The ideal $I=\{a_{0}+a_{1}x+\cdots+a_{k}x^{k} \in F[x]\mid a_{0}+a_{1}+\cdots+a_{k}=0\}$ equals $\langle x-1\rangle$?

Question

If $I=\{a_{0}+a_{1}x+\cdots+a_{k}x^{k} \in F[x]\mid a_{0}+a_{1}+\cdots+a_{k}=0\}$ is an ideal in $F[x]$, is it equivalent to $I=\langle x-1\rangle$?

Think it is because $1$ is obviously a root of the polynomial, and $I$ is equal to $\langle x-1\rangle$ since $F$ is a field and hence $F[x]$ is a PID.

Need your thoughts. Thanks!

Answer 1

Yes, exactly. Because $1$ is a root of every $f(x) \in I$, you get that every $f(x) \in I$ factors as $f(x) = (x-1) g(x)$ for some $g(x)$. Therefore $I \subseteq (x-1)$. Since evidently $x - 1 \in I$, you get $I = (x-1)$.