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Show that $K = \mathbb{Q}(\sqrt{p} \ | \ p\text{ is prime} )$ is a algebraic and infinite extension on Q.

For every prime $p$, consider the polynomial $f(x)=x^2−p$. Then $f(x) \in \mathbb{Q}(\sqrt{p}\,∣\,p \text{ is prime})[x]$. Therefore, the extension is algebraic.

Moreover, taking $B=\mathbb{Q}(\sqrt{p} \ | \ p\text{ is prime} )$ is a infinite basis for this space.

Is that correct?

Sally G
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Junior Soares
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  • See here: http://en.wikipedia.org/wiki/Normal_extension. It is a normal extension of $\mathbb{Q}$ since, as you noted, it is the splitting field of that family of polynomials. The fact that it is infinite is essentially because the number of prime numbers $p$ is infinite. If there were a finite basis, you could probably deduce that there are a finite number of primes. – Moya May 17 '15 at 00:37
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    That is not a basis. Linear combinations of elements of $B$ don't contain elements like $\sqrt{pq}$ where $p \neq q$ are distinct primes, for example. As a warmup you might try to compute the degree of the extension $\mathbb{Q}(\sqrt{2}, \sqrt{3})$. – Qiaochu Yuan May 17 '15 at 00:38
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    But yet, @QiaochuYuan, his set is $\Bbb Q$-linearly independent, which is enough to show that the field is infinite over the rationals. On the other hand, the linear independence would need to be shown. – Lubin May 17 '15 at 01:30
  • The fact that the extension has infinite degree follows from this Q&A. Alternatively you can use a bit of algebraic number theory: the prime $p$ ramifies in the extension $\Bbb{Q}(\sqrt p)/\Bbb{Q}$, but in a finite extension only finitely many primes can ramify (namely those that appear as factors of the discriminant). – Jyrki Lahtonen May 17 '15 at 12:11
  • Thanks for the warning that B above is not basis. – Junior Soares May 18 '15 at 11:38

1 Answers1

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Your argument that it's an algebraic extension is fine. To show that it is an extension of infinite degree, certainly it would suffice to find an infinite basis. As pointed out in the comments, the set $\{\sqrt{p} \ | \ \text{p is prime} \}$ is not a valid basis, but its elements are linearly independent (which does suffice). To avoid bases, here is another way of going about this:

Suppose that $K = \mathbb{Q}[\sqrt{p} \ | \ \text{p is prime} \}]$ were an extension of finite degree. Then a simple implication of Galois correspondence tells us that $K$ has only finitely many subfields of degree $2$ over $\mathbb{Q}$. Notice that $\mathbb{Q}[\sqrt{p}] \subset K$ for all primes $p$, and $\mathbb{Q}[\sqrt{p}]$ is of degree $2$ over $\mathbb{Q}$. Since there are an infinite number of primes but only a finite number of subfields of degree $2$, then we must be able to find a pair of distinct primes $p_1$ and $p_2$ such that $\mathbb{Q}[\sqrt{p_1}] \cong \mathbb{Q}[\sqrt{p_2}]$.

Let $\phi$ be an isomorphism between these fields. Note that $\phi$ necessarily fixes $\mathbb{Q}$, and we have $\phi(\sqrt{p_1}) = a + b\sqrt{p_2}$ for some $a, b \in \mathbb{Q}$. Further, $a \neq 0$ since otherwise $\phi(\sqrt{p_1})^2 = \phi(p_1) = b^2p_2$, which is not possible since $\phi$ fixes $\mathbb{Q}$, and certainly $p_1 \neq b^2p_2$. Similarly $b \neq 0$ since otherwise $\phi(\sqrt{p_1})^2 = \phi(p_1) = a^2$, which is again not possible for the same reason.

Hence, we must have $\phi(\sqrt{p_1}) = a + b\sqrt{p_2}$ for some $a, b \neq 0$. From this, we get $\phi(p_1) = a^2 + p_2b^2 + 2ab\sqrt{p_2}$, and since $\phi$ fixes $\mathbb{Q}$, we have $p_1 = a^2 + p_2b^2 + 2ab\sqrt{p_2}$. With some rearranging, we see that this implies that $\sqrt{p_2}$ is rational--a contradiction!

Kaj Hansen
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  • The argument in your last paragraph needs slightly more elaboration - as written, it doesn't use the hypothesis that $p_1 \neq p_2$ (and of course it is false without this hypothesis). You need to know that $a, b \neq 0$. – Qiaochu Yuan May 17 '15 at 06:01
  • @QiaochuYuan, I added a decent bit more about my thought process. I didn't fill in every little detail, but the argument should now be more complete. – Kaj Hansen May 17 '15 at 06:26
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    This is more highbrow, but you could observe that $Q(\sqrt{p_1})$ and $Q(\sqrt{p_2})$ have different ramification. – Spooky May 17 '15 at 06:50
  • @KajHansen Why does the proof of algebraic extension is fine? We need to prove for the arbitrary number in K not just sqrt(p). – Cauchy Dec 20 '17 at 14:15