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I understand that this result has been proven before, but I am trying to prove it in another way that I can't seem to find on SE.

The question goes as follows:

Let $E$ be a quadratic extension of a field $F$. Let $S(E)$ be the set of all $a\in F$ such that $a\neq 0$ and $a$ is a square in $E$. Show that $E\cong E' \Leftrightarrow S(E)=S(E')$. Use this result to show that for any odd prime $p$, there is exactly one field of order $p^2$ up to isomorphism.

I have proven the forward implication of the first statement, but not the reverse implication. The proof for the forward implication uses the fact that isomorphism is surjective and preserves squares. Apart from that, I feel totally stuck, and I don't even know why this result helps me prove that there is only one field of order $p^2$. I haven't learned about Galois extensions yet, so I'm trying to avoid them in my proof.

Ultimately, my questions are:

  1. How do I show that $S(E)=S(E')\Rightarrow E\cong E'$?
  2. How does this result show that there is only one field of order $p^2$ for odd primes $p$?
  3. Why does the oddness of $p$ matter?
IAAW
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    Can you use the following? If $L/K$ is a quadratic extension, then $L=K(\alpha)$ for any $\alpha\in L\setminus K$. Furthermore, the minimal polynomial of such an element $\alpha$ is a quadratic. Then the upshot: The quadratic formula gives the roots of a quadratic in terms of a square root of an element of $K$. The caveat: the quadratic formula works only when the characteristic of $K$ is $\neq2$, do you see why? Basically, unless $p=2$, $L=K(\sqrt{a})$ for some element $a\in K$. – Jyrki Lahtonen Mar 03 '23 at 07:15
  • I don't see why the quadratic formula works only when the characteristic is not 2. I also don't see how $L=K(\sqrt{a})$ shows that $S(L) = S(L')$ implies $L\cong L'$. Sorry for the confusion. – IAAW Mar 03 '23 at 17:20
  • $S(E)$ is a subgroup of $\Bbb{F}p^$. Because it contains all the quadratic residues modulo $p$, there are actually only two possibilities for it. Either the quadratic residues or all of $\Bbb{F}_p^$ for the former is an index two subgroup of the latter. And the earlier reasoning rules out one of these possibilities. Sorry, it's not actually clear to me what the poser of this question had in mind. I have simply used this argument to show that every element of $\Bbb{F}_p$ has a square root in $\Bbb{F}{p^2}$ when $p>2$. – Jyrki Lahtonen Mar 04 '23 at 04:50
  • (cont'd) Looks like the author has a variation in mind. Likely they are also leading up to Kummer theory with this, or some such motive. Anyway, what goes wrong if you try to apply the quadratic formula in a field where $2=0$? – Jyrki Lahtonen Mar 04 '23 at 04:51
  • Show 1 using the contrapositive ($E \neq E'$ implies $S(E) \neq S(E')$. Suppose $\alpha\in E$ and $\alpha\notin E'$ with quadratic minimal polynomial $f(X) = (x-b)^2 + c$ where $b,c\in F$ (we complete the square). Then $0=f(\alpha)=(\alpha-b)^2+c$ implies $-c=(\alpha-b)^2$. Since $\alpha \notin E'$, neither is $\alpha-d.$ Thus $-c \in S(E)$ but not in $S(E')$ whence $E\neq E'.$ Item 2 is justified by https://math.stackexchange.com/q/1175471 The prime $p$ must be odd in the link or else $2(p-1)$ doesn't divide $p^2-1=(p-1)(p+1).$ – user460693 Mar 06 '23 at 13:24

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