Why $\mathbb{Q} [x]/(x^{4}+1) \simeq \mathbb{Q}(i,\sqrt{2})$?

Question

I do not understand why $\mathbb{Q} [x]/(x^{4}+1) \simeq \mathbb{Q}(i,\sqrt{2})$.

I know that $x^{4}+1$ is irreducible $(f(x + 1) = (x + 1)^4 + 1$ is Eisenstein at $2$) and it has the roots: $\pm \frac{1\pm i}{\sqrt{2}}$. But I do not understand that isomorphism. Any theorem I am missing? A good explanation will help, I am kind of lost with this Galois theory

Thanks

Answer 1

Since you know that $x^4+1$ is irreducible over $\mathbb Q$ it follows that $K=\mathbb{Q} [x]/(x^{4}+1)$ is a field. Moreover, it contains at least a root of $x^4+1$, say $a=x\bmod(x^4+1)$, and $K=\mathbb Q(a)$. Another root is obviously $-a$ and this also belongs to $K$. But it is not hard to see that $\pm a^3$ are the other two roots and thus $K$ is a splitting field of $x^4+1$.

On the other hand, by computing the roots of $x^4+1$ we find these are $\pm\frac{1\pm i}{\sqrt{2}}$, so the splitting field of $x^4+1$ over $\mathbb Q$ is $\mathbb Q(\pm\frac{1\pm i}{\sqrt{2}}) $ which is easily seen equal to $\mathbb Q(i,\sqrt2)$.

Now use the uniqueness of the splitting field. (Maybe this is the theorem you missed!)

Answer 2

Let $u$ be a root of $x^4+1$ in $\mathbb{C}$, for instance $u=\frac{1+i}{\sqrt{2}}$. By the fundamental theorem on homomorphisms, there is a homomorphism of fields $\mathbb{Q}[x]/(x^4+1) \to \mathbb{C}$ mapping $\overline{x} \mapsto u$. As such, it has to be injective. The image is $\mathbb{Q}[u]$, the smallest subring of $\mathbb{C}$ containing $u$. But this equals $\mathbb{Q}(u)$ since $u$ is algebraic. Hence, $\mathbb{Q}[x]/(x^4+1) \cong \mathbb{Q}(u)$. Thus, we only have to show that $$\mathbb{Q}(u) = \mathbb{Q}(i,\sqrt{2}).$$ The inclusion $\subseteq$ is clear. For $\supseteq$, observe that $u^2=i$ and hence $\sqrt{2}=(u^2+1)/u$.