# Confusion about division with remainder in a multivariable polynomial ring

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?

• For polynomials in several variables the "quotients" and the remainder are not unique. Commented May 13, 2021 at 17:02
• So it can happen that f has the form $f = \sum_{i=1}^k q_i g_i + r$ with a non-zero r? Commented May 13, 2021 at 17:27
• 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. Commented May 13, 2021 at 17:32

$$a)$$ 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 $$\text{Le}(g_i) \mid \text{Le}(f)$$.