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$p$ is a prime. Let $ f_1, f_2 \in Z / p Z [t]$ both of degree 2 and irreducible. Show that they have isomorphic splitting fields.


My approach was let $ K_1 = F(\alpha_1, \beta_1) / F$ be the splitting field of $f_1$, and let $K_2 = F( \alpha_2, \beta_2) / F$ be the splitting field of $ f_2$. Then I have trouble finding any relations between $ \alpha$'s and $\beta$'s. Any help is appreciated.

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  • $\begingroup$ Hint: It is possible to find a polynomial $f$ (of degree $p^2$) such that both fields you are considering are minimal splitting fields of $f$. $\endgroup$ – Sebastian Schoennenbeck Apr 17 '14 at 7:37
  • $\begingroup$ Try finding an automorphism of $\Bbb{Z}/p\Bbb{Z}[t]$ mapping $f_1$ to $f_2$: What do automorphisms of $\Bbb{Z}/p\Bbb{Z}[t]$ look like? P.S. Note that $F(\alpha_i,\beta_i)=F(\alpha_i)=F(\beta_i)$ for both $i\in\{1,2\}$. $\endgroup$ – Servaes Apr 17 '14 at 9:39
  • $\begingroup$ @Servaes I think I have something. So I only need to show that $ F(\alpha_1)$ is isomorphic to $F(\alpha_2)$. Since both fields are finite, do I always get an isomorphism by sending $ \alpha_1 $ to $\alpha_2$? $\endgroup$ – user112564 Apr 17 '14 at 16:27
  • $\begingroup$ You get an isomorphism if and only if $f_1=f_2$. But I prefer DonAntonio's approach to my own. $\endgroup$ – Servaes Apr 18 '14 at 9:12
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Some ideas (hopefully you've already studied this stuff's details):

For a prime $\;p\;$ and for any $\;n\in\Bbb N\;$, prove that the set of all the roots of the polynomial $\;f(x):=x^{p^n}-x\in\Bbb F_p[x]\;$ in (the, some) algebraic closure $\;\overline{\Bbb F_p}\;$ of $\;\Bbb F_p:=\Bbb Z/p\Bbb Z\;$ , with the usual operations modulo $\;p\;$ , is a field with $\;p^n\;$ elements, which we denote by $\;\Bbb F_{p^n}\;$

From the above it is immediate that $\;\Bbb F_{p^n}\;$ is the minimal field which contains all the roots of $\;f(x)\;$ and is thus this polynomial's splitting field over $\;\Bbb F_p\;$ .

Now, apply the above to the particular case $\;n=2\;$ and deduce at once your claim.

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  • $\begingroup$ Thanks. I guess to continue I just apply the argument to the $n =2$ case and find a field with $p^2$ elements which is the splitting field of $ x^{p^2} -x $. But why is this field a splitting field for both $ f_1$ and $f_2$? $\endgroup$ – user112564 Apr 17 '14 at 15:52
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    $\begingroup$ Because (1) either all the three fields (the splitting field I proposed and the two polynomials') are one and the same field (the only one with $\;p^2\;$ elements) , or (2) because all three fields are two-dimensional spaces over the same base field $\;\Bbb F_p\;$ ... $\endgroup$ – DonAntonio Apr 17 '14 at 16:03
  • $\begingroup$ Ok, so that's because finite fields with same number of elements are isomorphic? But if so why can't we directly say both the splitting fields of $f_1$ and $f_2$ have $p^2$ elements thus are isomorphic? $\endgroup$ – user112564 Apr 17 '14 at 16:13
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    $\begingroup$ Yes, and also yes: we can say that. My argument above is designed just to help you say easily why it is so, since if you already knew this then what's this question's point? $\endgroup$ – DonAntonio Apr 17 '14 at 16:56

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