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

How would one describe elements from $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ and $\mathbb{Q}[\sqrt{2},\sqrt{3}]$?

I was under the mistaken impression, that if $\mathbb{Q}[\sqrt{2}]$ elements are $a+b\sqrt{2} \mid a,b \in \mathbb{Q}$, then $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ elements would be be $a+b(\sqrt{2}+\sqrt{3}) \mid a,b \in \mathbb{Q} $ , but that is not correct.

Similarly, I am unsure, what exactly $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ means. Surely, it does not mean the obvious $a+b\sqrt{2}+c\sqrt{3}) \mid a,b,c \in \mathbb{Q} $?

From what little I understand, the correct answer is that both sets are described by $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \mid a,b,c,d \in \mathbb{Q}$ and this is basis, but how do we get there?

This was an extra credit assignment in Number Theory course as the lecturer did not cover the subject.

EDIT: We use Ireland and Rosen's A Classical Introduction to Modern Number Theory, where on p.69 it is stated that we denote by $\mathbb{Q}[\alpha]$ the ring of polynomials in $\alpha$ with rational coefficients and then immediately it is proven that when $\alpha$ is in a ring of algebric integers $\mathbb{Q}(\alpha)$ = $\mathbb{Q}[\alpha]$ That still leaves me unclear on what $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ means, because $\sqrt{2},\sqrt{3}$ can not be an algebraic number, it is a tuple, a vector but not a single number.

Thanks for the answers guys, but it looks like a very similar question was posted last year: Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$? so I suppose my question should be merged and I should make a more basic question about meaning of $\mathbb{Q}[\sqrt{2},\sqrt{3}]$

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marked as duplicate by Fabian, YACP, Ittay Weiss, Alexander Gruber, Davide Giraudo Jan 3 '13 at 10:00

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All of the objects you describe are rings, so they need to be closed under multiplication. –  Qiaochu Yuan Jan 3 '13 at 0:17
    
Thanks, I see where $sqrt{6}$ comes from then, just I am unsure on what the actual definition of Q[a+b] and Q[a,b] then is. –  Sint Jan 3 '13 at 0:20
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4 Answers

Let $\alpha_1,\dots,\alpha_n$ be complex numbers. Then $\mathbb{Q}(\alpha_1,\dots,\alpha_n)$ can be defined as the intersection of all subfields of the reals that contain $\mathbb{Q}$ and $\alpha_1,\dots,\alpha_n$.

In our case, it is clear that $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ is a subfield of $\mathbb{Q}(\sqrt{2},\sqrt{3})$.

For the other direction, we can just play around. We want to show that $\sqrt{2}$ and $\sqrt{3}$ are in $\mathbb{Q}(\sqrt{2}+\sqrt{3})$.

If a field $K$ contains the rationals and $\sqrt{2}+\sqrt{3}$, then $K$ contains $(\sqrt{2}+\sqrt{3})^2$. So $K$ contains $5+2\sqrt{6}$, and therefore $2\sqrt{6}$, and therefore $\sqrt{6}$.

But then $K$ contains $\sqrt{6}(\sqrt{2}+\sqrt{3})$, that is, $2\sqrt{3}+3\sqrt{2}$. But then $K$ contains $3\sqrt{2}+2\sqrt{3}-2(\sqrt{2}+\sqrt{3})$, which is $\sqrt{2}$. Now it is easy to see that $K$ also contains $\sqrt{3}$, and we are finished.

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The OP was asking about $\mathbb{Q}[\alpha_1,\dots,\alpha_n]$ and not $\mathbb{Q}(\alpha_1,\dots,\alpha_n)$. It turns out the two are the same in this case because the $\alpha_i$ are algebraic, as explained in AndreaMori's answer. –  PatrickR Jan 3 '13 at 4:01
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There is a more general question about this sort of thing answered on this site, but you need to know a little Galois theory. This case can be dealt with more directly. The field $\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ contains $\frac{1}{\sqrt{2}+\sqrt{3}},$ which is $\sqrt{3}-\sqrt{2}.$ Hence it contains $2\sqrt{3} = (\sqrt{2}+\sqrt{3}) + (\sqrt{3}-\sqrt{2})$ and $2\sqrt{2} = (\sqrt{2}+\sqrt{3}) - (\sqrt{3}-\sqrt{2}).$ Hence both fields are equal, as it is clear that $\mathbb{Q}[\sqrt{2}+\sqrt{3}] \subseteq \mathbb{Q}[\sqrt{2},\sqrt{3}].$

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Since you mentioned some confusion about definitions, I can provide a ring-theoretic definition of the objects:

If $R$ is a commutative ring then we can define its polynomial ring $R[x]$ which is also a commutative ring. We can define $\sqrt{2}$ as a number $x$ such that $x^2-2=0$, and therefore, we can think of $\mathbb{Q}[\sqrt{2}]$ as $\mathbb{Q}[x]/(x^2-2)$, since we will want any multiple of $\sqrt{2}^2-2$ in the polynomials of $\sqrt{2}$ to be equal to zero.

It is easy to show that $R[x,y]$, the two-variable polynomial ring of $R$, is $R[x][y]$, the polynomial ring of the polynomial ring of $R$. So we can see that $\mathbb{Q}[\sqrt{2},\sqrt{3}]$ in this language becomes $\mathbb{Q}[x][y]/(x^2-2, y^2-3)$ for the same reasons.

$\mathbb{Q}[\sqrt{2}+\sqrt{3}]$ is syntactically a bit trickier, because it's not as clear what the relation should be but high school algebra can rescue us here: $(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}$, so this is a number $x$ such that $(x^2-5)^2-24=0$. Thus $\mathbb{Q}[\sqrt{2}+\sqrt{3}]=\mathbb{Q}[x]/(x^4-10x^2+1)$.

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Awesome, this makes it much clearer now! –  Sint Jan 3 '13 at 0:45
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Although it can be described as "the smallest subfield of $\Bbb C$ containing $\alpha$", the concrete description of the field $\Bbb Q(\alpha)$ strongly depends on the nature of the complex number $\alpha$.

  • If $\alpha$ is transcendental, meaning that $\alpha$ is not a root of any polynomial $P(X)$ with coefficients in $\Bbb Q$, then $\Bbb Q(\alpha)$ is set of complex numbers of the form $$ \frac{a_0+a_1\alpha+\dots+a_m\alpha^m}{b_0+b_1\alpha+\dots+b_n\alpha^n} $$ where the $a$'s and the $b$'s are rational numbers and the denominator is not $0$.
  • if $\alpha$ is algebraic, i.e. a root of some polynomial $P(X)$ with coefficients in $\Bbb Q$, then $\Bbb Q(\alpha)$ is set of complex numbers of the form $$ c_0+c_1\alpha+\dots c_{n-1}\alpha^{n-1} $$ where th $c$'s are again in $\Bbb Q$ and $n$ this time is fixed: it is the smallest degree of a polynomial of which $\alpha$ is root.

In the case under consideration $\alpha=\sqrt2+\sqrt3$ is algebraic and a root of a polynomial of degree 4, but not less. Thus $\Bbb Q(\alpha)$ is the set of complex numbers of the form $$ q_0+q_1\alpha+q_2\alpha^2+q_3\alpha^3, $$ where the $q$'s are in $\Bbb Q$. It's now matter of some elementary manipulation to show that these are the same numbers as those of the form $$ s_0+s_1\sqrt2+s_2\sqrt3+s_3\sqrt6, $$ where the $s$'s are in $\Bbb Q$.

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