Let $K $be a field and $f \in K[x]$. Then there exists a splitting field for $f$ over $K$

Question

Let $K $ be a field and $f \in K[X]$. Then there exists a splitting field for $f$ over $K$.

I don't understand what this means, I think I am interpreting it wrongly.

Take $x^2+1 \in \Bbb{Q}[X]$ then the theorem seems to me to say that there is a splitting field for $x^2+1$ over $\Bbb{Q}$ but clearly it can't split there since the roots of $x^2+1$ are not real.

So it actually saying we can find a field extension $\Bbb{Q}\subseteq L$ such that $L$ is the splitting field? (I think $L$ here would actually be $\Bbb{Q}(i)?$)

Answer 1

let $\alpha$ be a root of $f(x)=x^2+1$ in Extended field of $\mathbb{Q}\,$, therefore $$f(\alpha)=f(-\alpha)=\alpha^2+1=0$$ on the other hand $$-\alpha=-(\alpha)$$ as a result $$\mathbb{Q}(\alpha)=\{a+b\alpha|\,a,b\in\mathbb{Q}\}\cong \frac{{{\mathbb{Q}}}[x]}{({{x}^{2}}+1)}=a+bx+\left\langle {{x}^{2}}+1 \right\rangle$$

Answer 2

the signification of $ L $ is the splitting field of $f$ over $K$ is: in an algebraic closure $\Omega$ of $K$, the field $L$ is the smallest field containing K (as minimum degree of $L$ over $K$) and included in $\Omega$, in which $f$ split completely (that is has all roots in $L$)