# Division of multivariable polynomials by an ideal

Let $K$ be a commutative field and consider an ideal $I$ of $K[X_1,\dots,X_n]$.

Is there a well behaved "reduction modulo I", in the following sense :

Given a well-ordering $\leq$ on the set of monomials of $K[X_1,\dots,X_n]$, is there a map $$Red : K[X_1,\dots,X_n]\rightarrow K[X_1,\dots,X_n]$$ such that

(1) $Red(f)=0\iff f\in I$.

(2) $Red(f)\leqslant f$,

(3) $Red(f+g)=Red(f)+Red(g)$.

For $n=1$, there is the rest of the Euclidian division by the generator of $I$...

This seems the kind of things one worries about when dealing with Gröbner basis, but I couldn't find any thing related to condition (3), unless something like $Red(kf)=kRed(f)$ for all $k\in K$ and polynomial $f$.

• Do you mean $I$ in (1) instead of $P$? Besides that, some questions: Do you want the ordering to have some sort of compatibility with the ring operations? Usually in Gröbner basis theory, we only actually give a well-ordering to the set of all monomials. – Avi Steiner Feb 22 '16 at 21:27
• Yes I mean $I$, I just edited it. Concerning the ordering, I'd like to have $x_1$ minimal say, that's all. – Drike Feb 22 '16 at 21:32
• wouldn't (3), as well as $Red(kf)=kRed(f)$ follow directly from your $Red$ being ring homomorphism? – Dima Pasechnik Feb 22 '16 at 21:50
• @DimaPasechnik Sure, it would ! – Drike Feb 22 '16 at 21:54
• Unless I miss something trivial, the process of computing Grobner basis of $I$ with respect to your order $\leq$ will give you $Red$, and it will be ring homomorphism. – Dima Pasechnik Feb 22 '16 at 21:57

I assume that in condition (2) you meant to say that each monomial appearing in the $F$-support of $Red(f)$ is $\leq f$. (In other words, you should not assume reduction simply take monomials to monomials.)

Assuming this slight change, the answer to your question is yes there is such a function (and it is uniquely determined if you add the extra condition that scalars pull out, and monomials either are sent to themselves or sums of multiples of smaller monomials).

Existence: Consider an arbitrary monomial $m$. If $m$ occurs as the largest monomial (under your well-ordering) of some element $f\in I$, then we can write $f=m-\sum_{m'<m}\alpha_{m',m}m'$ for some constants $\alpha_{m',m}\in F$. Further, if some $m'<m$ is also a largest monomial of some element $g\in I$, then after subtracting $\alpha_{m',m}g$ from $f$, we can assume $\alpha_{m',m}=0$. After a finite number of iterations of this process (using the fact $\leq$ is a well-ordering) we may assume $\alpha_{m',m}=0$ whenever $m'$ is a "largest monomial" of some element in $I$. An easy argument now shows that this $f$ is uniquely determined (subject to this constraint), and we define $Red(m)=\sum_{m'<m}\alpha_{m',m}m'$. If $m$ is not the largest monomial of any element in $I$, set $Red(m)=m$. Finally, define $Red$ on an arbitrary element of $K[X_1,\ldots, X_n]$ by $F$-linearity on the monomials. (So your additional condition $Red(kf)=kRed(f)$ also holds!)

Condition (2) obviously holds, and condition (3) holds by linearity of the definition. It also isn't too hard to check that condition (1) holds. So this gives you the function you seek.

Uniqueness: (Very Brief Sketch) Show that $Red$ has to agree with the function defined above, by transfinite induction over the monomials.

Note 1: The answer to your question is no if you assume that reduction is also a multiplicative ring homomorphism. For example if $n=1$, $I=(X_1^2-X_1)$, and the monomials are well-ordered in the obvious way $1<X_1<X_1^2<\ldots$, then $Red(X_1)^2=X_1^2\neq X_1=Red(X_1^2)$. This shouldn't be surprising, as most reduction rules are not ring homomorphisms.

Note 2: One usually wants $Red$ to be "simplifying" so you don't get weird reductions like $Red(X_1)=X_1X_2$, and this requires some restrictions on your well-ordering (as hinted at by Avi Steiner).