# Kernel of evaluation map into field of quotients

Let $R$ be an integral domain and for $a \in R$ denote by $\text{eval}_a: R[X] \to R$ evaluation at $a$. It's well-known (and easy to see) that $$\ker(\text{eval}_a)=(X-a).$$ The next more complicated thing in this setting is to evaluate at an element $q$ of the quotient field $K$ of $R$: $\text{eval}_q: R[X] \to K,\,f \mapsto f(q)$.

Question 1: Is there an explicit description of the generators of $\ker(\text{eval}_q)$ ?

In particular, I wonder if

$$\ker(\text{eval}_q) =(\,bX-a \mid q=\frac{a}{b};\,a,b \in R\,)\qquad ?$$

I could solve the following special cases:

1. If $q=a\in R$ then $\ker(\text{eval}_q)=(X-a)$

2. If $q=1/b$ then $\ker(\text{eval}_q)=(bX-1)$

3. If $R$ is a GCD and $q=\frac{a}{b}$ with $a,b$ coprime then $\ker(\text{eval}_q)=(bX-a)$

According to 3. I wonder, if the GCD assumption is really needed:

Question 2: If $a, b\in R$ are coprime, i.e. $(a,b)=R$, is $\ker(\text{eval}_q)=(bX-a)$ for $q=\frac{a}{b}$ ?

 For a proof of 3. note that $bX-a\in R[X]$ is irreducible and hence prime (since $R$ is GCD, $R[X]$ is also GCD and irreducible elements in a GCD are prime). If $f \in R[X]$ annulates $q$, write $f=(X-q)h$ for some $h \in K[X]$. By clearing denominators, there is $r \in R$ and $\tilde{h}\in R[X]$ such that $rf =(bX-a)\tilde{h} \in (bX-a)$. Since $(bX-a)$ is prime and $r \not\in (bX-a)$ we finally obtain $f \in (bX-a)$.

Remark: I have asked the question on math.SE but didn't get any reply: https://math.stackexchange.com/questions/2718227/kernel-of-evaluation-map-into-field-of-quotients

• Consider $\phi = \frac{1 + \sqrt{5}}{2} \in \text{Frac}(\mathbb{Z}[\sqrt{5}])$. Apr 4, 2018 at 2:09

First, the answer to Question 2 is YES and it is not hard to show. For $q=\frac{a}{b}$, by assumption $(a,b)=R$, therefore by standard argument $(a^k,b)=R$ for any positive integer $k$. For any $f\in\mathrm{ker}(\mathrm{eval}_q)$ where $q=\frac{a}{b}$ we have $f(X)=(X-\frac{a}{b})g(X)$ where $g(X)\in K[X]$. Write $g(X)=\sum_{i=0}^{n}a_i X^i$ where $a_i\in K$. Since $f(X)\in R[X]$, by induction we see that $a^{i+1}a_i\in (b)$ for all $i\in\{0,1,\cdots,n\}$. We also have $a_n\in R$ as the leading coefficient of $f(X)$, which implies that $a_n=a_n(a^{n+1}x+by)=a^{n+1}a_nx+b a_n y\in (b)$ for some $x,y\in R$ since $(a^{n+1},b)=R.$ Now we have $a_{n-1}\in R$ since the coefficient of $X^n$ in $f(X)$ is in $R$ and that $a_n\in (b)$. Together with $a^n a_{n-1}\in(b)$ and $(a^n,b)=R$ we conclude that $a_{n-1}\in (b)$. Repeating this process, we see that $a_i\in(b)$ for all $i\in\{0,1,\cdots,n\}$. Therefore $f(X)\in (bX-a)$.

Second, it is true (as shown by proof of 3.) that if $R$ is GCD, then under the assumption that $\mathrm{gcd}(a,b)=1$ (i.e. weak version of being coprime), we also have $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$. However one cannot simply argue which special case is $stronger''$: on the one hand, confirmation of Question 2 requires only property of the element $q$ while Point 3. requires GCD property of the whole ring $R$; on the other hand, confirmation of Question 2 requires that $a,b$ are "strongly" coprime, i.e. $(a,b)=R$ instead of "weakly" coprime, i.e. $\mathrm{gcd} (a,b)=1$. While for PID they are equivalent, the former one is much stronger than the latter in general.

Third, a very rough answer to Question 1 is NOT REALLY- and NO to the In particular'' part: For example, consider a number ring case where $R=\mathbb{Z}[\sqrt{5}]$ and $q=\frac{\sqrt{5}+1}{2}$. Then one can compute to see explicitly $$\mathrm{ker}(\mathrm{eval}_q)=\Big(2X-(\sqrt{5}+1),(\sqrt{5}-1)X-2, X^2-X-1\Big)$$ which is not principal any more. In this case, we know that $R$ is not integrally closed, let alone GCD.

Last but not least, on the one hand if $R$ is GCD then $\mathrm{ker}(\mathrm{eval}_q)=(bX-a)$ for any $a,b\in R$ such that $\mathrm{gcd}(a,b)=1$ and $\frac{a}{b}=q$; on the other hand if $R$ is not GCD then there is an element $q\in K=\mathrm{Frac}(R)$ which fails to have such simple property. To see this point, find two nonzero elements $a,b\in R$ which do not have a $\mathrm{gcd}$. Let $q=\frac{a}{b}$ and we claim that $\mathrm{ker}(\mathrm{eval}_q)$ is not a principal ideal. Since $bX-a$ is in the kernel, if it is principal then it must be generated by a degree one polynomial $\beta X-\alpha$ satisfying $r\alpha=a$ and $r\beta=b$ from some element $r\in R$. Since $r$ is not a $\mathrm{gcd}$ of $a$ and $b$, there is another common divisor $d$ of $a$ and $b$ so that $d$ does not divide $r$. Write $b=db_0$ and $a=da_0$ and one sees that $b_0 X-a_0$ is also in the kernel, hence generated by $\beta X-\alpha$, whence $d$ divides $r$, contradiction. QED.

The following claim characterizes circumstances under which $$\ker(\text{eval}_q)$$ for $$q \in K = \text{Frac}(R)$$ is a principal ideal of $$R[X]$$. In particular, this answers question $$2$$ in the positive and provides a recipe to construct non-principal ideals of the form $$\ker(\text{eval}_q)$$.

Claim 1. Let $$R$$ be an integral domain and let $$q = \frac{a}{b} \in \text{Frac}(R)$$ . Then the following are equivalent:
$$(i)$$ The ideal of denominators of $$q$$, that is $$\mathfrak{d} \Doteq \{d \in R \,\vert\, dq \in R\}$$, is the principal ideal of $$R$$ generated by $$b$$.
$$(ii)$$ For every $$c \in R$$, if $$b$$ divides $$ac$$ then $$b$$ divides $$c$$, in other words $$Ra \cap Rb = Rab$$.
$$(iii)$$ The ideal $$\ker(\text{eval}_q)$$ is the principal ideal of $$R[X]$$ generated by $$bX - a$$.

Edit: This characterization is well-known under a slightly different form, see Addendum at the bottom.

Note that if $$a, b \in R$$ are co-prime in the sense that $$Ra + Rb = R$$, then $$(ii)$$ is satisfied. If $$R$$ is a Prüfer domain, e.g., a Dedekind domain, the condition $$(ii)$$ is equivalent to $$Ra + Rb = R$$. The condition $$\text{gcd}(a, b) = 1$$ doesn't imply $$(ii)$$ in general, but If $$R$$ is a pre-Schreier domain, then $$(ii)$$ is equivalent to $$\text{gcd}(a, b) = 1$$. You can find in [1] an example of a Schreier domain which is not a GCD domain.

Proof of Claim 1. $$(i) \Rightarrow (ii)$$ is immediate. Let us show that $$(ii) \Rightarrow (iii)$$. To do so, let us consider $$f(X) = \sum_{i = 0}^n a_i X^i \in \ker(\text{eval}_q)$$ with $$n > 0$$. There is $$g(X) = \sum_{i = 0}^{n - 1} b_i X^i \in K[X]$$ such that $$f(X) = (bX - a)g(X)$$. We deduce from the latter identity that $$b_{n - 1 - i} = \frac{ab_{n - i} + a_{n - i}}{b}$$ for every $$0 \le i \le n - 1$$, agreeing that $$b_n = 0$$. The identity $$f(q) = 0$$ can be re-written as $$a_n a^n + a_{n- 1}a^{n - 1}b + \cdots + a_1 ab^{n - 1} + a_0b^n = 0.$$ Hence it follows from $$(ii)$$ that $$b$$ divides $$a_n$$, so that the above equality is equivalent to $$a^{n - 1}(ab_{n - 1} + a_{n - 1}) + a_{n - 2}a^{n - 2}b + \cdots + a_1 ab^{n - 2} + a_0b^{n - 1 } = 0$$ where $$b_{n - 1} = \frac{a_n}{b} \in R$$. Substituting $$ab_{n - 1} + a_{n - 1}$$ with $$bb_{n -2}$$, dividing the left-hand side by $$b$$ and using $$(ii)$$ again yields $$b_{n - 2} \in R$$. By repeating this process we eventually obtain that $$g(X) \in R[X]$$. Therefore $$f(X) \in R[X](bX - a)$$, which establishes $$(iii)$$. We will complete the proof of Claim $$1$$ by showing that $$\neg (i) \Rightarrow \neg (iii)$$. By hypothesis we can find $$d \in \mathfrak{d}$$ such that $$b$$ doesn't divide $$d$$. As result $$dX - dq \in \ker(\text{eval}_q) \setminus R[X](bX - a)$$.

The next claim underlines the special rôle of GCD domains with respect to OP's questions.

Claim 2. Let $$R$$ be an integral domain. Then the following are equivalent:

$$(i)$$ For every $$q \in \text{Frac}(R)$$, the ideal of denominators of $$q$$ is a principal ideal of $$R$$.
$$(ii)$$ $$R$$ is a GCD domain.

If $$q = \frac{a}{b}$$, then $$\mathfrak{d} = \frac{1}{a}(Ra \cap Rb)$$. Thus the ideal of denominators of $$q$$ is principal if and only $$Ra \cap Rb$$ is principal, that is $$\text{lcm}(a,b)$$ exists. In this case, we know that $$\text{gcd}(a, b)$$ exists and is such that $$\text{lcm}(a,b)\text{gcd}(a, b) = ab$$ up to multiplication by a unit.

Proof of Claim 2. Assertion $$(i)$$ is equivalent to the fact that $$\text{lcm}(a,b)$$ exists for every $$a, b \in R$$. The latter is equivalent to the fact that $$\text{gcd}(a,b)$$ exists for every $$a, b \in R$$.

The following consequence is immediate.

Corollary. Let $$R$$ be an integral domain. Then the following are equivalent:

$$(i)$$ For every $$q \in \text{Frac}(R)$$, the ideal $$\ker(\text{eval}_q)$$ is a principal ideal of $$R[X]$$.
$$(ii)$$ $$R$$ is a GCD domain.

In general, we observe that $$\{ bX - a \,\vert \, q = \frac{a}{b} \} = \{ dX - dq \,\vert \, d \in \mathfrak{d} \setminus \{0\}\}$$ is the set of polynomials in $$\ker(\text{eval}_q)$$ with degree $$1$$. The question as to whether the latter set generates $$\ker(\text{eval}_q)$$ was answered in the negative by a user44191's comment: this doesn't hold when $$R = \mathbb{Z}[\sqrt{5}]$$ and $$q = \frac{1 + \sqrt{5}}{2} \in \text{Frac}(R)$$. Clearly, the ideal $$\mathfrak{d}$$ of denominators of $$q$$ contains $$2$$ and $$1 - \sqrt{5}$$. It is easy to check that the ideal $$\mathfrak{m} = R \cdot 2 + R \cdot (1 - \sqrt{5})$$ is a non-principal maximal ideal of $$R$$ of index $$2$$. Thus $$\mathfrak{d} = \mathfrak{m}$$ is not principal. As $$X^2 - X -1 \in \ker(\text{eval}_q)$$, any polynomial in $$\ker(\text{eval}_q)$$ is of the form $$f(X)(X^2 - X - 1) + g(X)$$ with $$f(X), g(X) \in R[X]$$, $$g(X) = 0$$ or $$\deg(g(X)) \le 1$$. Therefore $$\ker(\text{eval}_q) = R[X]((1 - \sqrt{5})X + 2) + R[X]((2X - (1 + \sqrt{5})X) + R[X](X^2 - X - 1).$$

But we can get a positive answer for a class of rings which are not necessarily Noetherian. A domain $$R$$ is called a locally GCD domain if every localization $$R_{\mathfrak{m}}$$ of $$R$$ at a maximal ideal $$\mathfrak{m}$$ of $$R$$ is a GCD domain.

Claim 3. Let $$R$$ be a locally GCD domain, e.g., a Prüfer domain, and let $$q \in \text{Frac}(R)$$. Then $$\ker(\text{eval}_q)$$ is generated as an ideal of $$R[X]$$ by the set $$\{ bX - a \,\vert \, q = \frac{a}{b} \}$$.

I ignore if the converse of Claim 3 holds true.

Proof of Claim 3. Let $$i \ge 1$$ and let $$\mathfrak{c}_i$$ be the ideal of $$R$$ generated the leading coefficients of the polynomials in $$\ker(\text{eval}_q)$$ with degree $$i$$. We have in particular $$\mathfrak{c}_1 = \mathfrak{d}$$, the ideal of denominators of $$q$$. We observe first that $$\ker(\text{eval}_q)$$ is generated by polynomials of degree at most $$n \ge 1$$ if and only if $$\mathfrak{c}_i = \mathfrak{c}_n$$ for every $$i > n$$. We shall establish that $$\mathfrak{c}_i = \mathfrak{c}_1$$ for every $$i > 1$$. Let $$\mathfrak{m}$$ be a maximal ideal of $$R$$ and let $$\ker(\text{eval}_q)_{\mathfrak{m}}$$ be the kernel of the evaluation map $$f \mapsto f(q)$$ from $$R_{\mathfrak{m}}[X]$$ to $$\text{Frac}(R)$$. Since $$\mathfrak{c}_i = \left( \frac{b}{a^i} (Ra + Rb)^{i - 1} \right)\cap R$$, the ideal generated by the leading coefficients of the polynomials in $$\ker(\text{eval}_q)_{\mathfrak{m}}$$ with degree $$i$$ is $$\mathfrak{c}_i R_{\mathfrak{m}}$$. As $$R_{\mathfrak{m}}$$ is a GCD domain by hypothesis, we deduce from Claim 2, that $$\mathfrak{c}_i R_{\mathfrak{m}} = \mathfrak{c}_1 R_{\mathfrak{m}}$$ for every $$i > 1$$. As this holds for every maximal ideal $$\mathfrak{m}$$, we deduce that $$\mathfrak{c}_i = \mathfrak{c}_1$$ holds for every $$i > 1$$.

Addendum. I discovered that Claim 1 above is known under the following form:

Claim 4 [2, Exercise 17.2]. Let $$R$$ be a commutative integral domain and $$a \in R, b \in R \setminus \{0\}$$ and let $$q = \frac{a}{b} \in F(R)$$. Then the following are equivalent:

• $$(i)$$ The sequence $$(b, a)$$ is a regular.
• $$(ii)$$ The cohomology group $$H^1(K(b,a))$$ of the Koszul complex of $$(b, a)$$ is trivial, that is $$(Rb: Ra)/Rb = \{0\}$$.
• $$(iii)$$ The polynomial $$(bX - a)$$ is a prime element of $$R[X]$$.
• $$(iv)$$ The ideal $$\ker(\text{eval}_q)$$ is the principal ideal of $$R[X]$$ generated by $$bX - a$$.

A sequence $$(a_1, \dots, a_n)$$ of elements in a ring $$R$$ is said to be regular if for each $$i$$ the element $$a_i$$ is a regular element of $$R/(Ra_1 + \cdots Ra_{i - 1})$$. Given two ideals $$I, J$$ of $$R$$, we used above the following notation: $$(I : J) \Doteq \{ r \in R \,\vert\, rJ \subseteq I \}$$.

[1] P. M. Cohn, "Bezout rings and their subrings", 1968.
[2] D. Eisenbud, "Commutative Algebra with a View Toward Algebraic Geometry", 1995.