# $F$ is finite field with $1+1=0$ and reducibility of polynomials.

$$F$$ 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.

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 $$f$$.

• 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 $f(x)\in F[x]$. – Batominovski Mar 25 '20 at 11:13
• @WETutorialSchool Absolutely right! I have corrected this error. – Servaes Mar 25 '20 at 11:20

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.