Let $R$ be a commutative ring, $a_1, ..., a_n$ its elements and

$\phi: R[x_1, ..., x_n] \to R$

defined by

$ \phi(f(x_1, ..., x_n)) = f(a_1, ... ,a_n)$

a ring homomorphism.

Prove: $\ker \phi = (x_1-a_1, ..., x_n-a_n)$

It is obvious that $(x_1 - a_1, ..., x_n -a_n) \subseteq \ker \phi$. I'm not sure how to prove the converse.

At this point I don't know any division algorithms for multivariable polynomials, only for the ones in $R[x]$(and the book from where I taken the exercise doesn't assume the reader to know something beyond basics of rings and ideals, and the division algorithm for $R[x]$). Though I know this could be solved for $i = 1$ by dividing by $x-a$:

Let $f(x) \in \ker \phi$, divide by $x-a: f(x) = q(x)(x-a) + r, f(a) = r = 0$, so $f(x) = q(x)(x-a) \in (x-a)$.

  • $\begingroup$ Hint: $f(x_1,\cdots, x_n)+(x_1-a_1, \cdots, x_n-a_n)=f(a_1,\cdots, a_n)+(x_1-a_1, \cdots, x_n-a_n)$. Also since $(x_1-a_1, \cdots, x_n-a_n)\subset \ker\phi$ we have an induced map $R[x_1, \cdots, x_n]/(x_1-a_1, \cdots, x_n-a_n)\to R$, what is this map? Is it surjective? Was $\phi$ surjective? What can you say about $\ker\phi$ now? $\endgroup$
    – Hamed
    Jul 10, 2016 at 1:16
  • $\begingroup$ You say you can do the $n=1$ case, but really every polynomial ring is a polynomial ring in one variable since $R[x_1, \ldots, x_{n-1}, x_n] = (R[x_1, \ldots, x_{n-1}])[x_n]$. I wonder if you can factor your evaluation map into one-variable evaluation maps and extend your proof... $\endgroup$ Jul 10, 2016 at 5:16
  • 2
    $\begingroup$ I found math.stackexchange.com/questions/1352606 $\endgroup$
    – Alphonse
    Feb 7, 2017 at 13:42

2 Answers 2


In problems like this, instead of explicitly showing that every element of the kernel is in $I=(x_1-a_1,\dots,x_n-a_n)$, it is easier to think about the quotient ring $S=R[x_1,\dots,x_n]/I$. Since $I\subseteq\ker(\phi)$, $\phi$ induces a surjective homomorphism $\tilde{\phi}:S\to R$ such that $\phi$ is equal to the composition of $\tilde{\phi}$ with the quotient map $\psi:R[x_1,\dots,x_n]\to S$. Let us now show that $\tilde{\phi}$ is injective, so it is actually an isomorphism and $\ker(\phi)=\ker(\psi)=I$.

To prove this, note that in $S$, $\psi(x_i)=\psi(a_i)$ for each $i$. Since $\psi$ is a homomorphism, it follows that given any polynomial $f(x_1,\dots,x_n)$, $\psi(f(x_1,\dots,x_n))=\psi(f(a_1,\dots,a_n))$. This means that every element of $S$ is actually $\psi(r)$ for some $r\in R$. But for $r\in R$, $\tilde{\phi}(\psi(r))=\phi(r)=r$. In particular, $\tilde{\phi}(\psi(r))=0$ iff $r=0$. It follows that the kernel of $\tilde{\phi}$ consists of only $\psi(0)=0$, so $\tilde{\phi}$ is injective.


This proof is pretty notation-heavy for which I apologize, but the idea is not complicated: a polynomial $f$ is usually written as a Taylor series centered at $(0,\ldots, 0)$. We use the Binomial Theorem to re-center it at $(\alpha_1, \ldots, \alpha_n)$ and show that the constant term of this series is $f(\alpha_1, \ldots, \alpha_n)$.

Lemma: Let $R$ be a unital commutative ring and $R[x_1, \ldots, x_n]$ be the $n$-variable polynomial ring over $R$. For $\alpha_1, \ldots, \alpha_n \in R$, let \begin{align*} \varphi = \text{eval}_{\alpha_1, \ldots, \alpha_n}: R[x_1, \ldots, x_n] &\to R\\ x_1, \ldots, x_n &\mapsto \alpha_1, \ldots, \alpha_n \end{align*} be the evaluation homomorphism. Then $\ker(\varphi) = (x_1-\alpha_1, \ldots, x_n-\alpha_n)$ and the induced map $\overline{\varphi}: R[x_1, \ldots, x_n]/(x_1-\alpha_1, \ldots, x_n-\alpha_n) \to R$ is an isomorphism.

Proof: Certainly $(x_1-\alpha_1, \ldots, x_n-\alpha_n) \subseteq \ker(\varphi)$ so it remains to show the reverse inclusion. Given $f \in \ker(\varphi)$, we may write $f(x_1, \ldots, x_n) = \sum_{\gamma} a_\gamma {x_1}^{\gamma_1} \cdots {x_n}^{\gamma_n}$ where $a_\gamma = 0$ for all but finitely many multi-indices $\gamma$. Writing $x_i = (x_i - \alpha_i) + \alpha_i$ and expanding by the binomial theorem, we have \begin{align*} f(x_1, \ldots, x_n) &= \sum_{\gamma} a_\gamma {x_1}^{\gamma_1} \cdots {x_n}^{\gamma_n} = \sum_{\gamma} a_\gamma ((x_1 -\alpha_1) +\alpha_1)^{\gamma_1} \cdots ((x_n -\alpha_n) +\alpha_n)^{\gamma_n}\\ &= \sum_{\gamma} a_\gamma \left(\sum_{\kappa_1=0}^{\gamma_1} \binom{\gamma_1}{\kappa_1} (x_1 -\alpha_1)^{\kappa_1}\alpha_1^{\gamma_1 - \kappa_1}\right) \cdots \left(\sum_{\kappa_n=0}^{\gamma_n} \binom{\gamma_n}{\kappa_n} (x_n -\alpha_n)^{\kappa_n}\alpha_n^{\gamma_n - \kappa_n}\right)\\ &= \sum_{\gamma} a_\gamma \sum_{\kappa_1=0}^{\gamma_1} \cdots \sum_{\kappa_n=0}^{\gamma_n} \binom{\gamma_1}{\kappa_1} \cdots \binom{\gamma_n}{\kappa_n}\alpha_1^{\gamma_1 - \kappa_1} \cdots\alpha_n^{\gamma_n - \kappa_n} (x_1 -\alpha_1)^{\kappa_1} \cdots (x_n -\alpha_n)^{\kappa_n}\\ &= \sum_{\gamma} a_\gamma \sum_{\kappa} \binom{\gamma_1}{\kappa_1} \cdots \binom{\gamma_n}{\kappa_n}\alpha_1^{\gamma_1 - \kappa_1} \cdots\alpha_n^{\gamma_n - \kappa_n} (x_1 -\alpha_1)^{\kappa_1} \cdots (x_n -\alpha_n)^{\kappa_n}\\ &=\sum_{\kappa} \underbrace{\left(\sum_{\gamma}a_\gamma\binom{\gamma_1}{\kappa_1} \cdots \binom{\gamma_n}{\kappa_n}\alpha_1^{\gamma_1 - \kappa_1} \cdots\alpha_n^{\gamma_n - \kappa_n}\right)}_{b_\kappa} (x_1 -\alpha_1)^{\kappa_1} \cdots (x_n -\alpha_n)^{\kappa_n}\\ &=\sum_{\kappa} b_\kappa (x_1 -\alpha_1)^{\kappa_1} \cdots (x_n -\alpha_n)^{\kappa_n}. \end{align*} Note that the constant term $b_{(0,\ldots, 0)}$ is \begin{align*} b_{(0,\ldots, 0)} = \sum_{\gamma} a_\gamma\binom{\gamma_1}{0} \cdots \binom{\gamma_n}{0}\alpha_1^{\gamma_1} \cdots\alpha_n^{\gamma_n} = f(\alpha_1, \ldots,\alpha_n) = \varphi(f) = 0 \end{align*} since $f \in \ker(\varphi)$. Now all other terms in the sum $\sum_{\kappa} b_\kappa (x_1 -\alpha_1)^{\kappa_1} \cdots (x_n -\alpha_n)^{\kappa_n}$ have a factor of $(x_i - \alpha_i)^{\kappa_i}$ for some $i$ with $\kappa_i \geq 1$, hence belong to $(x_1 - \alpha_1, \ldots, x_n - \alpha_n)$. Thus $f(x_1, \ldots, x_n) = \sum_{\kappa} b_\kappa (x_1 -\alpha_1)^{\kappa_1} \cdots (x_n -\alpha_n)^{\kappa_n} \in (x_1 - \alpha_1, \ldots, x_n - \alpha_n)$.

  • $\begingroup$ Would the downvoter care to comment? $\endgroup$ Jul 28, 2020 at 4:34
  • 1
    $\begingroup$ I like your answer as it is very straightforward. One possible way that could make it less notation-heavy is as follows: You may start with a monomial $m$ first, and use the binomial theorem to show that $m=p_0+m(\alpha_1,\ldots,\alpha_n)$ where $p_0\in(x_1-\alpha_1,\ldots,x_n-\alpha_n)$. Then because every polynomial $f$ is a linear combination of monomials, we shall directly have $f=p+f(\alpha_1,\ldots,\alpha_n)$ for some $p\in(x_1-\alpha_1,\ldots,x_n-\alpha_n)$. If also $f\in\ker(\phi)$, then $f=p\in(x_1-\alpha_1,\ldots,x_n-\alpha_n)$. $\endgroup$ Dec 8, 2022 at 18:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .