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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.

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  • $\begingroup$ 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
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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.

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