Is kernel of evaluation map on $R[x]$ always principal if $R$ is an integral domain?

Question

Suppose $R$ is an integral domain and $E \supseteq R$ is a ring extension of $R$. Let $\alpha \in E$ commute with all of $R$, and consider the evaluation homomorphism $\phi_\alpha : R[x] \to E$ given by $\phi_\alpha(p(x)) = p(\alpha)$. Is the kernel of this map necessarily principal?

I was thinking if $E$ is also an integral domain, then you can consider the fields of fractions $F$ of $R$ and $G$ of $E$, and then you could take the monic generator $f(x)$ of the kernel of the analogous evaluation map $\psi_\alpha : F[x] \to G$ given by $\psi_\alpha(p(x)) = p(\alpha)$, then multiply by some constant $c \in R$ to find $cf(x) \in R[x]$ generates $\ker(\phi_\alpha)$. However, then you'd also have to specify that $c$ is one of all such "integralizing" constants that is minimal by divisibility. Like, for example, if $R = \mathbb{Z}$, $E = \mathbb{Q}$, and $\alpha = \frac{1}{2}$, then $f(x) = x - \frac{1}{2}$, so you'd have to use $c = \pm 2$ and not, say, $4$ to get a generator of the kernel of $\ker(\phi_\alpha)$.

This leads me to the question, how do I know that such constant that is minimal by divisibility exists? Do I need to suppose additionally that $R$ is Noetherian? And even if such a $c$ exists, does $cf(x)$ necessarily give a principal generator of the kernel, or is there some example where it still doesn't work? Then there's still the case where $E$ is not an integral domain...

Any help is appreciated, thanks.

Edit: I have realized that to use the $c$ argument, you need $c$ to be an lcm of all the denominators of $f(x)$ (when it is written in expanded form with irreducible fraction coefficients), so I think it already fails if $R$ is not a gcd domain. For that argument, you can suppose that $R$ is a gcd domain, but I am still also interested more generally in a proof/counterexample for the case that $R$ is just an integral domain.

Answer 1

Here's a counterexample . . .

Let $R=K[t^2,t^3]$, where $K$ is a field and $t$ is an indeterminate.

Now let $E=K[t]$.

Then the kernel of the evaluation homomorphism $\phi_t:R[x]\to E$ is the ideal $$ (x^2-t^2,x^3-t^3) $$ which is not a principal ideal of $R[x]$.

The above is essentially an outline.

Here's a more complete explanation . . .

The ring $R=K[t^2,t^3]$ is the subring of $K[t]$ containing $K$ and generated by $t^2$ and $t^3$.

Explicitly we have $$ R=c_0+\sum_{i=2}^\infty c_it^i $$ where $c_0,c_2,c_3,c_4,...\in K$, with at most finitely many of them not equal to zero.

Note that $t^2$ and $t^3$ are irreducible in $R$, hence from the factorizations $x^6=(x^2)^3$ and $x^6=(x^3)^2$, it follows that $R$ is not a UFD.

Next consider the ideal $I$ of $R[x]$ given by $I=(x^2-t^2,x^3-t^3)$.

For each integer $n\ge 2$ let $f_n=x^{n-2}(x^2-t^2)$.

Then $f_n\in I$ and $f_n$ is monic of degree $n$ in $x$.

Next we show $\ker(\phi_t)=I$.

The inclusion $I\subseteq\ker(\phi_t)$ is immediate.

For the reverse inclusion, suppose instead that the set $G=\ker(\phi_t){\setminus}I$ is nonempty.

Choose $g\in G$ of least degree, $n$ say, in $x$, and secondarily, whose leading coefficient $a\in R$ has least degree, $m$ say, in $t$.

If $n\ge 2$, then $g-af_n\in G$, contrary to the minimality of $n$.

But we can't have $n=0$ since $\phi_t$ fixes $R$.

Hence we must have $n=1$, so $g=ax-b$ with $a,b\in R$, $b=at$, and where $a$ has degree $m$ in $t$.

Note that no element of $R$ has degree $1$ in $t$.

Hence we can't have $m=0$, else $b$ would have degree $1$ in $t$, and we can't have $m=1$, else $a$ would have degree $1$ in $t$.

Thus $m\ge 2$.

From the identities $$ (x^3-t^3)-x(x^2-t^2)=t^2x-t^3\\ (x^2+t^2)(x^2-t^2)-x(x^3-t^3)=t^3x-t^4 $$ we get $t^2x-t^3,t^3x-t^4\in I$.

Hence for all nonnegative integers $u$ we have $t^{2u}(t^2x-t^3),t^{2u}(t^3x-t^4)\in I$.

It follows that there exists $f=a'x+b'\in I$ with $a',b'\in R$, $b'=a't$, and where $a'$ has degree $m$ in $t$.

Hence for some $c\in K$, the degree of $a-ca'$ in $t$ is less than $m$.

But then $(a-ca')x-(b-cb')=g-cf\in G$, contrary to the minimality of $m$.

Thus $\ker(\phi_t)=I$.

Next suppose $I$ is a principal ideal, say $I=(h)$.

Since $I$ has elements which are monic in $x$ (e.g., $x^2-t^2$), we can assume $h$ is monic in $x$, and since $I$ has elements which have degree $1$ in $x$ (e.g., $t^2x-t^3$), it follows that the degree of $h$ in $x$ is at most $1$.

But since $h$ is monic, if the degree of $h$ in $x$ is $0$, then $h=1$, contrary to $1\not\in\ker(\phi_t)$, and if the degree of $h$ in $x$ is $1$, then $h=x-t$, contrary to $t\not\in R$.

Hence $I$ is not a principal ideal.

Answer 2

If $R$ is a unique factorization domain, we can take $G$ to just be any ring containing both $E$ and $F$, or at least $\alpha$ and $F$ (I am just going to assume such a ring can necessarily be constructed even if $E$ is not an integral domain). Then, since $f(x)$ generates $K_1 = \ker(\psi_\alpha)$, and $K_2 = \ker(\phi_\alpha) \subseteq K_1$, we find that every $h(x) \in K_2$ is of the form $h(x) = f(x)g(x)$ for some $g(x) \in F[x]$. Then, by Gauss's lemma, we can find some $k_1, k_2 \in F$ and associates $\hat{f}(x) = k_1f(x)$ and $\hat{g}(x) = k_2g(x)$ in $F[x]$ that are also in $R[x]$, such that $h(x) = \hat{f}(x)\hat{g}(x)$. Then, if you use $c$ as an lcm as stated in the question, then we will necessarily have $cf(x) \mid k_1f(x)$ in $R[x]$, because to have $k_1f(x) \in R[x]$, we must have $k_1$ be a common multiple of all the denominators of $f(x)$ (in expanded, reduced form), so $c \mid k_1$. Note in fact $k_1 \in R$ because we took $f(x)$ to be monic (otherwise leading coefficient of $k_1f(x)$ would not be in $R$). Thus, $cf(x)$ should indeed be generator of $K_2$ at least in the case that $R$ is unique factorization domain.

Someone please correct me if I did something wrong here.