What is the ideal of leading terms?

Question

Fix a monomial ordering on the polynomial ring $\Bbb{k}[x_1, \dots, x_n] = R$ over a field. What exactly is $LT(I)$ for an ideal $I$ of $R$? How is it defined and does it form an ideal?

Answer 1

$LT(I)$ is the ideal generated by the leading terms of all polynomials in $I$. By definition, it is an ideal.

As an aside, note that if an ideal is generated by polynomials $f_1, \dots, f_k$, it is not necessarily the case that $LT(I)$ is generated by the leading terms of the $f_i$. If it is, then the $f_i$ form a Groebner basis of $I$.

Answer 2

If you've fixed a monomial ordering then every polynomial has a leading term, for example in the lexicographic ordering $f = 2x_1^2x_2 + x_3$ has leading term $2x_1^2x_2$ so we write $\mathrm{LT}(f) = 2x_1^2x_2$.

Now $$\mathrm{LT}(I) = \{\mathrm{LT}(f) \ | \ f \in I\}$$ is the set of leading terms of polynomials in $I$. It is not an ideal because every element of $\mathrm{LT}(I)$ is a single term, so it's not closed under addition. But it is closed under multiplication by monomials.

If you want the ideal of leading terms then what you want is the ideal $\langle\mathrm{LT}(I)\rangle$ generated by $\mathrm{LT}(I)$.