# Factor Theorem for Multivariate Polynomials [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 12 at 16:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

## 1 Answer

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 $$f\in\langle t_1-u_1,\ldots,t_n-u_n\rangle$$.

• You are right! Sorry, I mis-stated (and simplified) the theorem actual theorem. (I have now edited the question appropriately.) – John Don Feb 12 at 14:20
• 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 $f(\pmb{u}) = 0$. – John Don Feb 12 at 14:27
• 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? – Servaes Feb 12 at 14:34
• 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 $F[t_1,..., t_n]$? – John Don Feb 12 at 14:47
• The same algorithm works in $F[t_1,\ldots,t_n]$, and in fact in any unique factorization domain. – Servaes Feb 12 at 15:02