposed field that has kernel exactly $(X^2-D)$. $\endgroup$ – Hayden Jan 29 '15 at 23:51
  • $\begingroup$ FYI, this question has nothing to do with finite fields, so I removed the tag. If you are unsure about using a tag, please read the tag excerpt (or the tag wiki). If there are words in there that you do not understand, then it is a near certainty that the tag is not appropriate. In finite-fields the attribute finite refers to the set. But surely $\Bbb{Q}$ already is an infinite set, so the same applies to its extension, some of which may be finite extensions, when finite refers to the dimension of the extension field as a vector space over the rationals. $\endgroup$ – Jyrki Lahtonen Jan 30 '15 at 9:06
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In $\mathbb Q[X]/(X^2-D)$, every element is of the form $c+dX$ since $X^2$ reduces to $c+dX$ where $c=D$ and $d=0$, and then $X^3$ reduces to $X^2 X = (c+dX)X = cX+dX^2 = cX+dD$, etc.

So ask yourself: which element $c+dX$ should correspond to the matrix $\begin{pmatrix} a & bD \\ b & a \end{pmatrix}$? You have a pair $(a,b)$ as input and a pair $(c,d)$ as output. What should $(c,d)$ be as a function of $(a,b)$ in order to make the mapping an isomorphism?

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$\begin{pmatrix} a & bD \\ b & a \end{pmatrix} \mapsto a\cdot 1 + b \cdot \overline{X}$ is an isomorphism of rings from $\lbrace\begin{pmatrix} a & bD \\ b & a \end{pmatrix} \mid a,b\in\mathbb{Q}\rbrace$ to $\mathbb{Q}[X]/(X^2-D)$, where $\overline{X}$ is the image of $X$ in $\mathbb{Q}[X]/(X^2-D)$ by the canonical "quotient" map $\mathbb{Q}[X]\to\mathbb{Q}[X]/(X^2-D)$. As $\mathbb{Q}[X]/(X^2-D)$ is a field, the ring $\lbrace\begin{pmatrix} a & bD \\ b & a \end{pmatrix} \mid a,b\in\mathbb{Q}\rbrace$ is also a field.

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  • $\begingroup$ I am still pretty new to this algebra stuff. What does the Xbar mean in this context? $\endgroup$ – user1593858 Jan 29 '15 at 23:50
  • $\begingroup$ @user1593858 Robert Green likely means the image of $X$ under the quotient map $\mathbb{Q}[X]\rightarrow \mathbb{Q}[X]/(X^2-D)$. $\endgroup$ – Hayden Jan 29 '15 at 23:52
  • $\begingroup$ @user1593858 Edited my answer with a complement... $\endgroup$ – Olórin Jan 29 '15 at 23:54
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    $\begingroup$ @Hayden Or maybe the image of $-X$ ;-) $\endgroup$ – Olórin Jan 29 '15 at 23:54
  • $\begingroup$ @RobertGreen haha, indeed. $\endgroup$ – Hayden Jan 29 '15 at 23:55
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You can define a ring-homomorphism from $\mathbf Q[x]$ into the (commutative) ring generated by the matrix $$U=\begin{pmatrix}0&D\\1&0 \end{pmatrix}$$ sending $1$ to $I_2$ and $x$ to $U$. This homomorphism is surjective and its kernel is generated by $x^2-D$ since $U^2=D I_2$. Hence the ring is isomorphic with $\mathbf Q[x] /(x^2-D)$.

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One way to see this isomorphism is to consider how left multiplication by an element works in $\mathbb{Q}[X]/(X^2 - D)$; the fancy name for this is the left regular representation. I think this approach shows how one might arrive naturally at this isomorphism without knowing of its existence beforehand.

First note that $\mathbb{Q}[X]/(X^2 - D) \cong \mathbb{Q}(\sqrt{D})$ is a degree $2$ extension of $\mathbb{Q}$. In particular, then it is a $2$-dimensional vector space over $\mathbb{Q}$ with basis $\{1, \overline{X}\}$, where $\overline{X}$ is the image of $X$ under the quotient map $\mathbb{Q}[X] \to \mathbb{Q}[X]/(X^2 - D)$.

Given $f(\overline{X}) = a + b \overline{X} \in \mathbb{Q}[X]/(X^2 - D)$, let $L_f$ be left multiplication by $f$, i.e., \begin{align*} L_f : \mathbb{Q}[X]/(X^2 - D) &\to \mathbb{Q}[X]/(X^2 - D)\\ g(\overline{X}) &\mapsto f(\overline{X}) g(\overline{X}) \, . \end{align*} One can show that this map is linear, so, as with all linear maps, we can compute its matrix with respect to our basis. To do so, we apply $L_f$ to the basis vectors: \begin{align*} L_f(1) &= (a + b \overline{X}) \cdot 1 = a \cdot 1 + b \cdot \overline{X}\\ L_f(\overline{X}) &= (a + b \overline{X}) \overline{X} = a\overline{X} + b \overline{X}^2 = bD + a \overline{X} = bD \cdot 1 + a \cdot \overline{X} \end{align*} since $\overline{X}^2 = D$. Thus the matrix of $L_f$ with respect to the basis $\{1, \overline{X}\}$ is $[L_f] = \begin{pmatrix} a & bD\\ b & a \end{pmatrix} $ which looks very familiar! Letting $R$ be the ring of matrices from the OP, define $\Phi : \mathbb{Q}[X]/(X^2 - D) \to R$ by $\Phi(f) = [L_f]$. Can you show that $\Phi$ is an isomorphism?

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