Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Theorem 5.27 in Polytopes, Rings, and K-Theory (Bruns, Gubeladze - 2009 - Springer SMM) claims:

Let $K$ be a field and $I\!\unlhd\!K[x]= K[x_1,\ldots,x_n]$ and $J\!\unlhd\!K[y]=K[y_1,\ldots,y_{n'}]$ be monomial ideals with $x_i\!\notin\!I$ and $y_j\!\notin\!J$ for all $i$ and $j$. Then t.f.a.e.:

  1. $\exists$bijection $\varphi\!:\{x_1,\ldots,x_n\}\!\rightarrow\!\{y_1,\ldots,y_{n'}\}$ whose induced 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.

Question: How to prove this using only elementary commutative algebra (no fancy shmancy Borel's theorem, like the mentioned book uses). I just need (3)$\Rightarrow$(1).

(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 and $I$ and $J$ don't contain any variables, $\overline{f}$ induces an isomorphism of $K$-modules $f: K\{x_1,\ldots,x_n\}\to K\{y_1,\ldots,y_{n'}\}$ (therefore $n=n'$) that determines $\overline{f}$ and is given by an invertible matrix $A=[\alpha_{i,j}]\in K^{n\times n}$. This induces an isomorphism $f:K[x]\to K[y]$ with $f(I)=J$. Now I don't know how to change $f$ so that each column of $A$ has only one nonzero entry. The map $f$ need not be a permutation yet, e.g. $K[x,y,z]\to K[x,y,z]$ given by $(x,y,z)\mapsto(x+2y+3z,y,z)$ maps $I=\langle yz\rangle$ to $J=\langle yz\rangle$.

Let's say that $I=\langle x^{a_1},\ldots,x^{a_k}\rangle$ and $J=\langle y^{b_1},\ldots,y^{b_l}\rangle$. W.l.o.g. we can assume that $a_i\nleq a_j$ and $b_i\nleq b_j$ for all $i,j$, where $\leq$ is the partial order $c\leq d :\Leftrightarrow \forall i: c_i\leq d_i$ (otherwise remove the unnecessary generators, since $c\leq d$ iff $x^c| x^d$). Hence $x^{a_1},\ldots,x^{a_k}$ and $y^{b_1},\ldots,y^{b_l}$ are minimal generating sets, which are unique. Thus we have $$\langle f(x^{a_1}),\ldots,f(x^{a_k})\rangle = \langle y^{b_1},\ldots,y^{b_l}\rangle=J.$$ Since this is a monomial ideal, every monomial appearing in every $f(x^{a_i})$ is also contained in $J$. For any $a\in\mathbb{N}^n$ we have $f(x^a)=f(x_1^{a_1}\cdots x_n^{a_n})=(\sum_j\alpha_{1,j})^{a_1}\cdots(\sum_j\alpha_{n,j})^{a_n}$, and each monomial of this has degree $|a|=a_1+\ldots+a_n$.

share|improve this question
    
Also on MSE. –  Leon Lampret Oct 6 '13 at 4:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.