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Let $F$ be a Field with $\mathrm{char}(F) > 0$. $E_1$ and $E_2$ are finite extensions of $F$ of same degree. Prove that $E_1$ and $E_2$ may not be isomorphic. In Fields of characteristic zero, I get the obvious counter example namely $\mathbf{Q}(\sqrt2)$ and $\mathbf{Q}(\sqrt3)$. In finite (Perfect) Fields, the result is true. But I am unable to get a counter example for an infinite Field of prime characteristic. I need help from someone please!!

Ashwin Iyengar
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Lawrence Mano
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1 Answers1

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Edit: sorry misread the question! For instance take $\mathbf{F}_{p^2}((t))/\mathbf{F}_p((t))$ and $\mathbf{F}_p((t^{1/2}))/\mathbf{F}_p((t))$. These are both degree 2 extensions, but are not isomorphic: in particular the second one is isomorphic to $\mathbf{F}_p((t))$ itself, which is not isomorphic to $\mathbf{F}_{p^2}((t))$.

Let's show that these are degree 2 extensions. Note $\mathbf{F}_{p^2}/\mathbf{F}_p$ is a degree $2$ extension, given by adjoining a root of an irreducible polynomial $f(x) \in \mathbf{F}_p[x]$ of degree $2$ over $\mathbf{F}_p$ (for instance, if $p$ is odd, take $x^2 - r$, where $r \in \mathbf{F}_p$ is not a quadratic residue mod $p$). Then the extension $\mathbf{F}_{p^2}((t))/\mathbf{F}_p((t))$ is given by, again, adjoining a root of $f(x)$, this time viewed as a polynomial in $\mathbf{F}_p((t))[x]$: note it's still irreducible in this ring, and thus the extension is of degree $2$. On the other hand, you could adjoin a root of the polynomial $x^2 - t \in \mathbf{F}_p((t))[x]$ and then you get the extension $\mathbf{F}_p((t^{1/2}))$: this is again of degree $2$ since the polynomial has degree $2$.

More generally this illustrates the distinction between ramified and unramified extensions of equal characteristic local fields.

You could also do this starting with $\mathbf{F}_p(t)$, the field of rational functions in one variable over $\mathbf{F}_p$, instead of $\mathbf{F}_p((t))$ the field of Laurent series in one variable over $\mathbf{F}_p$, it wouldn't change anything above, and you could come up with more examples. I just chose Laurent series because the theory of finite field extensions of a local fields is a bit simpler in the end.

Ashwin Iyengar
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    I don’t understand. You have given one Field. What is the other non isomorphic Field? – Lawrence Mano Jan 08 '20 at 23:52
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    Surely rational functions $\mathbb{F}_p(t)$ would suffice, so no need for Laurent series $\mathbb{F}_p((t))$. | @LawrenceMano Well, this is a degree $2$ extension. Can you think of a different non-isomorphic degree $2$ extension? – anon Jan 08 '20 at 23:57
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    Yeah global fields would also work, but I just really like local fields. – Ashwin Iyengar Jan 09 '20 at 02:56
  • I don’t understand how to calculate the degree of the Field of rational functions over a Field F. If you could please give an example I will be thankful to you. – Lawrence Mano Jan 09 '20 at 14:24
  • Edited: added some explanation. – Ashwin Iyengar Jan 09 '20 at 17:36
  • Remember that the degree of a field extension is the dimension of the big field as a vector space over the little. The degree of the rational function field over a constant field $F$ is thus infinite. Here’s an infinite $F$-linearly independent set, if the variable is $t$: ${1,t,t^2,t^3,\cdots}$. – Lubin Jan 10 '20 at 03:55
  • Right good point, I guess i should have made it clear that my base field is $\mathbf{F}_p((t))$ and I'm taking extensions of this field, rather than the finite prime field $\mathbf{F}_p$. – Ashwin Iyengar Jan 10 '20 at 04:21