Find a ring that contain $\mathbb{Q}$ as a group and has solution for $x^2 \equiv 2$ but no solution to $x^2 \equiv 3$

Question

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.

Answer 1

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 $\sqrt 3$.

Answer 2

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.

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Case 1: $a = 0$

I'll leave this to you.

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Case 2: $b = 0$

I'll leave this to you.

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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.

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