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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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    @DavidWard I just don't understand why you treat homework questions differently from non-homework ones. If you think giving an answer to a homework question soon is pedagogically harmful, the same can be said for a non-homework question. IMO, some(or many) people don't want to give away an answer to a homework question just because they feel like being used, not because they are particularly interested in pedagogy. While I understand their sentiment, it may have negative effects in this site(e.g. only obscure answers are given).2012-09-19

1 Answers 1

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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 ?2015-08-09