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

  • 6
    $\begingroup$ Consider $\phi = \frac{1 + \sqrt{5}}{2} \in \text{Frac}(\mathbb{Z}[\sqrt{5}])$. $\endgroup$
    – user44191
    Commented Apr 4, 2018 at 2:09

2 Answers 2


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.


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