Finding the splitting field of a polynomial.

I am confused on how to find the splitting field of a polynomial. For example, consider the polynomial $$p(x)=x^2+2.$$ I know that the splitting field is the smallest field that contains the roots of the polynomial, here $\pm \sqrt{2} i$. I know that the splitting field here is $\mathbb{Q}(\sqrt{2}, i)$ but shouldn't it be $\mathbb{Q} (\sqrt{2} i)$ ?

• The splitting field is ${\mathbb Q}(\sqrt{2}i)$ and not ${\mathbb Q}(\sqrt{2},i)$. – Magdiragdag Jun 13 '18 at 19:49
• I have generally seen cases where we have roots like $\zeta^i a$ for $\zeta$ being the nth root of unity and i is an integer where the splitting field is something like $\mathhbb{Q}(\zeta, a)$. Is this wrong? – Dimtsol Jun 13 '18 at 20:07
• I often find it is less confusing to write that the splitting field is $\Bbb{Q}(\sqrt{-2})$ in these situations! – mathphys Jun 13 '18 at 20:44

I don't know how you "know that the splitting field is $\mathbf Q(\sqrt 2, i)$" but it isn't. You may be confusing what the splitting field is in the general case:

The splitting field of $x^n - a$ is $$\mathbf Q(\sqrt[n]{a}, \omega)$$

where $\omega = e^{2\pi i/n}$ is a primitive $n$-th root of unity. This is because the roots of $x^n - a$ are $\sqrt[n]{a}, \omega \sqrt[n]{a}, \dots ,\omega^{n-1} \sqrt[n]{a}$ and so if we have all the roots of $x^n - a$ then we also have

$$\omega = \frac{\omega \sqrt[n]{a}}{\sqrt[n]{a}}.$$

And conversely, if we have $\sqrt[n]{a}$ and $\omega$ then we have $\sqrt[n]{a}, \omega \sqrt[n]{a}, \dots ,\omega^{n-1} \sqrt[n]{a}$.

When $n = 2$ and $a = - 2$, then $\omega = e^{\pi i} = -1$ (not $\omega = i$). Therefore the splitting field of $x^2 + 2$ is just $\mathbf Q(\sqrt{-2})$ but you can also verify that this is correct without looking at the general case.

• I have seen that $\mathbb{Q}(\sqrt{2},i) is the splitting field in my homework solutions that where given to me, but as it turns out it was wrong. – Dimtsol Jun 13 '18 at 20:17 A splitting field of a polynomial$f(x) \in F[x]$is defined as follows: Suppose that the polynomial splits over the a field extension$K/F$(equivalently,$K$contains the roots of$f(x)$). Then the splitting field of$f(x)$is the minimal subfield of$K$containing$F$such that$f(x)$splits. If we have such a field extension$K$, and$\alpha_1,\cdots,\alpha_n$are the roots of$f(x)$, then a splitting field will be$F(\alpha_1,\cdots,\alpha_n)$. One can show that for any field and polynomial, there is a splitting field and all splitting fields are isomorphic (hence, we refer to "the" splitting field of a polynomial). So in your problem you have$x^2+2 \in \Bbb{Q}[x]$. Your polynomial splits over$\Bbb{C}$and has roots$\pm i\sqrt{2}$. So the splitting field will be$\Bbb{Q}(i\sqrt2,-i\sqrt2)=\Bbb{Q}(i\sqrt2)\$