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Is it true that $\forall a,b \in \mathbb F_9$: $a \cdot b$ is a square $\iff$ $a \cdot \overline b$ is a square ?

2 Answers2

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Since your group of units is cyclic, you can argue very similarly to the case of $F_p$, and the "Legendre symbol" analogue here is still a homomorphism.

In particular, multiplying by $a$ doesn't matter. You can reduce to $$b \; \mathrm{square} \Leftrightarrow \overline{b} \; \mathrm{square}.$$ This is true, because for nonzero $b$ $$b \; \mathrm{square} \Leftrightarrow b^4 = 1 \Leftrightarrow (\overline{b})^4 = \overline{b^4} = 1 \Leftrightarrow \overline{b} \; \mathrm{square}.$$

Cocopuffs
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  • I think you could refine the last sentence by $b=t^2\iff \overline b=\overline t^2$. And, as a consequence of your proof, this holds for every finite field $F_q$, not only for $q=3$. Thanks for the answer. – awllower Jun 11 '13 at 17:22
  • @awllower This is a good point - that would have been more efficient. – Cocopuffs Jun 11 '13 at 17:26
  • I was about to apply that in an answer, but I dwelled upon the presence of $a$, for not realising the function of the Legendre symbol. What a surprise looking at your answer! :P – awllower Jun 11 '13 at 17:27
  • You need to justify "multiplying by $a$ doesn't matter", else the proof is not complete. Can you please provide the details. – Key Ideas Jun 11 '13 at 18:12
  • @KeyIdeas If we define $\psi(a) = 1$ if $a \in F_9^{\times}$ is a square and $\psi(a) = -1$ if it isn't, then $f$ is the same as the quotient map $F_9^{\times} \rightarrow F_9 / (F_9^{\times})^2 \cong \mathbb{Z}/2\mathbb{Z}$, which is a homomorphism. So $f(ab) = f(a)f(b) = f(a)f(\overline{b}) = f(a\overline{b})$. The special case where $a$ or $b$ is zero is easy to handle – Cocopuffs Jun 11 '13 at 18:31
  • Some typos in the above comment. Replace $f$ with $\psi$, or vice versa, and $F_9$ with $F_9^{\times}$. – Cocopuffs Jun 12 '13 at 20:11
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Adding a different approach to the one in Cocopuffs +1 answer.

The lone conjugate of any element $b$ in $F=\mathbb{F}_{p^2}$ is the Frobenius conjugate $\overline{b}=b^p$ (Caveat: The overline may mean something other than changing the sign of the coefficient of $\sqrt{-1}$. If $p\equiv3\pmod4$ we get the quadratic extension by adjoining $i$, but if $p\equiv1\pmod4$ another square root needs to be used instead).

But then we see that $$ \frac{\overline{b}}{b}=\frac{b^p}b=(b^{(p-1)/2})^2 $$ is always a square, and the claim follows trivially from this.

Jyrki Lahtonen
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