# Problem of the Week #273 - Jul 03, 2018

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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Let $F$ be the field of fractions of a unique factorization domain $A$, and let $L$ be an algebraic extension field of $F$. Fix $c\in L$. Prove that the kernel of the evaluation map $\operatorname{ev}_c : A[x] \to L$ is principal.

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Since $L$ is an algebraic extension of $F$, $c$ has a minimal polynomial $p(x) \in F[x]$. Fix $f \in \operatorname{ker}(\operatorname{ev}_c)$. By the Euclidean algorithm, $f(x) = p(x) q(x) + r(x)$ for some polynomials $q(x)$ and $r(x)$ in $F[x]$ with $\operatorname{deg} r < \operatorname{deg} q$. Now $r(c) = f(c) - p(c)q(c) = 0 - 0q(c) = 0$, so by minimality of $p$, $r \equiv 0$. Thus $f(x) = p(x)q(x)$ in $F[x]$. Write $p(x) = u p’(x)$ where $u$ is nonzero in $F$ and $p’(x)\in A[x]$ is primitive. By Gauss’s lemma, $f$ is in the principal ideal (in $A[x]$) generated by $p’$. As $p’(c) = 0$, $\operatorname{ker}(\operatorname{ev}_c) = (p’)$.