1

I want to prove the following: From the Nullstellensatz, deduce that any linear polynomial vanishing on the zero set of a set of linear varieties is a K-linear combination of them.

Let $f_1, \ldots, f_r$ be linear polynomials in $n$-variables over a field $K$, i.e., they belong to $K[X_1, \ldots, X_n]$ and they are linear. Let $J$ be the ideal generated by them. Let $f$ be a linear polynomial such that $f$ vanishes on the zero set of $J$, i.e., $f \in I(V(J))$. By Hilbert's Nullstellensatz $I(V(J)) = \sqrt{J}$, so $f \in \sqrt{J}$. So $f^N = \sum h_i f_i$ where $h_i \in K[X_1, \ldots, X_n]$. By degree comparisons we can write $f^N = \sum (h_i)_{N-1} f_i$ where $(h_i)_{N-1}$ is the degree $N-1$ homogeneous component of $h_i$. I can't proceed further and deduce that $N$ is actually $1$, so that $h_i$ are all constants. In other words I want to prove that $f$ is a $K$-linear combination of the $f_i$. Please help.

Vanya
  • 629
  • 3
  • 9
  • https://math.stackexchange.com/questions/2496792/ideals-in-polynomial-rings-generated-by-linear-polynomials – leoli1 May 25 '25 at 16:44
  • By the reasoning at the linked duplicate, via a change of coordinates we may arrange that $f_i=X_i$, so $J=(X_1,\cdots,X_r)$ which is radical. So $f\in(X_1,\cdots,X_n)$ is linear, and it is a linear combination of the $X_i$, finishing the problem. – KReiser May 25 '25 at 17:23
  • @KReiser Thank you. I got it. – Vanya May 26 '25 at 08:16

0 Answers0