# Sum of initial ideals

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 $\{in_<f:f\in I\}$)

• Can you say what's $in_{<}$? (at least in words, to be searchable, if the definition is standard)
– YCor
Jan 23 '18 at 22:27
• could you say what's $in_{<}f$?
– YCor
Jan 23 '18 at 23:01
• @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. Jan 24 '18 at 7:16
• Removed the tag "local rings" as deemed irrelevant. I wish to add the tag "computational commutative algebra/algebraic geometry" if there were such one. Jun 24 '18 at 22:42
• @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?) Jun 25 '18 at 4:27

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.