# Math Help - Inseparable extension of rational function field

1. ## Inseparable extension of rational function field

I am to construct an extension E of k:=F(x), where F=Z/2, such that E/k is not separable. I saw in an article by Keith Conrad that t^2-x in k[t] is not separable, because if a is a root, then a^2=x so that t^2 -x = t^2 - a^2 = (t-a)^2 (remember we're in characteristic 2). Can I use this somehow to construct the desired extension? I mean, k[t]/k itself is not an extension, a field extension that is, which I can only assume is what is meant. I was thinking k(t) at first, but I really don't know. I would really appreciate help with this one, thanks.

2. ## Re: Inseparable extension of rational function field

Originally Posted by spudwish
I am to construct an extension E of k:=F(x), where F=Z/2, such that E/k is not separable. I saw in an article by Keith Conrad that t^2-x in k[t] is not separable, because if a is a root, then a^2=x so that t^2 -x = t^2 - a^2 = (t-a)^2 (remember we're in characteristic 2). Can I use this somehow to construct the desired extension? I mean, k[t]/k itself is not an extension, a field extension that is, which I can only assume is what is meant. I was thinking k(t) at first, but I really don't know. I would really appreciate help with this one, thanks.
The field extension you want is $k(x^{\frac{1}{2}}) \cong k[t]/(t^2-x)$ (I omitted the /k for notational convenience). First, you need to see that $x\in k$ is not a square. Then observe that the polynomial $f(t)=t^2-x \in k[t]$ has derivative $f'(t)\equiv 0$ (since you have characteristic 2). Therefore $f(t)$ is not separable. It is easy to see that $f(t)$ is the minimal polynomial for a root $x^{\frac{1}{2}}$ in the extension $k[t]/(t^2-x)$. To conclude, $f(t)$ is an inseparable minimal polynomial for $x^{\frac{1}{2}}\in k(x^{\frac{1}{2}}) \cong k[t]/(t^2-x)$, so the extension is inseparable.

3. ## Re: Inseparable extension of rational function field

Thank you! Getting really abstract here...