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Let $R$ be a commutative ring with unity.

So I was given a polynomial $f(x_1,\dots,x_n)$ in $R[x_1,x_2,\dots,x_n]$. I was given the root of $f$ as $(a_1,\dots,a_n)$.

My claim was we can write $f(x_1,\dots,x_n) = \sum_{i=1}^{n} (x_i - a_i)q_i(x_1,\dots,x_n)$ using the division alogrithm. Now I realised division algorithm doesn't make sense in multivariate polynomial rings. So I tried to prove my claim using induction, showing this holds for $m=1$, assumed it's true for $m= n-1$ (induction hypothesis) and try to prove it now for $m=n$. Here $m$ is number of variables in the polynomial. Is that a correct approach or do there exist a simpler way out?

Thanks in advance.

KReiser
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Caratheodory_Enthusiast
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1 Answers1

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That is the correct approach. I'll supply a proof of the following claim behind a spoiler barrier so that you can try to prove it yourself if you'd like. We will need the following. See here for some more details about this fact.

Proposition(Division Algorithm): Let $A$ be a commutative ring $f, g \in A[x]$ are degree $d < d'$ polynomials respectively. Assume $f$ is monic. Then, we may write $g = fq + r$ where $q \in A[x]$ is some polynomial, and $r \in A[x]$ has degree less than $d$.

Claim: Let $\phi: R[x_1, \dots, x_n] \to R$ be $f \mapsto f(0)$. Then, $\ker \phi = (x_1, \dots, x_n)$.

Proof: Clearly, $(x_1, \dots, x_n) \subset \ker \phi$ so we focus on the other direction. Let $f$ be an element of the kernel, so that $f(0) = 0$. We divide $f$ by $x_n$ in $R[x_1, \dots, x_{n - 1}][x_n]$ to obtain $f = q x_n + g$ for some $q \in R[x_1, \dots, x_n]$ and $g \in R[x_1, \dots, x_{n - 1}]$. Since $g(0)$ must also be zero, $g \in (x_1, \dots, x_{n - 1})$ by induction, which proves that $f \in (x_1, \dots, x_n)$ as required.

Your claim follows from mine, and a change of variables.

Daniel
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