I'm teaching soon an introductory (undergraduate) class in Grobner bases. I am planning to use the below as a motivating example, but my Algebra is a bit rusty, can someone confirm that I am not saying anything stupid?

Consider a field $k$ and some polynomials $f_1,..., f_n, g \in k[x_1,..., x_m]$.

Question 1: If $f_1,..., f_n, g$ are linear, then the following are equivalent:

  • (i) $g \in \langle f_1,f_2, \ldots , f_n \rangle$ where $ \langle f_1,f_2, \ldots , f_n \rangle$ is the ideal generated by those polynomials.
  • (ii) $g \in \mbox{Span} \{ f_1,f_2, \ldots , f_n \}$
  • (iii) $g=0$ on the variety defined by $f_1,...,f_n$.

(ii)$\Rightarrow$ (i) $\Rightarrow$ (iii) is obvious, while $(iii) \Rightarrow (ii)$ can be proven via standard row reduction. (Am I making any dumb mistake here?)

Question 2: Is it true that the equivalence (i) $\Leftrightarrow$ (iii) is still true if $f_1,..., f_n$ are linear, but $g$ is not?

I suspect that this is true by Nullstellensatz, since the ideal generated by linear polynomials should be radical (is it?). They do not know Nullstellensatz, but I am planning to briefly cover a version of it at the end of the class.

Of course, (ii) is false in this case, (i) $\Rightarrow$ (iii) is true with no restrictions on $f's$ and $g's$, while in general $(iii) \Rightarrow (i)$ only holds for radiacal ideals.

  • $\begingroup$ I'm not seeing that you need the assumption about linearity. $\endgroup$
    – Wuestenfux
    Commented Dec 18, 2021 at 14:07
  • $\begingroup$ @tkf yes. Span means span as a vector space over $k$. $\endgroup$
    – N. S.
    Commented Dec 18, 2021 at 16:36
  • $\begingroup$ @Wuestenfux The assumptions about linearity are important. For example $(x_1-1)^2 \in \langle x_1-1 \rangle$ but $(x_1-1)^2 \notin \mbox{Span}\{ x_1-1 \}$. Also $g=x_1-1$ is zero on the variety definied by $f_1=(x_1-1)^2$ but does not belong to the ideal generated by $f_1$. $\endgroup$
    – N. S.
    Commented Dec 18, 2021 at 16:38
  • $\begingroup$ @tkf I see, I think in my argument for (iii) implies (ii) I was using the fact that the variety defined by $f_1,..., f_n$ is non empty... So the claim is probably true for homogeneous linera poly, but not interesting. $\endgroup$
    – N. S.
    Commented Dec 18, 2021 at 16:46
  • 1
    $\begingroup$ @tkf I have a slightly different argument in mind, which I think works over every field: (iii) implies that the system given by $f_1=0,...,f_n=0$ and the system given by $f_1=0,...,f_n=0, g=0$ has exactly the same solution. If this is non-trivial (I was missing this), then, when written in matrix form, the two matrices must have same rank, which gives (ii). $\endgroup$
    – N. S.
    Commented Dec 18, 2021 at 16:54

1 Answer 1


As stated we do not have (i)$\implies$ (ii). For a counterexample take $$g=x_2, \qquad f_1=x_1,\,\,f_2=x_1+1.$$

Then $$g=x_2f_2-x_2f_1,$$ so $g\in\langle f_1,f_2\rangle$. However $g\notin {\rm Span}_k\{f_1,f_2\}$.

This example only works because the variety defined by the $f_i$ is empty. As long as this variety is non-empty your argument for (i)$\implies$ (ii) is valid:

If $g$ is not in the span of the $f_i$ and the variety defined by the $f_i$ is non-empty, then throwing $g$ into the system adds a pivot, which reduces the dimension of the variety (or makes it empty). This makes the new variety a proper subset of the original, which contradicts (iii).

It is a nice result - definitely worth mentioning to your students. In my example the $f_i$ are multiplied by linear polynomials, and the degree $2$ terms cancel yielding a linear polynomial not in the span. It is surprising to me that this can only happen in the restricted case that the variety is empty. Otherwise the linear polynomials obtained through this kind of cancellation will always happen to lie in the $k$-linear span of the $f_i$.


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