Let $R=k[X_1,...,X_n]$ be a polynomial ring, where $k$ is a field.

Suppose we have a Gröbner basis $G = (g_1, g_2, ... , g_n), g_i \in R$ for the ideal $I = \langle f_1, f_2, ..., f_m \rangle, f_i \in R$ and that we have using the division algorithm for polynomials in several variables that $f^G = 0$ for $f \in R$ (remainder from division by $G$ is $0$).

Since $f \in I$ if and only if $f^G = 0$ we know that $f \in I$.

Now we want to write $f= a_1f_1 + \cdots + a_nf_n$ for $a_i \in R$ as a $k$-linear combination of the polynomials generating $I$. We know we can do this, since we assumed that $f^G = 0$.

We can easily write $f$ as a $k$-linear combination of the polynomials in $G$, since the remainder is unique and $0$ for every permutation of $G$. So let $f = b_1 g_1 + \cdots + b_n g_n$.

Can I use this to easily find a way to write $f= a_1f_1 + \cdots + a_nf_n$ ?

It can be tiredsome to to write $f$ as a $k$-linear combination of polynomials $f_i$, since there are several path to take during the algorithm.

  • $\begingroup$ A $k$-linear combination means that the coefficients are in $k$, but in your question they are in $R$. $\endgroup$
    – user26857
    Dec 12, 2013 at 21:11

1 Answer 1


If you found $G$ using Buchberger's algorithm then you constructed them as linear combinations of the $f_i$. You can then substitute these expressions into $f = \sum b_ig_i$ to get $f$ as a linear combination of the $f_i$.

  • $\begingroup$ Why ? I start with a set of polynomials $F=(f_1, f_2, ..., f_n)$ and using Buchbergers algorithm I add $S(f_i, f_j)^F$ to $F$ to get $F^{'}=(f_1,f_2,...,f_n,S(f_i, f_j)^F)$ where $S(f_i, f_j)^F$ denote the remainder of the $S$-polynomial $S(f_i, f_j)$ after division by $F$. Is it because I can write $$S(f_i, f_j) = a_1 f_1 + a_2 f_2 + ... + a_n f_n + S(f_i, f_j)^F \Rightarrow S(f_i, f_j)^F = -a_1 f_1 - a_2 f_2 - ... - a_n f_n + S(f_i, f_j)$$ and $S(f_i, f_j)$ is a linear combination of $f_i$ and $f_j$ thus getting $S(f_i, f_j)^F$ written purely in terms of $F$? $\endgroup$
    – Shuzheng
    Dec 12, 2013 at 16:32
  • $\begingroup$ Yes, that is exactly why. $\endgroup$
    – Jim
    Dec 12, 2013 at 17:54
  • $\begingroup$ Thanks a lot Jim, this makes my work easier when I need to express $f$ in term of the ideal $I$ after confirmed that $f$ actually lies in $I$ :). $\endgroup$
    – Shuzheng
    Dec 12, 2013 at 18:47

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