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$F$ is a finite field. Show that $1+1=0$ in $F$ iff for every $f \in F[x]$ with $\deg(f)\geq 1$, $f(x^2)$ is reducible polynomial in $F[x]$.

I didn't know what test that I can use to show $f(x^2)$ is reducible.

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If $f(x^2)$ is reducible for every $f\in F[x]$, then consider the linear polynomials; it means that for every $a\in F$ the polynomial $x^2-a$ is reducible. In other words, every element of $F$ is a square. Again put differently; the group homomorphism $$F^{\times}\ \longrightarrow\ F^{\times}:\ a\ \longmapsto\ a^2,$$ is an isomorphism. Here $F^{\times}$ is a finite cyclic group because $F$ is a finite field, and this shows that $|F^{\times}|$ is coprime to $2$. That means $|F|$ is even, so the characteristic of $F$ is $2$, meaning that $1+1=0$ in $F$.

Conversely, if $1+1=0$ in $F$ then the above shows that every element of $F$ is a square, from which it follows that $$f(x^2)=(g(x))^2,$$ where the coefficients of $g$ are the square roots of the coefficients of $f$.

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    $\begingroup$ This is true only when $F=\{0,1\}$. For a field with $2^k$ elements where $k>1$, the equality $\big(f(x)\big)^2=f(x^2)$ doesn't hold for some $f(x)\in F[x]$. $\endgroup$ – Batominovski Mar 25 '20 at 11:13
  • $\begingroup$ @WETutorialSchool Absolutely right! I have corrected this error. $\endgroup$ – Servaes Mar 25 '20 at 11:20
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To show that $f(x^2)$ is reducible when $1+1 =0$ one first has to recall:

  • in such a field $(a+b)^2 = a^2 + b^2$, the "missing" $2ab$ is $0$.

  • in such a field every element admits a square-root, that is, for each $a \in F$ there is some $b\in F$ such that $b^2 =a$. This is for example because $x\mapsto x^2 $ is injectif in this case and thus surjectif (as the field is finite).

Thus for $f (x)=\sum_{i=1}^d a_i x^i $ one has that $f(x^2) =g(x)^2$ with $g (x)=\sum_{i=1}^d b_i x^i $ with $b_i$ as squareroot of $a_i$.

For the converse directions it suffices to consider linear polynomials. Considering the question for $f(x)= x-a$, the reducibility of $f(x^2)= x^2 -a$ amounts to deciding whether $a$ admits a squareroot. If $1+1 \neq 0$, then not every element of $F$ admits a squareroot. This is because $x \to x^2$ is not injectif and thus not surjectif. Thus there is a linear polynomial $f(x)$ for which $f(x^2)$ is irreducible.

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