let F be a field, let ${h(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}}$
belongs to F[x], and let $g(x)=x-1$.
show that $g(x)$ divides $h(x)$ iff ${a_{n}+a_{n-1}+...+a_{1}+a_{0}=0}$.
Can someone help me proving this, please!!
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1Did you mean to say $g(x)$ divides $h(x)$ iff $a_n+a_{n-1}+\ldots+a_0=0$? – morrowmh Apr 04 '21 at 18:57
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I edited it, thank you!! – Hoho Jojo Apr 04 '21 at 19:06
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Summarized from Dummit and Foote page 307 (Proposition 9):
Let $F$ be a field and let $p(x)\in F[x]$. Then $(x-\alpha)$ divides $p(x)$ if and only if $p(\alpha)=0$.
Proof: If $(x-\alpha)$ divides $p(x)$ then we can write $p(x)=q(x)(x-\alpha)$ for some $q(x)\in F[x]$. Then $p(\alpha)=q(\alpha)(\alpha-\alpha)=0$.
For the other direction, suppose $p(\alpha)=0$. Then by the Division Algorithm we have $p(x)=q(x)(x-\alpha)+r$ where $r$ is a constant polynomial. Then since $p(\alpha)=0$, it follows that $r=0$ (show this), and thus $(x-\alpha)$ divides $p(x)$.
Remark: to obtain your result, just take $\alpha=1$.
morrowmh
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