# Does the Frobenius endomorphism apply to polynomials?

I know that if $$F$$ is a field of characteristic $$p$$, then for all $$a,b \in F$$, $$(a+b)^p=a^p+b^p$$.

But what if now we have a polynomial say $$x+a \in F[x]$$. Is $$(x+a)^p=x^p+a^p$$?

• Yes, all the binomial coefficients $\binom{p}{k}$, for $k=1,2,...,p-1$, are multiples of $p$ in that case.
– user752802
Mar 1 '20 at 19:39
• If $R$ is a commutative unital ring where $p=0$ then $$(a+b)^p= a^p+b^p$$ You can replace unital by $\forall a\in R,pa=0$ it works too. When $R$ is non-commutative it fails, eg. in $M_2(\Bbb{F}_p)$. Mar 1 '20 at 19:46

Yes. Note that $$(x+a)^p=\sum_{r=0}^p\binom{p}{r}x^ra^{p-r}=x^p+a^p$$ because $$p$$ divides $$\binom{p}{r}$$ except when $$r=0,p$$.