Find a ring that contain $\mathbb{Q}$ as a group and has solution for $x^2 \equiv 2$ but no solution to $x^2 \equiv 3$ I have the following question :
Find a ring that contain $\mathbb{Q}$ as a group and has solution for $x^2 \equiv 2$ but no solution to $x^2 \equiv 3$
Hint : Start from $\mathbb{Q}[x]$
I really don't know how to approach this, how to force that $x^2 \equiv 3$ has not solution???
This is what I did - I don't know if its right
Maybe if I choose $I=x^2-2$? I think its an ideal since no root since there is no such $x\in \mathbb {Q}$ such that $x^2-2=0$, 
Yet how that implies solution for $x^2 \equiv 2$? what about no solutions for $x^2 \equiv 3$?
Any help will be appreciated.
 A: Your attempt is indeed fruitful. If you define $R=\Bbb Q[x]/I$, where $I=(x^2-2)$, then $$(x+I)^2 = x^2+I = (x^2+I) - (x^2-2+I) =2+I$$ so you have solution to $x^2=2$. Note that $\Bbb Q$ is embedded in $R$ via $q\mapsto q + I$ and that this is "sort of cheating". We literally defined $R$ so it will have this root. 
To show that there are no solutions to $x^2=3$, first note that every element of $R$ can be uniquely written as $ax + b+I$, $a,b\in\Bbb Q$. Existence follows from Euclidean division: if you have $f(x)+I\in R$, then $f(x) = q(x)(x^2-2) + r(x)$, with $\deg r \leq 1$ and $f(x)+I = r(x)+I$. Uniqueness follows from degree argument:
$$ax + b + I = a'x + b' + I \iff (a-a')x+(b-b')\in I$$
but there are no polynomials of degree $0$ or $1$ in $I$, thus $(a-a')x+(b-b)$ is zero polynomial.
Now,
$$(ax+b+I)^2 = a^2x^2+2abx+b^2 + I = 2abx+b^2+2a^2 + I$$
and you can easily check that system
\begin{align}
2ab &= 0\\
2a^2+b^2 &= 3
\end{align}
has no rational solutions.
Alternatively, you can prove that $R\cong \Bbb Q[\sqrt 2] = \{a+b\sqrt 2\mid a,b\in\Bbb Q\}$, which contains $\sqrt 2$, but does not contain $\sqrt 3$.
A: You have the right idea. The ideal $(x^2 - 2) \subseteq \mathbb{Q}[x]$ is maximal and so $R = \frac{\mathbb{Q}[x]}{(x^2-2)}$ is a field. 
Note that $R \cong \{ a + b \sqrt{2} \ | \ a,b \in \mathbb{Q} \}$ and $\mathbb{Q} \subseteq R$.
Suppose that $\sqrt{3} \in R$. Then there exist $a,b \in \mathbb{Q}$ such that
$$\sqrt{3} = a + b\sqrt{2}.$$
We now seek a contradiction.

Case 1: $a = 0$
I'll leave this to you.

Case 2: $b = 0$
I'll leave this to you.

Case 3: $a,b \neq 0$
If you square both sides of that equation and rearrange, you get:
$$ 3 - a^2 - 2b^2 = 2ab \sqrt{2}.$$
This implies that $$\sqrt{2} = \frac{3-a^2-2b^2}{2ab} \in \mathbb{Q}_.$$
So we arrive at a contradiction. 

Thus $\sqrt{3} \notin R$. Obviously neither is $-\sqrt{3}$. So $x^2-3$ has no roots in $R$.
