Does $(x-a)^n\in K[x]$ imply that $a\in K$?

Question

Let $F$ be an extension field of $K$. Let $a\in F$ and $n$ be a positive integer. It is also given that the polynomial $(x-a)^n$ has all of its coefficients in $K$, i.e. $(x-a)^n\in K[x]$. Does it follow that $a\in K$ ?

Thank you

Answer 1

Not in general.

Let $F$ be the field with $p$ elements, $p$ a prime.

Let $L = F(a)$ be a field of rational functions in the indeterminate $a$, and let $K = F(a^{p})$.

Then $(x - a)^{p} = x^{p} - a^{p} \in K[x]$, but it is easy to show that $a \notin K$. (Try and write $a = f(a^{p})/g(a^{p})$, with $f, g \in F[x]$, multiply out, and you'll see that the degrees do not match.)

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However, when the characteristic is zero, your question has a positive answer, as $$ (x - a)^{n} = x^{n} - n a x^{n-1} + \dots, $$ so that $n a \in K$, and thus $a \in K$.

In characteristic $p$, it still holds true if $K$ is a finite fields, as $a \mapsto a^{p}$ is an automorphism of $K$.

In fact, the (standard) example above is based on a(n infinite) field $L$ of positive characteristic $p$ with $L \ne L^{p}$.

Answer 2

It is true if $char(F)=0$

As $(x-a)^n=x^n-{n \choose 1}ax^{n-1}+....$

Then we must have ${n \choose 1}a \in K$ and since ${n \choose 1}\neq0 \implies a\in K$.