Questions about $\Bbb Q[\sqrt{p}]$ and $\Bbb Q(\sqrt{p})$

Question

I studied this part where they talk about $\Bbb{Q}(\sqrt{2})$ and $\Bbb{Q}[\sqrt{2}]$ and I really start to get confused.

Definitions: $$ \Bbb{Q}[\sqrt{2}] = \left\{ a + b \sqrt{2} \mid a,b \in \Bbb{Q} \right\} \\ \Bbb{Q}(\sqrt{2}) = \left\{ \tfrac{a + b \sqrt{2}}{c + d \sqrt{2}} {\Large \mid} a,b,c,d \in \Bbb{Q} \text{ and } (c, d) \ne (0, 0) \right\} $$

I have a few questions that I might not be able to prove, but I'll provide my intuition.

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(Q1) Is $\Bbb{Q}(\sqrt{2})$ ring or group isomorphic to $\Bbb{Q}(\sqrt{3})$?

(Q2) Is $\Bbb{Q}[\sqrt{2}]$ ring or group isomorphic to $\Bbb{Q}[\sqrt{3}]$?

(Q3) Find order of $[\Bbb{Q}(\sqrt{p})(\sqrt{q}): \Bbb{Q}]$.

(Q4) Is $\Bbb{Q}(\sqrt{p})(\sqrt{q}) = \Bbb{Q}(\sqrt{p} + \sqrt{q})$?

I think (Q1) is not ring isomorphic. When they say group isomorphism in a field, I don't know whether we consider the additive group or multiplicative as $\Bbb{Q}(\sqrt{2})$ is a field.

(Q2) Not ring, but group isomorphic (considering addition)

(Q3) 4?

(Q4) Yes.

Please enlighten me on this. I am confused.

Answer 1

Fact: $\mathbb Q(a)=\mathbb Q[a]$ if $a$ is algebraic over $\mathbb Q$.

The reason is consequnces of the lemma:

Lemma $\mathbb Q(a)$ is a vector space over $\mathbb Q$ which spanned by $1,a,a^2,..,a^n$ where $a$ is algebraic over $\mathbb Q$.

Now, Let $\phi:\mathbb Q(\sqrt2)\to \mathbb Q(\sqrt3)$ be a nontrivial ring homomorphism.

Notice that $\phi(1)=1$ so if $2=\phi(2)=\phi(\sqrt2 )^2 \implies \phi(\sqrt2)=+-\sqrt 2$ which is not in $\mathbb Q(\sqrt3)$

But they are isomorphic as a group as both of them has $2$ dimension over same field space.

Q3: $|\mathbb Q(\sqrt p ,\sqrt q):\mathbb Q|=|\mathbb Q(\sqrt p ,\sqrt q):\mathbb Q(\sqrt p)||\mathbb Q(\sqrt p): \mathbb Q|$ and we have $|\mathbb Q(\sqrt p): \mathbb Q|=2$ and $|\mathbb Q(\sqrt p ,\sqrt q):\mathbb Q(\sqrt p)|\leq 2$ but notice that it can not be $1$ which means that they are equal which is not the case if $p,q$ are different prime. Since both of them $2$, answer is $4$.

Q4: It is clear that $L=\mathbb Q(\sqrt p +\sqrt q) \subseteq \mathbb Q(\sqrt p ,\sqrt q)=K$ And let's assume that L is a proper field of $K$. Since $|K:\mathbb Q|=4$ then we must have $|L:\mathbb Q| =2$ but it means that $\sqrt p +\sqrt q$ is a root of second degree polynomial $x^2+bx+c$ from there you can easily reach contradiction.

Answer 2

First of all $\Bbb{Q}[\sqrt{p}] \cong \Bbb{Q}(\sqrt{p})$, as rings. The latter notation is meant to suggest that these rings are, in fact, fields. Every nonzero element is a unit. It all boils down to this identity: $$ (c + d \sqrt{p})(c - d \sqrt{p}) = c^2 - pd^2 \in \Bbb{Q}. $$ This allows any rational expression in $\Bbb{Q}(\sqrt{p})$ to be written in a form that shows its membership in $\Bbb{Q}[\sqrt{p}]$: $$ \frac{a + b\sqrt{p}}{c + d\sqrt{p}} = \frac{(a + b\sqrt{p})(c - d\sqrt{p})}{(c + d\sqrt{p})(c - d\sqrt{p})} = \left( \frac{ac - pbd}{c^2 - pd^2} \right) + \left( \frac{bc - ad}{c^2 - pd^2} \right) \sqrt{p} \in \Bbb{Q}[\sqrt{p}], $$ provided that the only solution to $c^2 - pd^2 = 0$ is the trivial one $(c, d) = (0, 0)$.

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Does this help?

Answer 3

1) No, $\mathbb{Q}(\sqrt{2}) \not\cong \mathbb{Q}(\sqrt{3})$. Any ring homomorphism between them will fix $\mathbb{Q}$. Suppose $\varphi:\mathbb{Q}(\sqrt{2})\mapsto\mathbb{Q}(\sqrt{3})$ is some ring homomorphism. We will require $\varphi(\sqrt{2})=a+b\sqrt{3}$ for some $a,b\in\mathbb{Q}$. Carrying on with this, we will find $2 = \varphi(2) = \varphi(\sqrt{2}\sqrt{2})= \ldots = a^2+3b^2+2ab\sqrt{3} \not\in\mathbb{Q}$ which is a contradiction.

3) and 4). There is a need to be careful here, unless $p$ and $q$ are definitely prime (which you haven't specified). The reason is that $\mathbb{Q}(\sqrt{a}) = \mathbb{Q}(\sqrt{b})$ if there is a $c\in\mathbb{Q}$ such that $a = c^2b$. Other than this case, I think you're right.