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This question came up in the link below. Since it is not good etiquette to use someone else's question to ask a question, I created a new question. I am trying to build a linear isomorphism between the two structures $\mathbb{Q}(\sqrt\alpha$) and $ \mathbb{Q}(\sqrt\beta)$:

Would it be:

$T: = \begin{pmatrix} 1 & 0 \\ 0 & \sqrt\beta/\sqrt\alpha \end{pmatrix} $

The other question that arose from thinking about this problem was: Must the entries inside the matrix belong to the Field of scalars? I am thinking No.

Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?

Quester
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  • If the coefficients of the isomorphism were rational, then $ℚ(\sqrt α) = ℚ(\sqrt β)$. – k.stm Mar 25 '14 at 19:35
  • Would you be approaching this with elements in $\mathbb{Q}(\sqrt{\alpha})$ and treat them as vectors $(a,b\sqrt{\alpha})^T$, where $a,b\in\mathbb{Q}$? – Ellya Mar 25 '14 at 19:36
  • possible duplicate of Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$? – Dietrich Burde Mar 25 '14 at 19:39
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    OVer the same field, two vector spaces are isomorphic iff they have the same dimension, so $$\Bbb Q(\alpha)\cong\Bbb Q(\beta)\iff \dim_{\Bbb Q}\Bbb Q(\alpha)=\dim_{\Bbb Q}\Bbb Q(\beta)\iff \deg,Irr(\alpha,\Bbb Q)=\deg,Irr(\beta,\Bbb Q)$$ – DonAntonio Mar 25 '14 at 19:43
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    Any degree $d$ extension of $\mathbb{Q}$ is isomorphic to $\mathbb{Q}^d$ as a $\mathbb{Q}$-vector space, and all matrices of $\mathbb{Q}$-linear maps belong to a matrix ring over $\mathbb{Q}$, so your $T$ is not a morphism of $\mathbb{Q}$-vector spaces. All degree $d$ extensions of $\mathbb{Q}$ are isomorphic, and with respect to a suitable choice of basis, the matrix of this isomorphism is just $I_d$. – ah11950 Mar 25 '14 at 19:49
  • @Don, the question is about finding that isomorphic linear matrix – Quester Mar 25 '14 at 20:14
  • @ah11950 how do you define $\mathbb{Q}^d$ – Quester Mar 25 '14 at 20:16
  • The field of rational numbers... – ah11950 Mar 25 '14 at 20:20
  • @ah11950, that's funny...I forgot the $d$. – Quester Mar 25 '14 at 20:21
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    $\mathbb{Q}^d = \mathbb{Q} \times \ldots \times \mathbb{Q}$. If you have an extension $\mathbb{Q}(\xi)$ of degree $d$, then $\mathbb{Q}(\xi) = \mathbb{Q}\cdot 1 \oplus \mathbb{Q}\cdot \xi \oplus \mathbb{Q}\cdot \xi^2 \oplus \ldots \oplus \mathbb{Q}\cdot \xi^{d-1}$ – ah11950 Mar 25 '14 at 20:25

1 Answers1

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No, your map $T$ isn't a linear map over ${\mathbb Q}$: all its entries must be rational. (And if you give a linear map as a matrix, you should indicate what bases you're using. In this case $1, \sqrt{a}$ and $1, \sqrt{b}$ are obvious candidates, but it's better to be explicit about this.)

What you could do is pick a basis for each of the two vector spaces (e.g. $1, \sqrt{\alpha}$ and $1, \sqrt{\beta}$) and use that to build a map by sending a basis element on the left to the corresponding basis element on the right (i.e., $1 \mapsto 1$, $\sqrt{\alpha} \mapsto \sqrt{\beta}$; or in full $x + y \sqrt{a} \mapsto x + y \sqrt{b}$). The matrix of this map, expressed on these bases, is of course the identity matrix.

Magdiragdag
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  • so do I define $T = \begin{cases} id & \forall x \in \mathbb{Q}\ \sqrt\beta & if \ x = \sqrt\alpha \end{cases} $ – Quester Mar 25 '14 at 20:07
  • That doesn't make sense. You should say what you want $T(x)$ to be for all $x \in {\mathbb Q}(\sqrt{\alpha})$. – Magdiragdag Mar 25 '14 at 20:18
  • but the problem would be that $T(\sqrt\alpha) = \sqrt\alpha \not\in \mathbb{Q}(\sqrt\beta)$ – Quester Mar 25 '14 at 20:19
  • Yes, so you define $T(x+y\sqrt{\alpha}) = x+y\sqrt{\beta}$ – ah11950 Mar 25 '14 at 20:21
  • so how do I write it as a matrix – Quester Mar 25 '14 at 20:21
  • @Quester What $T$ are you now talking about? The one in the question you cannot directly apply to $\sqrt{a}$: you must first pick bases. The map I specify in the answer maps $\sqrt{a}$ to $\sqrt{b}$. (I don't want to call it $T$ to avoid getting mixed up with your map). – Magdiragdag Mar 25 '14 at 20:21
  • @M I was trying to write your transformation as a Matrix. – Quester Mar 25 '14 at 20:23
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    @Quester To write a linear map as a matrix, pick bases left and right first. If you pick $1, \sqrt{a}$ and $1, \sqrt{b}$, then the map $x + y \sqrt{a} \mapsto x + y \sqrt{b}$ is represented by the 2 by 2 identity matrix. – Magdiragdag Mar 25 '14 at 20:23
  • got it. Thank you Both M and ah. – Quester Mar 25 '14 at 20:36