I am currently learning something about multivariable polynomial rings, and I'm stuck on this problem, or rather I am confused by it. Let $R = F[X_1, ... , X_n]$ be the polynomial ring in $n$ variables over a field $F$. Let $g_1, ... , g_k$ be non-zero elements in $R$, and let $I$ be the ideal generated by those $g_i$. We want to assume that the "leading coefficients" of these $g_i$ are all $1$, the "leading coefficient" for a multivariable polynomial of the form $f(X_1,...,X_n) = \sum_{\alpha\in \mathbb{N}^n} a_\alpha X^\alpha$ can be defined as $\text{Le}(f):= \max_\preceq \{ \alpha \in \mathbb{N}^n : a_\alpha \neq 0\}$, with $\preceq$ being a monomial order on $\mathbb{N}^n$. Now I have to show the equivalence of two statements, which are:

$a)$ If $f \in I \setminus\{0\},$ then $\text{Le}(g_i) \mid \text{Le}(f) $ for some $i$

$b)$ If $f \in I$, then for any division with remainder of $f$ by the $g_i$, the remainder vanishes

The "$\mid$" for two elements $a,b \in \mathbb{N}^n$ means that $a_i \leq b_i$ for all $i$. But I was wondering, because I think that $b)$ is always true and doesnt need $a)$, since if $f \in I$, it has the form $f = \sum_{i=1}^k r_i g_i$, with $r_i \in R$, but the division with remainder would look like $f = \sum_{i=1}^k q_i g_i + r$, so we could just directly say $r = 0$ and $q_i = r_i$ and are done, we don't need to use the $a)$ anywhere. I don't see what the flaw is, could someone help me?

  • 1
    $\begingroup$ For polynomials in several variables the "quotients" and the remainder are not unique. $\endgroup$
    – user26857
    Commented May 13, 2021 at 17:02
  • $\begingroup$ So it can happen that f has the form $f = \sum_{i=1}^k q_i g_i + r$ with a non-zero r? $\endgroup$ Commented May 13, 2021 at 17:27
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    $\begingroup$ Yes, a polynomial $f$ can belong to the ideal generated by $g_1,\dots,g_k$ and the remainder could be non-zero. This is why people introduced the Grobner bases. If $g_1,\dots,g_k$ form a Grobner basis this phenomenon can't occur. $\endgroup$
    – user26857
    Commented May 13, 2021 at 17:32

1 Answer 1


The following are equivalent:
$a)$ If $f \in I \setminus\{0\},$ then $\text{Le}(g_i) \mid \text{Le}(f)$ for some $i$.
$b)$ If $f \in I$, then for any division with remainder of $f$ by the $g_i$, the remainder vanishes.

$a)\implies b)$ Write $f = \sum_{i=1}^k q_i g_i + r$. Then $r\in I$ and therefore $\text{Le}(g_i) \mid \text{Le}(r) $ for some $i$. But this contradicts the definition of the remainder.

$b)\implies a)$ We can write $f = \sum_{i=1}^k q_i g_i$. Then $\text{Le}(f)=\text{Le}(q_ig_i)$ for some $i$. But $\text{Le}(q_ig_i)=\text{Le}(q_i)+\text{Le}(g_i)$, and therefore $\text{Le}(g_i) \mid \text{Le}(f)$.


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