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Math Help - Inseparable extension of rational function field

  1. #1
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    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.
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    Re: Inseparable extension of rational function field

    Quote Originally Posted by spudwish View Post
    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.
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    Re: Inseparable extension of rational function field

    Thank you! Getting really abstract here...
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