# Degree of splitting field extension of a polynomial over $K=\mathbb{F}(x)$

Let x be transcendent over $\mathbb{F}$, where $\mathbb{F}$ is a field with characteristic 2. I have a polynomial $f=t^3+xt+x\in K[t]:=\mathbb{F(x)}[t]$. Let L be the splitting field of $f$. I know that $f$ is irreducible by Eisenstein, so I know that the degree $[L:K]\in\{3,6\}$. In a larger proof I am working on I need that $[L:K]=6$ but I have no idea how to prove it. Thanks in advance.

4kiraL

• Have you checked if the usual approach via the discriminant still applies in characteristic $2$? (if you have and it doesn't, you should probably have commented on that in the question) – Hurkyl Jun 13 '17 at 0:51
• See this paper by Keith Conrad -> math.uconn.edu/~kconrad/blurbs/galoistheory/… He defines a resolvent quadratic valid in characteristic 2. He may even consider that cubic. – sharding4 Jun 13 '17 at 0:53
• See Example 2.5 pg 5 in previous reference. Galois Group is $S_3$ – sharding4 Jun 13 '17 at 0:58
• Yes I am doing something with discriminants. Sorry that I didn't mention it. I am trying to find a field and an irreducible polynomial $f=t^3+at+b$ for which $[L:K]=6$ and the discriminant $disc(f)=-4a^3-27b^2$ is a Square of an Element $\delta\in K$. That is why I choose $f$ and $K$ as above. Then Disc(f)=x^2. (I already proved that if $char K$ dies not equal 2, we have $[L:K]=3$ IFF $Disc(f)=\delta^2$. So now I am looking for a counterexample if $K=2$). – 4kiraL Jun 13 '17 at 1:40
• I don't understand the paper by Konrad since I haven't learned anything about Galois theory yet. – 4kiraL Jun 13 '17 at 1:42