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Let $p$ and $q$ be two distinct primes. Pick the correct statements from the following:

a. $Q(\sqrt p)$ is isomorphic to $Q(\sqrt q)$ as fields.

b. $Q(\sqrt p)$ is isomorphic to $Q(\sqrt{−q})$ as vector spaces over Q.

c. $[Q(\sqrt p,\sqrt q) : Q] = 4$.

d. $Q(\sqrt p,\sqrt q) = Q(\sqrt p + \sqrt q)$.

(c) & (d) are correct. (a) is not correct for fields but correct for vector spaces. (b) not sure. i think it is correct by the same arguement of (a).am i right?

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Is this a homework question, or just one for personal development, as this may effect the type of response that people give? – David Ward Sep 17 '12 at 8:16
@DavidWard I don't know why it should affect the type of response. If one should not give a full answer to a homework question, many good questions might have no full answer in this site. I think this is against the policy of this site. – Makoto Kato Sep 17 '12 at 18:13
@MakotoKato My reasoning is that if it is a homework question, then the answer itself is of importance to the person asking the question. However, if the question is not homework, then for a question as above, I would feel happy to give the answer, but then lead someone through the reasoning on how to get there. – David Ward Sep 18 '12 at 8:26
@DavidWard I'm sorry that I don't understand your reasoning. Could you explain why you don't want to give the answer to a homework question? – Makoto Kato Sep 18 '12 at 8:43
@MakotoKato My reasoning is as follows: Firstly suppose that the question asked is from an assessed piece of work. Then as the question is a multiple-choice question, the actual answer will attract some if not all of the marks (depending on the desire of the assessor). Thus knowledge of the answer is of critical importance. On the other hand, if the question asked is purely for personal development, whilst leading the member towards the solution, it can be beneficial to actually know where you are being led. It just gives some motivation for where the reasoning is heading. – David Ward Sep 18 '12 at 15:26

a. Suppose $\mathbb{Q}(\sqrt p)$ is isomorphic to $\mathbb{Q}(\sqrt q)$ as fields. Then $\mathbb{Q}(\sqrt q)$ has an element $\alpha$ such that $\alpha^2 = p$. Let $\alpha = a + b\sqrt q$, where $a, b \in \mathbb{Q}$. We denote the conjugate of $\alpha$ by $\alpha'$. Since $\alpha + \alpha' = 0$, $a = 0$. Since $\alpha^2 = p$, $p = b^2 q$. Hence $p = q$. This is a contradiction. Hence $a$. is not true.

b. Since the both fields have dimension 2 as vector spaces over $\mathbb{Q}$, $b.$ is true.

c. As we see in the above, $\sqrt p$ is not contained in $\mathbb{Q}(\sqrt q)$. Hence $[\mathbb{Q}(\sqrt p,\sqrt q) : \mathbb{Q}(\sqrt q)] = 2$. Hence $[\mathbb{Q}(\sqrt p,\sqrt q) : \mathbb{Q}] = 4$. Thus $c.$ is true.

d. Let $K = \mathbb{Q}(\sqrt p,\sqrt q)$. Clearly $K/\mathbb{Q}$ is Galois. Let $\sigma \in Gal(K/\mathbb{Q})$. Then $\sigma(\sqrt p) = \sqrt p$ or $-\sqrt p$, $\sigma(\sqrt q) = \sqrt q$ or $-\sqrt q$. Since $|Gal(K/\mathbb{Q})| = 4$ by $c.$, $Gal(K/\mathbb{Q}) = \{1, \sigma_1, \sigma_2. \sigma_3\}$, where

$$\sigma_1(\sqrt p) = \sqrt p,\ \ \ \sigma_1(\sqrt q) = -\sqrt q$$

$$\sigma_2(\sqrt p) = -\sqrt p,\ \sigma_2(\sqrt q) = \sqrt q$$

$$\sigma_3(\sqrt p) = -\sqrt p,\ \ \ \sigma_3(\sqrt q) = -\sqrt q$$

Let $\alpha = \sqrt p + \sqrt q$.

Clearly $\alpha, \sigma_1(\alpha) = \sqrt p - \sqrt q, \sigma_2(\alpha) = -\sqrt p + \sqrt q, \sigma_3(\alpha) = -\sqrt p - \sqrt q$ are distinct. Hence $K = \mathbb{Q}(\alpha)$. Thus $d.$ is true.

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please explain why α+α′=0? In the proof of a ? – GA316 Aug 9 '15 at 4:01

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