# Example of a field extension $K/F$ such that $K$ is the splitting field of some (non-separable) polynomial in $F$, but not Galois over $F$?

An extension $K/F$ is Galois if and only if $K$ is the splitting field of some separable polynomial over $F$. Is there an example of some non-Galois extension $K/F$ where a (necessarily) non-separable polynomial splits in $K$, but no separable polynomial splits in $K$?

• Any product of linear polynomials over $F$ will split in $K$, no matter what $K/F$ is. In particular, we will always be able to say $x^2-x$ is separable and splits in a field $K$. – anon Apr 23 '16 at 1:30
• In any case, to address the question in the title, if $K/F$ is the splitting field of an inseparable polynomial over $F$, then $K/F$ is automatically not Galois. – anon Apr 23 '16 at 1:36
• $\mathbb{Q}(i)$ is the splitting field of $(x^2+1)^2$ over $\mathbb{Q}$, a non-separable polynomial. However, clearly, $\mathbb{Q}(i)$ is also the splitting field of $(x^2+1)$, a separable polynomial. So, $\mathbb{Q}(i)$ is Galois. So, I don't see how your second claim holds. – cemulate Apr 23 '16 at 1:50
• Sorry, I was thinking irreducible on top of inseparable. In any case, you're familiar with the non-Galois inseparable extension $\Bbb F_q(t)/\Bbb F_q(t^p)$ right? – anon Apr 23 '16 at 1:51
• I see no requirement for separable polynomials in the definition of splitting field. If you're asking why we have that requirement to call the splitting field of an irreducible polynomial a Galois extension, it's because we want "enough" symmetries but there aren't "enough" symmetries if the polynomial is inseparable. Most advanced books or lecture notes on number theory, or field theory or even Galois theory should cover purely inseparable extensions. You can peruse Pete L. Clark's online notes, for example. – anon Apr 23 '16 at 1:59

## 1 Answer

Splitting fields can be taken with respect to any family of polynomials over a fixed field $K$. Fields $E$ that are splitting fields of families of polynomials over a field $K$ are precisely those fields such that $E/K$ is normal. Galois extensions are defined to be normal and separable. If the family is finite, say $f_1,\ldots,f_s$, then $E$ is the splitting field of $f=f_1\cdots f_s$, and $E/K$ (which is always normal) will be separable if and only if the irreducible factors of $f$ are separable.