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Let $G = \big\{a + b\sqrt2 \mid a,b \in\mathbb{Q}\big\}$.

Let $H = \bigg\{\begin{bmatrix} a & 2b \\ b & a \end{bmatrix} \biggm |a,b \in\mathbb{Q}\bigg\} $

And denote $0_{2\times 2} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

I have proven that both $G$ and $H$ are abelian/commutative because after some computations we have $G1 + G2 = G2 + G1$ and $H1 + H2 = H2+ H1$.

Now I have to show that $G$ and $H$ are isomorphic. I know that an isomorphism from $G1$ to $G2$ is a bijective homomorphism. We call $G1$ and $G2$ isomorphic, and write $G1 \cong G2$ if an isomorphism from $G1$ to $G2$ exists.

I am struggling how to construct such a proof.

Thanks in advance.

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  • $\begingroup$ What about $\varphi(a +b \sqrt2) = \begin{bmatrix} a & 2b \\ b & a \end{bmatrix}$? $\endgroup$ Sep 15, 2020 at 13:10
  • $\begingroup$ I was thinking of that, but not sure how to mathematically write a good proof $\endgroup$
    – Mathlover
    Sep 15, 2020 at 13:12
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    $\begingroup$ Try to prove that $\varphi$ satisfies isomorphism properties. $\endgroup$ Sep 15, 2020 at 13:13
  • $\begingroup$ I shall try it. Thanks. $\endgroup$
    – Mathlover
    Sep 15, 2020 at 13:14
  • $\begingroup$ For $G$ and $H$ to be isomorphic means that there exists an isomorphism $\varphi : G \to H$. To prove that there exists an isomorphism $\varphi : G \to H$ there are two steps. First, using your mathematical imagination and experience, write down an appropriate formula for $\varphi : G \to H$; this may require intuition, exploration, experimentaiton, digging around, etc. Second, once you think you have the appropriate formula for $\varphi : G \to H$, prove that it satisfies the definition of an isomorphism. $\endgroup$
    – Lee Mosher
    Sep 15, 2020 at 13:23

3 Answers 3

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Since$$\left(a+b\sqrt2\right)\left(c+d\sqrt2\right)=\color{red}{ac+2bd}+(\color{blue}{ad+bc})\sqrt2$$and since$$\begin{bmatrix}a&2b\\b&a\end{bmatrix}.\begin{bmatrix}c&2d\\d&c\end{bmatrix}=\begin{bmatrix}\color{red}{ac+2bd}&2(\color{blue}{ad+bc})\\\color{blue}{ad+bc}&\color{red}{ac+2bd}\end{bmatrix},$$simply take$$\psi\left(a+b\sqrt2\right)=\begin{bmatrix}a&2b\\b&a\end{bmatrix}.$$

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  • $\begingroup$ I put the bit about a general irreducible polynomial and its companion matrix. $\endgroup$
    – Will Jagy
    Sep 15, 2020 at 14:00
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If you begin with the rationals and some monic polynomial $f(x)$ of degree $n$ is irreducible, then we get two fields. One is $$ \mathbb Q [x] / (f(x)) $$

The other: take the companion matrix (or its transpose) $M.$ By Cayley-Hamilton, $f(M) = 0$ as matrices. We get a ring from all matrices of the form $$ a_0 I + a_1 M + a_2 M^2 + \cdots + a_{n-1} M^{n-1} $$ Plus, any polynomial expression (arbitrary degree) in $M$ is equal to such an expression. This ring of matrices is also a field. They are isomorphic as fields.

Your polynomial is $x^2 - 2$ and its companion matrix is $$ M = \left( \begin{array}{cc} 0 & 2 \\ 1 & 0 \\ \end{array} \right) $$

This is exactly the construction that gives the complex numbers (rational coefficients), polynomial $x^2 + 1,$

$$ M = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} \right) $$

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$\{1, \sqrt{2}\}$ is a basis for the space $\{a + \sqrt{2} b \mid a,b \in \mathbb Q\}$.

We can represent multiplication by $a + \sqrt{2} b$ as a matrix by noting how it acts on the basis vectors.

  • $(a + \sqrt{2} b) \cdot 1 = a + \sqrt{2} b$
  • $(a + \sqrt{2} b) \cdot \sqrt{2} = 2 b + \sqrt{2} a$

so $1$ gets mapped to $(a,b)$ and $\sqrt{2}$ gets mapped to $(2b,a)$.

So we can tabulate this into a matrix $$M = \left( \begin{array}{cc} a & 2b \\ b & a \\ \end{array} \right)$$

and indeed

  • $M(1,0) = (a,b)$
  • $M(0,1) = (2b,a)$

Secondly if $M$ represents $\alpha$ and $N$ represents $\beta$ it can be seen that the matrix product $MN$ represents $\alpha \beta$.

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