# 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!

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?

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.

• In the first link given above by 4-1er, a minimum requirement for using the remainder theorem seems to be that $R$ be an integral domain. This was not assumed in Rotman's exercise. However, as 4-ier pointed out the leading coefficient of the divisor must be a unit. – student Aug 26 '18 at 5:35
• If $R$ is a ring with identity and $f,g \in R[x] \setminus 0$ such that the leading coefficient of $g$ is a unit, then there exist $q,r \in R[x]$ such that $f = gq + r$. Since the leading coefficient of $x - a$ is a unit the remainder theorem holds in this case. – matt stokes Aug 26 '18 at 5:42
• Matt can you edit your answer so I can give you an up vote? Your argument now makes a lot of sense to me. The key fact is the coefficient of the divisor be a unit. This enables a version of the division algorithm to go through-even if $R$ is not an integral domain. – student Aug 26 '18 at 5:56