# How to prove $\mathbb{Q}[\alpha,\beta]=\mathbb{Q}[\gamma]$? [duplicate]

Possible Duplicate:
Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?

On Page 379, Algebra, Artin(1991)

Let $\mathbb{Q}[\alpha,\beta]$ denote the smallest subring of $\mathbb{C}$ containing $\mathbb{Q}$, $\alpha$=$\sqrt{2}$, $\beta$=$\sqrt{3}$, and let $\gamma=\alpha+\beta$. Probe that $\mathbb{Q}[\alpha,\beta]=\mathbb{Q}[\gamma]$.

To be precise, how to show $\alpha, \beta \in \mathbb{Q}[\gamma]$?It seems to me two subrings have different dimensions, how can they equal?

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## marked as duplicate by Rudy the Reindeer, BenjaLim, Zhen Lin, Matthew Pressland, lhfDec 13 '12 at 11:58

@MattN Positive; it can be added. –  user26857 Dec 13 '12 at 11:28

$\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is clear.

Now note that $$(\sqrt{2} + \sqrt{3})^{-1} = \frac{1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{2 - 3} = \sqrt{3} - \sqrt{2}$$ hence $\sqrt{3} - \sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\sqrt{2} + \sqrt{3} + \sqrt{3} - \sqrt{2} = 2 \sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Note that by a similar argument you get $\sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3})$.

As pointed out by YACP, if $a$ is algebraic over $K$ then $K[a]=K(a)$, see e.g. here.

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Thank you for your answer, but, I can't follow, why a subring must a field? –  Metta World Peace Dec 13 '12 at 11:07
@MattN Maybe it's worthy to say in your answer that Q[...]=Q(...) for algebraic elements in order to be complete. –  user26857 Dec 13 '12 at 11:22

You can check that $(\sqrt 2 + \sqrt 3)^3 = 11\sqrt 2 + 9\sqrt 3$, which is conveniently not a multiple of $\sqrt 2 + \sqrt 3$. So you can obtain $\sqrt 2$ by a suitable linear combination of $(\sqrt 2 + \sqrt 3)^3$ and $\sqrt 2 + \sqrt 3$ with rational coefficients.

In general, if $K$ is a field and $x$ is algebraic over $K$, $K[x] = K(x)$ : $K[x]$ is a finite dimensional $K$-vector space, and if you pick $y \in K[x]$, multiplication by $y$ is a $K$-linear endomorphism of $K[x]$. It is invertible if and only if it is injective, which is equivalent to $y$ being nonzero.

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More directly, without reference to theory of algebraic extensions: $(\sqrt 2 + \sqrt 3)^3 = 11 \sqrt 2 + 9 \sqrt 3 \in \mathbb Q[\gamma]$. And $(\sqrt 2 + \sqrt 3)^3 - 11(\sqrt 2 + \sqrt 3) = -2 \sqrt 3$, so $\sqrt 3 \in \mathbb Q[\gamma]$. Similary $\sqrt 2 \in \mathbb Q[\gamma]$.

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