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