Prove $\langle x-1\rangle = I$ where $I= \{ a_0 x ^0 +\dots +a_n x^n: a_0+\dots +a_n=0\}$ in $F[x]$, $F$ a field.
Proof
we need to show $\langle x-1\rangle \subseteq I$ and $I \subseteq \langle x-1\rangle$
($\langle x-1\rangle \subseteq I$)
Let $f(x) \in \langle x-1\rangle$ so $\exists g(x) \in F[x]:f(x)=(x-1)g(x)$. Let $n-1= \deg g(x)$. So, $$f(x)=a_0x^0+\dots+a_nx^n=(x-1)g(x) $$ Let $x=1$, so $$f(1)=a_0(1)^0+\cdots +a_n (1)^n=a_0+\cdots +a_n=(1-1)g(x)=0.$$ So, $f(x) \in I$.
($I \subseteq \langle x-1\rangle$)
Now, Let $f(x) \in I \subset F[x]$ $$ f(x)=a_0x^0+\cdots +a_n x^n \text{ s.t. } a_0 +\cdots + a_n=0$$ We need
Remainder theorem Let F be a field $f(x) \in F[x] $ and $a\in F$. The remainder when $f(x)$ is divided by the polynomial $x-a$ is f(a)
Note that $f(x)\in I \subset F[x]$, so $f(x) \in F[x]$ so we have also $1_F \in F[x]$ so dividing $f(x)$ by $(x-1)$ yields remainder $$f(1)=a_0+\cdots+a_n=0 $$ So, $x-1\mid f(x)\Leftrightarrow \exists g(x) \in F[x] \text{ s.t. } g(x)(x-1)=f(x)$ so $f(x) \in \langle x-1\rangle$
Making sure there are redundancy and I did not do something dumb. Did I make a mistake somewhere. Also, interested in other ways arriving to the same conclusion if possible and more elegant
Your proof looks good to me.
Another proof goes as follows: Let $f:\Bbb{F}[x]\to \Bbb{F}:p \mapsto p(1)$. Then $\ker(f)=I$ by definition. Now, we also know that a polynomial has $1$ as a root if and only if it is divisible by $x-1$. So $\ker(f)=<x-1>$. Thus $I=<x-1>$.