Is a homogeneous non-zero divisor on $R/\operatorname{in}_<(I)$ also a non-zero divisor on $R/I$?
问题内容
Let $R = k[x_1, \dots, x_n]$ be a polynomial ring over a field $k$ equipped with a standard grading, and let $<$ be a monomial order on $R$. Let $I$ be a homogenous ideal of $R$, and let $\operatorname{in}_<(I)$ denote the initial ideal of $I$ with respect to $>$. My question is
If $f\notin I$ is a homogeneous element of $R$ that is a non-zero divisor on $R/\operatorname{in}_<(I)$, then is $f$ a non-zero divisor on $R/I$?
I know it is true if $f$ is a monomial but not sure if $f$ is an arbitrary polynomial.
回答 (1)
The answer is no.
Let $R = k[x,y]$ with the lexicographic monomial order $x > y$.
Let
$
I = (x^2 - y^2) \subset R.
$
This ideal is homogeneous. Its initial ideal with respect to $x > y$ is
$$
\operatorname{in}_<(I) = (x^2),
$$
because the leading term of $x^2 - y^2$ is $x^2$.
Now take the homogeneous polynomial $f = x + y.$
Clearly, $f \notin I$, since $I$ contains only polynomials of degree at least $2$ (or zero).
We work in $$ R/\operatorname{in}_<(I) = k[x,y]/(x^2). $$ Every element here can be uniquely written as $p = a(y) + x\,b(y),$ with $a(y), b(y) \in k[y]$.
Suppose $f \cdot p \in (x^2)$. That is, $$ (x+y)(a(y) + x b(y)) \equiv 0 \pmod{x^2}. $$ Expanding modulo $x^2$, $$ y a(y) + x\bigl(a(y) + y b(y)\bigr) \equiv 0 \pmod{x^2}. $$ Since $k[x,y]/(x^2) \cong k[y] \oplus k[y]x$ as a $k[y]$-module, the coefficients of $1$ and $x$ must both be zero, that is, $$ y a(y) = 0, \qquad a(y) + y b(y) = 0. $$ Because $k[y]$ is a domain, the first equation gives $a(y) = 0$. Then the second gives $y b(y) = 0$, hence $b(y) = 0$. Therefore $p = 0$.
Thus multiplication by $f = x+y$ has zero kernel on $R/\operatorname{in}_<(I)$, so $f$ is a non-zero divisor on $R/\operatorname{in}_<(I)$.
In $R/I = k[x,y]/(x^2 - y^2)$, we have $$ (x+y)(x-y) = x^2 - y^2 \equiv 0 \pmod{I}. $$ But $$ x - y \notin I, $$ because $I$ contains only elements of degree at least $2$ (or zero), while $x-y$ is a nonzero linear polynomial.
Hence $x-y$ is a nonzero element of $R/I$ that is annihilated by $f = x+y$. Therefore $f$ is a zero divisor on $R/I$.