${\rm LT}(I \cap J) = {\rm LT}(I) \cap {\rm LT}(J)$: Is It Always True?

Hey guys! Ever wondered if the equation ${\rm LT}(I \cap J) = {\rm LT}(I) \cap {\rm LT}(J)$ always holds true for ideals in a polynomial ring? Well, buckle up because we're about to dive deep into the fascinating world of algebraic geometry and commutative algebra to find out! This question stems from a classic problem in the book Ideals, Varieties, and Algorithms by Cox, Little, and O'Shea, and it touches on some fundamental concepts that are super important for anyone studying these areas. So, let's get started and unravel this mystery together.

Understanding the Key Players

Before we can tackle the main question, let's make sure we're all on the same page with some key definitions. This will give us a solid foundation to build upon. Understanding these concepts thoroughly is extremely helpful.

What's an Ideal?

In the context of polynomial rings, an ideal is a special subset that behaves nicely under certain operations. More formally, an ideal $I$ in a polynomial ring $R$ (like $\mathbb{K}[x_1, \ldots, x_n]$ where $\mathbb{K}$ is a field) satisfies two crucial properties:

1. If $f, g \in I$, then $f + g \in I$. (It's closed under addition.)
2. If $f \in I$ and $h \in R$, then $hf \in I$. (It absorbs multiplication by any element from the ring.)

Think of ideals as special containers within the polynomial ring that keep these properties intact. They play a crucial role in defining algebraic varieties and studying polynomial equations.

What's a Leading Term (${\rm LT}$)?

To understand leading terms, we first need to talk about monomial orderings. A monomial ordering is a way to compare monomials (terms with coefficients of 1, like $x^2y$ or $z^5$) in a consistent manner. Common examples include lexicographic order (like in a dictionary) and graded reverse lexicographic order (which often behaves better computationally).

Given a monomial ordering and a polynomial $f$, the leading term ${\rm LT}(f)$ is simply the term in $f$ that is the largest according to that ordering. For example, if we use lexicographic order with $x > y$, then in the polynomial $f = x^2y + xy^2 + y^3$, the leading term is ${\rm LT}(f) = x^2y$.

The leading term provides a way to represent any polynomial with only one monomial, making it simple to compute Gröbner bases.

What's ${\rm LT}(I)$?

Now, when we talk about ${\rm LT}(I)$ for an ideal $I$, we mean the ideal generated by the leading terms of all the polynomials in $I$. In other words:

${\rm LT}(I) = \langle {\rm LT}(f) : f \in I \rangle$

This ideal is generated by all the leading terms. This concept is super important in computations and in understanding the structure of the ideal $I$.

What's $I \cap J$?

The intersection of two ideals $I$ and $J$, denoted $I \cap J$, is simply the set of all polynomials that belong to both $I$ and $J$. That is:

$I \cap J = \{ f : f \in I \text{ and } f \in J \}$

The intersection is also an ideal.

The Big Question: Is ${\rm LT}(I \cap J) = {\rm LT}(I) \cap {\rm LT}(J)$ Always True?

Now that we have our definitions down, let's tackle the main question: Is it always true that the leading term of the intersection of two ideals is equal to the intersection of their leading terms? In other words, does ${\rm LT}(I \cap J) = {\rm LT}(I) \cap {\rm LT}(J)$ hold for any ideals $I$ and $J$?

The short answer is: Nope, it's not always true!

This is one of those things in math that can be a bit surprising. You might expect the equality to hold, but it turns out that it doesn't. So, let's explore why.

Why It's Not Always True: A Counterexample

To show that the equality doesn't always hold, we need to find a specific example where it fails. Let's consider the polynomial ring $\mathbb{Q}[x, y]$ (polynomials in $x$ and $y$ with rational coefficients) and the ideals:

$I = \langle x^2, y \rangle$ $J = \langle x, y^2 \rangle$

We'll use lexicographic order with $x > y$. Now, let's break this down step by step.

Finding ${\rm LT}(I)$ and ${\rm LT}(J)$

First, let's find the leading term ideals of $I$ and $J$.

For $I = \langle x^2, y \rangle$, the leading terms are ${\rm LT}(x^2) = x^2$ and ${\rm LT}(y) = y$. Thus, ${\rm LT}(I) = \langle x^2, y \rangle$.

For $J = \langle x, y^2 \rangle$, the leading terms are ${\rm LT}(x) = x$ and ${\rm LT}(y^2) = y^2$. Thus, ${\rm LT}(J) = \langle x, y^2 \rangle$.

Finding ${\rm LT}(I) \cap {\rm LT}(J)$

Now, let's find the intersection of these leading term ideals. We have:

${\rm LT}(I) \cap {\rm LT}(J) = \langle x^2, y \rangle \cap \langle x, y^2 \rangle$

To find this intersection, we look for polynomials that are in both ideals. A polynomial in $\langle x^2, y \rangle$ has the form $f = a(x, y)x^2 + b(x, y)y$, and a polynomial in $\langle x, y^2 \rangle$ has the form $g = c(x, y)x + d(x, y)y^2$. For $f$ and $g$ to be the same, we need terms that are multiples of both $x^2$ and $x$, and $y$ and $y^2$. The simplest such terms are multiples of $x^2$, $xy$, $xy^2$, and $y^2$. The intersection is: ${\rm LT}(I) \cap {\rm LT}(J) = \langle x^2, xy, y^2 \rangle$

Finding $I \cap J$

Next, we need to find the intersection of the original ideals $I$ and $J$.

$I \cap J = \langle x^2, y \rangle \cap \langle x, y^2 \rangle$

Again, we are looking for polynomials that are in both $I$ and $J$. A polynomial in $I$ has the form $f = a(x, y)x^2 + b(x, y)y$, and a polynomial in $J$ has the form $g = c(x, y)x + d(x, y)y^2$. For $f$ and $g$ to be the same, we need terms that are multiples of both $x^2$ and $x$, and $y$ and $y^2$. After carefully considering the possible terms, we find that:

$I \cap J = \langle x^2, xy, y^2 \rangle$

Finding ${\rm LT}(I \cap J)$

Finally, we need to find the leading term ideal of $I \cap J$. Since $I \cap J = \langle x^2, xy, y^2 \rangle$, the leading terms are ${\rm LT}(x^2) = x^2$, ${\rm LT}(xy) = xy$, and ${\rm LT}(y^2) = y^2$. The leading term ideal is: ${\rm LT}(I \cap J) = \langle x^2 \rangle$

Comparing ${\rm LT}(I \cap J)$ and ${\rm LT}(I) \cap {\rm LT}(J)$

Now, let's compare what we found:

${\rm LT}(I \cap J) = \langle x^2 \rangle$ ${\rm LT}(I) \cap {\rm LT}(J) = \langle x^2, xy, y^2 \rangle$

Clearly, ${\rm LT}(I \cap J) \neq {\rm LT}(I) \cap {\rm LT}(J)$. In this example, $\langle x^2 \rangle$ is strictly contained in $\langle x^2, xy, y^2 \rangle$, so the equality does not hold.

Why Does This Happen?

The reason this equality doesn't always hold comes down to the fact that taking leading terms can sometimes