Quotient of a polynomial ring and leading terms

Question

Let $A=\mathbb C[x_1,\dots,x_n]$ be the polynomial ring in $n$ variables over the field of complex numbers and consider the ideal $I=\langle f \rangle\subseteq A$, where $f$ is a polynomial. Denote by $I':=\langle \mathrm{LT}(f)\rangle$, where $\mathrm{LT}(f)$ is the leading term of $f$ with respect to a monomial order. Is it true that $A/I$ is isomorphic to $A/I'$? Is there a natural isomorphism?

Answer 1

The two rings are not isomorphic in general.

For $n=1$, take $X^2-1$. Then $\mathbb{C}[X]/(X^2)$ has the nilpotent element $\bar{X}$. On the other hand, $\mathbb{C}[X]/(X^2-1) \cong \mathbb{C}[X]/((X-1)(X+1))$, which by the Chinese Remainder Theorem is isomorphic to $\mathbb{C}[X]/(X-1) \times \mathbb{C}[X]/(X+1)\cong \mathbb{C} \times \mathbb{C}$. This latter ring has no nilpotent elements.

For $n>1$, take $f=X_1^d+\dots+X_n^d+1$. Since $f$ is irreducible of degree $d$ (see Irreducible homogeneous polynomials of arbitrary degree), $I=(f)$ is a prime ideal, and so $A/I$ is an integral domain. On the other hand $A/I'$ has zero divisors.