Let $I,J\unlhd K[x_1,\ldots,x_n]=K[x]$ be monomial ideals and $f\!: K[x]\to K[x]$ a graded isomorphism (given by a matrix $A=[\alpha_{i,j}]\in K^{n\times n}$, i.e. $x_i\mapsto\sum_j\alpha_{i,j}x_j$ is $f$), such that $f(I)=J$, i.e. $K[x]/I\cong K[x]/J$.

Question 1: How can I prove: There exists a bijection $\varphi\!: \{x_1,\ldots,x_n\}\to\{x_1,\ldots,x_n\}$ whose induced isomorphism $\overline{\varphi}\!: K[x]\to K[x]$ satisfies $\overline{\varphi}(I)=J$?

I suspect that $f$ suffices as $\varphi$, i.e. each column of $A$ contains only one nonzero entry, but I have problems proving this. Can a linear isomorphism send a monomial ideal to a monomial ideal without being a permutation of the variables? Let $I=\langle x^{a_1},\ldots,x^{a_k}\rangle$ and $J=\langle x^{b_1},\ldots,x^{b_l}\rangle$, so that $\langle f(x^{a_1}),\ldots,f(x^{a_k})\rangle =\langle x^{b_1},\ldots,x^{b_l}\rangle$. Here $f(x^a)=f(x_1^{a_1}\cdots x_n^{a_n})=$ $(\sum_j\alpha_{1,j}x_j)^{a_1}\cdots(\sum_j\alpha_{n,j}x_j)^{a_n}$, so every monomial of $f(x^a)$ has degree $|a|$. Since $\langle f(x^{a_1}),\ldots,f(x^{a_k})\rangle$ is a monomial ideal, every monomial of every $f(x^{a_i})$ is contained in $f(I)$, thus $f(I)$ is actually generated by all monomials of all $f(x^{a_1}),\ldots,f(x^{a_k})$. But I don't know how to prove that each $f(x^{a_i})$ has only one monomial, i.e. $f$ is a permutation of variables.

Basically I'm asking for an elementary proof of Theorem 5.27 in Polytopes, Rings, and K-Theory (Bruns, Gubeladze - 2009 - Springer SMM).

EDIT: It turns out (see mbrown's answer below) that my claim was wrong, so I change the question to look for a positive result.

Question 2: How can I prove the following claim Thm.5.27 using only elementary commutative algebra (no fancy shmancy Borel’s theorem):

Let $I\!\unlhd\!K[x]$ and $J\!\unlhd\!K[y]$ be monomial ideals with $\forall i,j\!: x_i\!\notin\!I,\, y_j\!\notin\!J$. Then t.f.a.e.:

  1. $\exists$bijection $\varphi\!:\{x_1,\ldots,x_n\}\!\rightarrow\!\{y_1,\ldots,y_{n'}\}$ whose morphism $\overline{\varphi}\!: K[x]\!\rightarrow\! K[y]$ satisfies $\overline{\varphi}(I)\!=\!J$;
  2. $K[x]/I \cong K[y]/J$ as $K$-algebras;
  3. $K[x]/I \cong K[y]/J$ as $\mathbb{N}$-graded $K$-algebras.

(1)$\Rightarrow$(2) is clear.

(2)$\Rightarrow$(3) is shown here.

(3)$\Rightarrow$(1): Since the isomorphism $\overline{f}: K[x]/I\to K[y]/J$ is graded, it induces a $K$-linear isomorphism $f: K\{x_1,\ldots,x_n\}\to K\{y_1,\ldots,y_{n'}\}$ (so $n=n'$) that determines $\overline{f}$ and is given by an invertible matrix $A\in K^{n\times n}$. This induces an isomorphism $f:K[x]\to K[y]$ with $f(I)=J$. Now I don't know how to change $f$ to an isomorphism with $f(I)=J$, so that each column of $A$ has only one nonzero entry.

If there's an elementary proof in the literature, that would be a more than sufficient answer.

  • $\begingroup$ In THM 5.27, it is required that $I,J$ do not contain any of the $x_i$. Do you want to include this as well? $\endgroup$
    – user55407
    Oct 3 '13 at 3:03
  • $\begingroup$ BTW: this implies Combinatorial Invariance of Stanley-Reisner rings (Bruns & Gubeladze): $\Delta\cong \Delta'$ iff $K[\Delta]\cong K[\Delta']$, for any simplicial complexes $\Delta,\Delta'$. $\endgroup$
    – Leo
    Oct 3 '13 at 5:01
  • $\begingroup$ Also on MO. $\endgroup$
    – Leo
    Oct 6 '13 at 4:38

Say $n=2$. Define $f:k[x,y] \to k[x,y]$ by $x \mapsto x+y$, $y \mapsto y$. Say $I=J=(y^2)$ (I am squaring $y$ just in case you want to include the requirement I mentioned in the comments).

$f$ is a graded $k$-algebra automorphism, and it maps $I$ to $J$. But $f$ does not permute the variables.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.