Show $f(x)=x^2+1$ is irreducible over $\mathbb{Q} (i\sqrt{2})$

Question

I am wanting to show $x^2+1$ is irreducible over $\mathbb{Q} (i\sqrt{2})$, however, I'm a little stuck.

Here's my attempt: Suppose, for contradiction's sake, that $x^2+1$ is reducible, and factors as $(x-\alpha)(x-\beta)$. We can expand this as $x^2-(\alpha + \beta)x + \alpha\beta$, or in other words, obtain the system:

$$\alpha\beta = 1, \alpha + \beta = 0$$

Is this the right way to go? Is there a contradiction reached, and why?

Answer 1

Let $a+bi\sqrt2$ be a root of $x^2+1$, so $(a+bi\sqrt2)^2 = -1$, i.e. $(a^2-2b^2)+2abi\sqrt2 = -1$.

Equate coefficients and derive a contradiction.

Answer 2

Hint: $x^2+1$ has a unique factorization in an appropriate extension field, say $\Bbb Q(i)$. Then $x^2+1 = (x+i)(x-i)$.

You have to show that $i\not\in \Bbb Q(\sqrt 2 i)$.

Answer 3

You are on the right track assuming it's irreducible, but you have not reached a contradiction yet. This is because you need to use what elements in $\mathbb{Q}(i\sqrt{2})$ are. Notice that any element in $\mathbb{Q}(i\sqrt{2})$ can be written as $a + ib\sqrt{2} $ for $a,b \in \mathbb{Q}$ (why?).

So for an element in $a + ib\sqrt{2} \in \mathbb{Q}(i\sqrt{2})$ to be a solution to the equation, we must have

$-1= (a + ib\sqrt{2})^2= a^2-2b^2+iab\sqrt{2}$

From this equation, we first get that $ab = 0$. Secondly, if $b=0$, we have $-1=a^2$, which is impossible. So $a=0$. Therefore we get

$-1=-2b^2 \implies b^2=\frac{1}{2}$

which is impossible.