I am looking for neat ways of proving the following theorem:

Let $F$ be a field and let $f \in F[t_1, ..., t_n]$ be a polynomial. If $f(\pmb{u}) = 0$ for some $\pmb{u} \in F^n$, then $f$ lies in the ideal $\langle t_1 - u_1, ..., t_n - u_n \rangle$.

I know of one method, but it does not seem particularly direct or intuitive; I get the sense that this theorem admits more straightforward proofs...


marked as duplicate by Eric Wofsey abstract-algebra Feb 12 at 16:28

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  • $\begingroup$ @André3000 Yes it is! $\endgroup$ – John Don Feb 12 at 14:53

Your 'theorem' is false, as witnessed by the counterexample ${\bf u}=(1,1)\in F^2$ and $f=t_1t_2-1\in F[t_1,\ldots,t_n]$ satisfying $f({\bf u})=0$ without $t_i-1$ dividing $f$ for any $i$.

As for the edited question; note the ideal is the kernel of the ring homomorphism $$F[t_1,\ldots,t_n]\ \longrightarrow\ F:\ g\ \longmapsto\ g({\bf u}).$$ More concretely, repeated division with remainder of $f$ with the $t_i-u_i$ leaves remainder $f({\bf u})=0$:

Starting with $t_n-u_n$, the Euclidean algorithm gives us $q_n,r_n\in F[t_1,\ldots,t_n]$ such that $$f=q_n(t_n-u_n)+r_n,$$ and $\deg_{t_n}r_n<\deg_{t_n}(t_n-u_n)$. This means $r_n\in F[t_1,\ldots,t_{n-1}]$. Then we can divide $r_n$ by $t_{n-1}-u_{n-1}$ with the Euclidean algorithm to get $q_{n-1},r_{n-1}\in F[t_1,\ldots,t_{n-1}]$ such that $$r_n=q_{n-1}(t_{n-1}-u_{n-1})+r_{n-1},$$ and $\deg_{t_{n-1}}r_{n-1}<\deg_{t_{n-1}}(t_{n-1}-u_{n-1})$.

Repeating this $n$ times yields $q_n,\ldots,q_1\in F[t_1,\ldots,t_n]$ and $r_1\in F$ such that $$f=q_n(t_n-u_n)+q_{n-1}(t_{n-1}-u_{n-1})+\cdots+q_1(t_1-u_1)+r_1.$$ Plugging ${\bf u}$ into the above shows that $0=f({\bf u})=r_1$, and so $f\in\langle t_1-u_1,\ldots,t_n-u_n\rangle$.

  • $\begingroup$ You are right! Sorry, I mis-stated (and simplified) the theorem actual theorem. (I have now edited the question appropriately.) $\endgroup$ – John Don Feb 12 at 14:20
  • $\begingroup$ Is this not just re-stating the claim in a different way. I.e. For $f$ to be in the kernel is exactly the statement that $f(\pmb{u}) = 0$. $\endgroup$ – John Don Feb 12 at 14:27
  • $\begingroup$ It is. What constitutes a proof really depends on the reader; does the fact that the kernel is (contained in) the ideal $\langle t_1-u_1,\ldots,t_n-u_n\rangle$ require proof to you? $\endgroup$ – Servaes Feb 12 at 14:34
  • $\begingroup$ I guess so. The reverse inclusion is obvious, but this one isn't as much. I'm not quite sure what you mean by division of $f$ with/by $t_i - u_i$? In $F[x]$ we can just use the division algorithm, but what about in $F[t_1,..., t_n]$? $\endgroup$ – John Don Feb 12 at 14:47
  • $\begingroup$ The same algorithm works in $F[t_1,\ldots,t_n]$, and in fact in any unique factorization domain. $\endgroup$ – Servaes Feb 12 at 15:02

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