the kernel of the evaluation map Assume that $R$ is a commutative ring with a multiplicative identity element. Fix $a\in{R}$, and consider the evaluation map $e_{a} : R[x]\rightarrow{R}$ defined to be ${e_{a}}(f(x))=f(a)$. If $(x-a)$ is the principal ideal generated by $x-a$ in $R[x]$, it is clear that $(x-a)\subset \ker{e_{a}}$. Is the reverse inclusion always true? One can see immediately that it holds by the factor theorem if $R$ is a field.
This is Exercise 47 in J.J. Rotman's book Galois Theory (Second Edition), but I am not sure if $\ker{e_{a}}=(x-a)$ is true in an arbitrary ring. Thanks!
 A: By the remainder theorem, if $a \in R$ and $f \in R[x]$ with $f \in \ker(e_a)$ (i.e. $f(a) = 0$), then $f(x) = q(x)(x - a) + f(a) = q(x)(x-a)$ for some $q \in R[x]$.  But then $(x-a) \mid f(x)$ which implies that $f \in \langle x-a \rangle$.  Hence $\ker(e_a) \subseteq \langle x-a \rangle$.
Note that the remainder theorem relies on $R$ having unity and that the leading coefficient of $x-a$ is a unit.
A: Certainly $ker e_a$ is an ideal in $R[x]$ that contains $x-a$. The fact that this ideal equals $(x-a)$ in $R[x]$ if $R$ is a field depends on the division algorithm which then gives the factor theorem.
However, the division algorithm works when the leading coefficient of the divisor is a unit. (Just imagine doing polynomial division of $b(x)$ by $a(x)$, where you have to renormalize the leading coefficient of $a(x)$, so you can cancel out the term that you want to "get rid of".) So, it then follows by the normal proof that $(x-a) = \ker e_a$.
Related:
Division algorithm for polynomials in R[x], where R is a commutative ring with unity.
In general, when does a ring have a division algorithm?
