1
$\begingroup$

Let $S_1=k[x_1,\ldots,x_n]$ and $S_2=k[y_1,\ldots,y_m]$ be two polynomial ringsover a field $k$ and $I\subset S_1$ and $J\subset S_2$ be two ideals. Let $S=k[x_1,\ldots,x_n,y_1,\ldots,y_m].$

Question Can we say $in_<(IS+JS)=(in_<IS+in_<JS)$ (all monomial orders are degree reverse lex in repective rings)?

(where $in_<I$ is the ideal generated by $\{in_<f:f\in I\}$)

$\endgroup$
  • 2
    $\begingroup$ Can you say what's $in_{<}$? (at least in words, to be searchable, if the definition is standard) $\endgroup$ – YCor Jan 23 '18 at 22:27
  • $\begingroup$ could you say what's $in_{<}f$? $\endgroup$ – YCor Jan 23 '18 at 23:01
  • 2
    $\begingroup$ @YCor Presumably OP means that $\operatorname{in}_<(f)$ is the initial term of $f$: the smallest term with respect to $<$ that appears in $f$ with nonzero coefficient. $\endgroup$ – Zach Teitler Jan 24 '18 at 7:16
  • $\begingroup$ Removed the tag "local rings" as deemed irrelevant. I wish to add the tag "computational commutative algebra/algebraic geometry" if there were such one. $\endgroup$ – Fan Zheng Jun 24 '18 at 22:42
  • $\begingroup$ @Cusp Could you please clarify what you mean by $in_{<} f$? Is it the largest or smallest term of $f$? For example, if $f = x^2+y^3$, then is $in_{<} f$ equal to $x^2$ or $y^3$? (Some authors use “initial term” for the largest term, like $y^3$, others use it for the smallest term, like $x^2$. Which one do you want to ask about?) $\endgroup$ – Zach Teitler Jun 25 '18 at 4:27
2
$\begingroup$

Some authors write $\operatorname{in}_<(f)$ to mean the largest term of $f$, and other authors write $\operatorname{in}_<(f)$ to mean the smallest term of $f$. (Some authors say "leading term".) In your question you did not say which version of "initial term" you are using. The comments on the question, and on @Billy's answer, reflect some of this confusion.

For this answer let's suppose $\operatorname{in}_<(f)$ means the largest term of $f$. So for example, if $<$ represents degree reverse lexicographic order (or any degree, i.e., graded, order), and $f = x^2 + y^3$, then we are saying $\operatorname{in}_<(x^2+y^3) = y^3$.

In this case the answer to your question is yes. If $S_1 = k[x_1,\dotsc,x_n]$, $S_2 = k[y_1,\dotsc,y_m]$, $S = S_1 \otimes_k S_2 = k[x_1,\dotsc,x_n,y_1,\dotsc,y_m]$, and $I \subseteq S_1$, $J \subseteq S_2$ are ideals, then $$\operatorname{in}_<(IS+JS) = (\operatorname{in}_<(I))S + (\operatorname{in}_<(J))S,$$ where $<$ stands for degree reverse lexicographic order on each of $S_1$, $S_2$, and $S$.

Proof: Let $G$ be a Gröbner basis for $I$ and $H$ be one for $J$. That is, $\operatorname{in}_<(I) = (\operatorname{in}_<(g) \mid g \in G)$ and similarly for $H$. The claim is that $G \cup H$ is a Gröbner basis for $IS+JS$. This follows from results in section 2.9 of the book by Cox, Little, O'Shea:

Cox, David A.; Little, John; O’Shea, Donal, Ideals, varieties, and algorithms. An introduction to computational algebraic geometry and commutative algebra, Undergraduate Texts in Mathematics. Cham: Springer (ISBN 978-3-319-16720-6/hbk; 978-3-319-16721-3/ebook). xvi, 646 p. (2015). ZBL1335.13001.

Specifically there is a notion of "standard representation" and these results:

  1. A basis $G = \{g_1,\dotsc,g_t\}$ for an ideal $I$ is a Gröbner basis if and only if $S(g_i,g_j)$ has a standard representation w.r.t. $G$, for all $i \neq j$. This is Theorem 3 in section 2.9 of that book.
  2. Given any finite set $G$, if $f,g \in G$ are such that $\operatorname{in}_<(f)$ and $\operatorname{in}_<(g)$ are relatively prime (i.e., the monomials have no common variables) then $S(f,g)$ has a standard representation w.r.t. $G$. This is Proposition 4 in section 2.9 of that book.

Applying these to our $G \cup H$ we see that for every $f,g \in G \cup H$, $S(f,g)$ has a standard representation w.r.t. $G \cup H$, because one of the following happens: either $f,g \in G$, and apply the standard representation result for the Gröbner basis $G$; looking at the definition of standard representation (which I am not going to say), a standard representation w.r.t. $G$ is also a standard representation w.r.t. $G \cup H$. Or $f,g \in H$, same reasoning. Or finally one is in $G$ and the other is in $H$, in which case, they involve separate variables, one has $x$'s and the other has $y$'s. So $\operatorname{in}_<(f)$ and $\operatorname{in}_<(g)$ are relatively prime, and $S(f,g)$ has a standard representation w.r.t. $G \cup H$.

Since every $S$-polynomial of a pair of elements in $G \cup H$ has a standard representation w.r.t. $G \cup H$, then $G \cup H$ is a Gröbner basis for the ideal generated by $G \cup H$, which is nothing other than $IS + JS$. Therefore $$ \begin{split} \operatorname{in}_<(IS+JS) &= (\operatorname{in}_<(f) \mid f \in G \cup H) \\ &= (\operatorname{in}_<(f) \mid f \in G) + (\operatorname{in}_<(f) \mid f \in H) \\ &= (\operatorname{in}_<(I))S + (\operatorname{in}_<(J))S. \end{split} $$ This completes the proof. $\square$

I don't know if there's a way to get this with "bare knuckles" Gröbner bases and Buchberger algorithm, and nothing more. (I.e., without invoking "standard representations".) You would have to show that for $f \in G$ and $g \in H$, upon computing the $S$-polynomial $S(f,g)$, the remainder $$ \overline{S(f,g)}^{G \cup H} $$ is zero. That's apparently true (since we know that $G \cup H$ is a Gröbner basis) but if you wanted to prove this directly, in order to prove that $G \cup H$ is a Gröbner basis...? It seems messy. (Perhaps someone knows how to see it easily.)

What if $\operatorname{in}_<(f)$ is supposed to mean the smallest term of $f$? Then I think the answer is yes, but I am less familiar with this side of things, so I don't really know. You might be able to find some answers in books such as:

Kreuzer, Martin; Robbiano, Lorenzo, Computational commutative algebra. I, Berlin: Springer. x, 321 p. (2000). ZBL0956.13008.

which includes discussion of both local and global orders.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.