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


*

*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.

*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.
