# Prove that $\mathbb{Q}(\sqrt{2}, i)$ is a splitting field over $\mathbb{Q}$.

Prove that $\mathbb{Q}(\sqrt{2}, i)$ is a splitting field over $\mathbb{Q}$.

I know basic Group, Ring, and Field theory, and I've read the definition of a splitting field, yet I still have no idea where to start on this one.

Is it asking me to show that any polynomial in $\mathbb{Q}$ will split over $\mathbb{Q}(\sqrt{2}, i)$?

• No, the splitting field for a polynomial $f(x)$ is the smallest field in which $f(x)$ splits. A splitting field $E$ is a field extension of a field $F$ such that there is some polynomial $f(x)$ for which $E$ is the splitting of $f$. – Thomas Andrews Jan 5 '17 at 16:17
• So the question is asking you to show that there is a polynomial $f(x)\in\mathbb Q[x]$ for which $\mathbb Q(\sqrt2,i)$ is the smallest field in which $f$ splits. – Thomas Andrews Jan 5 '17 at 16:18

It is the splitting field of $X^4-2$. Let $F$ be the splitting field of $X^4-2$. It contains $\sqrt2$ and $i\sqrt2$ which are roots of $X^4-2$. this implies that it contains $(\sqrt2)^3$ and $i\sqrt2(\sqrt2)^3=2i$, so it contains $Q(\sqrt2,i)$. Since $Q(\sqrt2,i)$ contains all the roots of $X^4-2$, it is its splitting field.
• Thanks! So in general, to solve a problem like this, it looks like you would need to know what the polynomial is. How did you know it was $x^4 - 2$? – setholopolus Jan 5 '17 at 16:22
• @setholopolus If an extension $L/K$ of fields is a normal extension generated as $L = K(\alpha_1, \alpha_2, \ldots, \alpha_k)$, then $L/K$ is the splitting field of the product $f_1 f_2 \ldots f_k$, where each $f_i$ is the minimal polynomial of $\alpha_i$ over $K$. In this case, we don't need the $X^2 + 1$ factor coming from $i$. – Ege Erdil Jan 5 '17 at 16:27