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Problem 1.19 in Eisenbud's Commutative Algebra asks the following.

Given $R = k[x,y,z,w]$ and $I = (yw - z^2, xw - yz, xz - y^2)$, show that $R/I$ is free as an $k[x,w]$-module, and exhibit a basis.

It is simple to show that under the desired substitutions any monomial can be reduced to $\{1, y, z\}$ with some $k[x,w]$ monomial scalar. Linear independence is implicitly a kind of converse: we must demonstrate the limits of the substitution rules, often by finding an invariant or canonical form. I've handled problems like this before using the following techniques:

  1. Prove that $z$ (or maybe $x, w$) is not a zero divisor. This would be a good start, as all monomials are of the form $w^a x^b z^{\{0, 1, 2\}}$.
  2. Apply some kind of counting trick on the grading of the ring (since the ideal is homogeneous, the $R$ grading does descend). For example, if the parity of the total degree in $y$ and $z$ were preserved by the relations, then that could be used to show $y$ and $y^2$ are distinct, for example. I tried a number of such tricks, but there doesn't seem to be any obvious "invariant" of the substitutions like degree, or difference of degrees, etc.
  3. Can we reduce to the case of showing linear independence of the form $m_1 + m_2y + m_3 z = 0$ with $m_i$ monomials with scalar 1? This is true of pairwise linear independence, as the elements would have to cancel pairwise in each degree, so the general case of pairwise linear independence reduces to comparing two monomials.
  4. While the problem does not seem to be looking for it, there may well be a sophisticated argument using dimensionality arguments or something else algebraic-geometric. The problem is trying to conclude the ideal is prime, so a geometric argument could show the corresponding projective variety to be irreducible.

Does one of the above techniques work? Is there an elementary proof of this fact? A high-tech slick proof?

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  • $\begingroup$ As a starting point, note that the given ideal defines a twisted cubic in $\mathbb P^3$ (so is prime). $\endgroup$ Aug 8 '18 at 9:24
  • $\begingroup$ That is certainly true. Is there a geometric interpretation of the coordinate ring being free like this? $\endgroup$
    – C.D.
    Aug 12 '18 at 16:16
  • $\begingroup$ Sorry, I suppose my comment wasn't much help. I noticed that $\{1,y,z\}$ were a basis over each of the two affine patches $w=1,x=1$ and so they glue together -- but that's probably not helpful as a general approach! $\endgroup$ Aug 13 '18 at 14:09
  • $\begingroup$ That is helpful for this problem though, thank you. $\endgroup$
    – C.D.
    Aug 13 '18 at 16:41
  • $\begingroup$ Is there some kind of locally free => free argument needed to make the gluing work? Does this argument translate to something about proving freeness in the localizations at w and z? $\endgroup$
    – C.D.
    Aug 13 '18 at 17:36
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This is not as expansive as the other answer by Z Wu, but perhaps that will be a strength.

First I claim that we can write every monomial in $R/I$ as $x^aw^d$, $x^aw^dy$, or $x^aw^dz$ by applying the given relations: given a monomial of the form $x^ay^bz^cw^d$, we may apply the relations $z^2=yw$ and $y^2=xz$ if either $b$ or $c$ is greater than one to reduce $b+c$, and by repeated applications we may assume $b,c\leq 1$. By the relation $yz=xw$, we may assume they are not both $1$, leading to the claim. This shows that $\{1,y,z\}$ is a spanning set.

Now I claim that $\{1,y,z\}$ is linearly independent. Suppose we have a linear dependence relation $p(x,w)+q(x,w)y+r(x,w)z=0$ in $R$, and apply the homomorphism $R/I\to k[t,u]$ by $x\mapsto t^3$, $y\mapsto x^2u$, $z\mapsto tu^2$, $w\mapsto u^3$. Examining the residue class of the exponent of $x$ modulo $3$, we find that all the monomials coming from $p(x,w)$ have $x$-exponent $0\pmod 3$, all the monomials coming from $q(x,w)y$ have $x$-exponent $2\pmod 3$, and all the monomials coming from $r(x,w)z$ have $x$-exponent $1\pmod 3$. So there can be no cancellation between these terms, and so $p=q=r=0$.

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  • $\begingroup$ (Indeed, you are showing that the relations given by the OP form a confluent rewriting system, and they in fact form a Gröbner basis!) $\endgroup$
    – Pedro Tamaroff
    Jun 20 '21 at 22:10
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I have a solution for a specific set of ideals s.t. the basis is of the from $\{x_1^{t_1}\cdots x_m^{t_m}|\ \forall 1\leq i\leq m,\ 0\leq t_i <n_i\}$. Let me know in the comments if you have seen a similar result in any published reference.

Let's start with a simple case:

Theorem 1: Let $R$ be a commutative ring, $n$ be a positive integer, and $f(x)\in R[x]$ be a polynomial of degree less than $n$. Then the quotient ring $$\frac{R[x]}{(x^n-f(x))}$$ is a free $R$-module with the basis $\{x^i|\ 0\leq i\leq n-1\}$.

Proof: Clearly $\{x^i|\ 0\leq i\leq n-1\}$ is an $R$-linearly generating set of $\frac{R[x]}{(x^n-f(x))}$ by a simple induction or contradiction argument. It remains to show linearly independence. Assume there exists a vector $(r_0,...,r_{n-1})\in R^n$ s.t. $$\sum_{i=0}^{n-1} r_i x^i\equiv 0\text{ in }\frac{R[x]}{(x^n-f(x))}.$$ Then there exists a polynomial $g(x)\in R[x]$ s.t. $$\sum_{i=0}^{n-1} r_i x^i=g(x)\cdot (x^n-f(x))$$ Let $c_m x^m$ be the highest non-zero term of $g(x)$ (if it exists), then the highest non-zero term on RHS is $c_m x^{m+n}$ and the highest non-zero term of LHS has degree $\leq n-1$, contradiction. So $g(x)$ is the zero polynomial. It follows that $r_i=0$ for all $i$. The result follows.$\square$

We can generalize the previous theorem as following:

Theorem 2: Let $R$ be a commutative ring and $m\in \mathbb{N}_+$. Let $\mathop{\mathrm{ord}}(\cdot)$ be a monomial order on the polynomials of $R[x_1,...,x_m]$. For each $1\leq i\leq m$, let $n_i\in \mathbb{N}_+$ and $f_i\in R[x_1,...,x_m]$ be a polynomial s.t. $\mathop{\mathrm{ord}}(f_i)< \mathop{\mathrm{ord}}(x_i^{n_i})$. Then the quotient ring $$\frac{R[x_1,...,x_m]}{(x_1^{n_1}-f_1,...,x_m^{n_m}-f_m)}$$ is a free $R$-module with the basis $\{x_1^{t_1}\cdots x_m^{t_m}|\ \forall 1\leq i\leq m,\ 0\leq t_i <n_i\}$.

The previous theorem can be shown by proving a stronger statement:

Theorem 3: Let $R$ be a commutative ring, $j\leq m$ be positive integers. Let $\mathop{\mathrm{ord}}(\cdot)$ be a monomial order on the polynomials of $R[x_1,...,x_m]$. For each $1\leq i\leq j$, let $n_i\in \mathbb{N}_+$ and $f_i\in R[x_1,...,x_m]$ be a polynomial s.t. $\mathop{\mathrm{ord}}(f_i)< \mathop{\mathrm{ord}}(x_i^{n_i})$. Then the quotient ring $$S_j:=\frac{R[x_1,...,x_m]}{(x_1^{n_1}-f_1,...,x_j^{n_j}-f_j)}$$ is a free $R$-module with the basis $\{x_1^{t_1}\cdots x_m^{t_m}|\ \forall 1\leq i\leq j,\ 0\leq t_i <n_i \text{ and } \forall i>j,\ t_i\geq 0\}$.

Remark 1: A monomial order on a polynomial ring $R[x_1,...,x_m]$ is a total order on the set of monomials s.t.

  1. If $u\leq v$ and $w$ is any other monomial, then $uw\leq vw$.
  2. If $u$ is any monomial then $u\geq 1$.

The order of a polynomial is defined to be the order of its leading term under this order. On $R[x]$, the only monomial order is $x^i<x^j\Leftrightarrow i<j$.

Proof of theorem 3: We fix $m$ and prove by induction on $j$. When $j=1$, we would like to use theorem 1 on the quotient ring $$\frac{R[x_1,...,x_m]}{(x_1^{n_1}-f_1)}=\frac{R[x_2,...,x_m][x_1]}{(x_1^{n_1}-f_1)}.$$ It suffices to show the degree of $f_1$ w.r.t. $x_1$ is less than ${n_1}$. Let $x_1^{c_1}\cdots x_m^{c_m}$ be the leading monomial of $f_1$, if $c_1\geq n_1$ then $\mathop{\mathrm{ord}}(x_1^{c_1}\cdots x_m^{c_m})\geq \mathop{\mathrm{ord}}(x_1^{n_1})$, contradication. So we must have $c_1<n_1$ and we can apply theorem 1. Thus the quotient ring $$\frac{R[x_2,...,x_m][x_1]}{(x_1^{n_1}-f_1)}$$ is a free $R[x_2,...,x_m]$-module with the basis $\{x_1^{t_1}|\ 0\leq t_1<n_1\}$. Hence it is a free $R$-module with the basis $\{x_1^{t_1}\cdots x_m^{t_m}|\ 0\leq t_1 <n_1 \text{ and } \forall i>1,\ t_i\geq 0\}$.

Now assume the result holds for $j-1\geq 1$.

The part of being a generating set

Denote $\Gamma_{j}:=\{x_{1}^{t_{1}}\cdots x_{m}^{t_{m}}|\ \forall1\leq i\leq j,\ 0\leq t_{i}<n_{i}\text{ and }\forall i>j,\ t_{i}\geq0\}$. For each element $\overline{f}\in S_{j}$, define the the order of it as $$\mathrm{ord}(\overline{f}):=\min\{\mathrm{ord}(g)|\ g\in f+(x_{1}^{n_{1}}-f_{1},...,x_{j}^{n_{j}}-f_{j})\}.$$

Let $S$ be the subset of $S_{j}$ not generated by $\Gamma_{j}$. Assume $S\neq\emptyset$. Then there exists an element $\overline{f}\in S$ with a minimal order. WLOG assume $\mathrm{ord}(\overline{f})=\mathrm{ord}(f)$. Remove all terms of $f$ in $\Gamma_{j}$ and call it $f'$. Then $\overline{f'}\in S$ and by minimality of $\overline{f}$ we have $$\mathrm{ord}(\overline{f'})=\mathrm{ord}(f')=\mathrm{ord}(f)=\mathrm{ord}(\overline{f}).$$ Hence the leading monomial $x_{1}^{c_{1}}\cdots x_{m}^{c_{m}}$ of $f'$ is not in $\Gamma_{j}$, so there exists $1\leq k\leq j$ s.t. $c_{k}\geq n_{k}$. Then by reduction $x_{k}^{n_{k}}-f_{k}$, we have $f'\equiv g+(x_{1}^{n_{1}}-f_{1},...,x_{j}^{n_{j}}-f_{j})$ for some $g$ s.t. $\mathrm{ord}(g)<\mathrm{ord}(f)$ (we use the fact $\mathrm{ord}(x_{1}^{c_{1}}\cdots x_{k}^{c_{k}-n_k}\cdots x_{m}^{c_{m}}\cdot f_k)< \mathrm{ord}(x_{1}^{c_{1}}\cdots x_{k}^{c_{k}-n_k}\cdots x_{m}^{c_{m}}\cdot x_k^{n_k})$). So $$\mathrm{ord}(\overline{f'})=\mathrm{ord}(\overline{g})\leq\mathrm{ord}(g)<\mathrm{ord}(f)=\mathrm{ord}(\overline{f'}),$$ contradiction. Hence $S$ is empty and $\Gamma_{j}$ is a generating set.

The part of linear independence

Assume for each $I$ indexed by $\Gamma_{j}$ there exists $r_{I}\in R$ s.t. only a finite number of them are non-zero and $$\sum_{I}r_{I}x^{I}\equiv0\text{ in }\frac{R[x_{1},...,x_{m}]}{(x_{1}^{n_{1}}-f_{1},...,x_{j}^{n_{j}}-f_{j})}.$$ Hence there exists polynomials $g_{i}\in R[x_{1},...,x_{m}]$ s.t. $$\sum_{I}r_{I}x^{I}=g_{1}\cdot(x_{1}^{n_{1}}-f_{1})+\cdots+g_{j}\cdot(x_{j}^{n_{j}}-f_{j})$$ Pass to $S_{j-1}:=\frac{R[x_{1},...,x_{m}]}{(x_{1}^{n_{1}}-f_{1},...,x_{j-1}^{n_{j-1}}-f_{j-1})}$, we have $$\sum_{I}r_{I}x^{I}\equiv g_{j}\cdot(x_{j}^{n_{j}}-f_{j})\text{ in }\frac{R[x_{1},...,x_{m}]}{(x_{1}^{n_{1}}-f_{1},...,x_{j-1}^{n_{j-1}}-f_{j-1})}$$ For each element $\overline{f}\in S_{j-1}$ with a representation $f$, by induction hypothesis $\overline{f}$ has a unique representation $\mathrm{re}(f)$ by a linear combination of monomials in $\Gamma_{j-1}$. Clearly there exists an algorithm which reduces $f$ to $\mathrm{re}(f)$ in finite steps and the order is non-increasing, so $\mathop{\mathrm{ord}}\mathrm{re}(f)\leq\mathop{\mathrm{ord}} f$.

WLOG assume $g_{j}=\mathrm{re}(g_{j})$, so we have $g_{j}\cdot x_{j}^{n_{j}}=\mathrm{re}(g_{j}\cdot x_{j}^{n_{j}})$ and $$ \mathop{\mathrm{ord}}\mathrm{re}(g_{j}\cdot f_{j}) \leq\mathop{\mathrm{ord}} (g_{j}\cdot f_{j})<\mathop{\mathrm{ord}} (g_{j}\cdot x_{j}^{n_{j}}). $$ We have $$\sum_{I}r_{I}x^{I}\equiv g_{j}\cdot x_{j}^{n_{j}}-\mathrm{re}(g_{j}\cdot f_{j})\text{ in }\frac{R[x_{1},...,x_{m}]}{(x_{1}^{n_{1}}-f_{1},...,x_{j-1}^{n_{j-1}}-f_{j-1})}$$ By induction hypothesis, we actually have $$\sum_{I}r_{I}x^{I}=g_{j}\cdot x_{j}^{n_{j}}-\mathrm{re}(g_{j}\cdot f_{j})$$ Let $c_{J}x^{J}$ be the highest non-zero term of $g_{j}$ (if it exists), then $c_{J}x^{J}\cdot x_{j}^{n_{j}}$ must be the highest non-zero term of RHS due to the fact $\mathop{\mathrm{ord}}\mathrm{re}(g_{j}\cdot f_{j})<\mathop{\mathrm{ord}} (g_{j}\cdot x_{j}^{n_{j}})$. But there is no such term on LHS, contradiction. So $g_{j}=0$ and $\text{RHS}=0$. So $r_{I}=0$ for all $I$. The result follows.$\square$

Remark 2: It's tempting to generalize theorem 2 further by requiring $f_i$ to be a linear combination of the potential basis $\{x_1^{t_1}\cdots x_m^{t_m}|\ \forall 1\leq i\leq m,\ 0\leq t_i <n_i\}$ (if it fails then try $n_i=n$ for all $i$). But sadly I have found counter examples when I tried to prove it:

Generating set counter example: consider $\frac{R[x,y]}{(x^2-xy,y^2-xy)}$ with a potential basis $\{1,x,y,xy\}$, but acutally its basis is $\{x^i|\ i\geq 0\}\cup \{y\}$.

Linear independence counter example: consider $\frac{R[x,y]}{(x^2-xy,\ y^2-xy+1)}$ with a potential basis $\{1,x,y,xy\}$, then $$ \begin{align} 0&\equiv y^2\cdot (x^2-xy)+xy\cdot (y^2-xy+1)\\ &\equiv x^2y^2-xy^3+xy^3-x^2y^2+xy\\ &\equiv xy \end{align} $$ So $\{1,x,y,xy\}$ is not linearly independent.


To answer the specific problem in this question, I attempted to prove the following conjecture, which is essentially imposing a further relation $x^J-f_J$ where $x^J$ is one of the basis elements:

Conjecture 1: In theorem 2, denote $\Gamma:=\{x_1^{t_1}\cdots x_m^{t_m}|\ \forall 1\leq i\leq m,\ 0\leq t_i <n_i\}$. Let $x^J\in \Gamma\backslash \{1\}$ and $f_J$ be a polynomial of $R[x_1,...,x_m]$ with an order less than $x^J$, then the quotient ring $$S_J:=\frac{R[x_1,...,x_m]}{(x_1^{n_1}-f_1,...,x_m^{n_m}-f_m,x^J-f_J)}$$ is a free $R$-module with the basis $\Gamma_J:=\{x^I\in \Gamma|\ x^J\nmid x^I\}$.

It's easy to prove the part of being a generating set:

Proof of being a generating set: Denote $x^{J}=x_{1}^{d_{1}}\cdots x_{m}^{d_{m}}$.

For each element $\overline{f}\in S_{J}$, define the the order of it as $$\mathrm{ord}(\overline{f}):=\min\{\mathrm{ord}(g)|\ g\in f+(x_{1}^{n_{1}}-f_{1},...,x_{m}^{n_{m}}-f_{m},x^{J}-f_{J})\}.$$ Let $S$ be the subset of $S_{J}$ not generated by $\Gamma_{J}$. Assume $S\neq\emptyset$. Then there exists an element $\overline{f}\in S$ with a minimal order. WLOG assume $\mathrm{ord}(\overline{f})=\mathrm{ord}(f)$. Remove all terms of $f$ in $\Gamma_{J}$ and call it $f'$. Then $\overline{f'}\in S$ and by minimality of $\overline{f}$ we have $$\mathrm{ord}(\overline{f'})=\mathrm{ord}(f')=\mathrm{ord}(f)=\mathrm{ord}(\overline{f}).$$ Hence the leading monomial $x_{1}^{c_{1}}\cdots x_{m}^{c_{m}}$ of $f'$ is not in $\Gamma_{J}$, so either there exists $1\leq k\leq m$ s.t. $c_{k}\geq n_{k}$ (i.e. $x_{1}^{c_{1}}\cdots x_{m}^{c_{m}}\notin\Gamma)$ or $d_{i}\leq c_{i}\leq n_{i}$ for all $1\leq i\leq m$. In the first case we use the reduction $x_{k}^{n_{k}}-f_{k}$, then we have $f'\equiv g+(x_{1}^{n_{1}}-f_{1},...,x_{j}^{n_{j}}-f_{j},x^{J}-f_{J})$ for some $g$ s.t. $\mathrm{ord}(g)<\mathrm{ord}(f)$. So $$\mathrm{ord}(\overline{f'})=\mathrm{ord}(\overline{g})\leq\mathrm{ord}(g)<\mathrm{ord}(f)=\mathrm{ord}(\overline{f'}),$$ contradiction. In the second case we use the reduction $x^{J}-f_{J}$ and we get a contradiction with a similar argument. Hence $S$ is empty and $\Gamma_{J}$ is a generating set.$\square$

And later I find a counter example when I tried to prove the linear independence:

Consider $\frac{R[x]}{(x^3-x,x^2)}$ with a potential basis $\{1,x\}$, we can see that $x\equiv -(x^3-x)+x\cdot x^2\equiv 0$.

So we have to answer the specific problem specifically.

Proof: Denote $S=k[x,w]$, by theorem 2, the quotient ring $$\frac{S[y,z]}{(y^2-x\cdot z,z^2-w\cdot y)}$$ with the graded lexicographic order (i.e. one monomial is larger than another if the first has higher total degree or if they have equal total degree and the first is larger in the lexicographic ordering) is a free $S$-module with the basis $\{1,y,z,yz\}$. Now we claim $\{1,y,z\}$ is a basis of the $S$-module $$\frac{S[y,z]}{(y^2-x\cdot z,z^2-w\cdot y,yz-xw)}.$$ The part of being a generating set is proved as above. It remains to show linear independence. Assume there exists elements $s,\ s_{y},\ s_{z}\in S$ s.t. $$s+s_{y}\cdot y+s_{z}\cdot z\equiv0\text{ in }\frac{S[y,z]}{(y^{2}-x\cdot z,z^{2}-w\cdot y,yz-xw)}.$$ Hence there exists a polynomial $g\in S[y,z]$ s.t. $$s+s_{y}\cdot y+s_{z}\cdot z\equiv g\cdot(yz-xw)\text{ in }\frac{S[y,z]}{(y^{2}-x\cdot z,z^{2}-w\cdot y)}$$ We know there exists $d,\ d_{y},\ d_{z},\ d_{yz}\in S$ s.t. $g\equiv d+d_{y}\cdot y+d_{z}\cdot z+d_{yz}\cdot yz$ in $\frac{S[y,z]}{(y^{2}-x\cdot z,z^{2}-w\cdot y)}$. We have $$\begin{align} s+s_{y}\cdot y+s_{z}\cdot z & \equiv g\cdot(yz-xw)\\ & \equiv(d+d_{y}\cdot y+d_{z}\cdot z+d_{yz}\cdot yz)(yz-xw)\\ & \equiv(d\cdot yz+d_{y}xw\cdot y+d_{z}xw\cdot z+d_{yz}xw\cdot yz)\\ & -(dxw+d_{y}xw\cdot y+d_{z}xw\cdot z+d_{yz}xw\cdot yz)\\ & \equiv-dxw+d\cdot yz \end{align}$$ in $\frac{S[y,z]}{(y^{2}-x\cdot z,z^{2}-w\cdot y)}$. So $d=0$ and $s=s_{y}=s_{z}=0$. The result follows.$\square$

When I finished it, I felt that I have just done a small math project, feels good man.

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  • $\begingroup$ Is $R/I$ from the question of the form you propose us? $\endgroup$
    – user26857
    Jun 12 '21 at 22:17
  • $\begingroup$ @user26857 No, but $R/I'$ is, where $I'\subset I$, then $R/I$ is a simple quotient of $R/I'$. I will add a corollary for this situation. $\endgroup$
    – Z Wu
    Jun 12 '21 at 22:29
  • $\begingroup$ @user26857 The corollary turns out to be wrong, I changed it into a conjecture with a counter example and provided a proof for this question. $\endgroup$
    – Z Wu
    Jun 13 '21 at 1:05
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Let me try to make the argument the go to tool here are Gröbner bases/rewriting systems.

Associate a monomial $z^ay^bx^cw^d$ to the tuple $t=(a,b,c,d)$, and let us say $t<t'$ if:

  • either $r_t = (a+b,c,d) < r_{t'} = (a'+b',c',d')$ in the lexicographical order of $\mathbb N^3$ or
  • $r_t = r_{t'}$ and $t<t'$ in the lexicographical order of $\mathbb N^4$.

This is known as an elimination order or multigraded lexicographical order (see Definition 7.2.3.8 here). It has the effect of making $y^2 > zx$ since $(2,0,0,0)$ yields $(2,0,0)$ while $(1,0,1,0)$ yields $(1,1,0)<(2,0,0)$, thus orienting your relations as follows:

$$z^2 \leadsto yw, \quad yz \leadsto wx, \quad y^2 \leadsto xz.$$

These leading terms lead to the overlappings $yyz$ and $zzy$ and hence to the following $S$-polynomials:

$$S_1 = y(z^2-yw) - z(yz-wx) = zwx - \underline{yy}w\leadsto zwx-xzw = 0$$

$$S_2 = y(yz-wx) - z(y^2-xz) = x\underline{zz} - wxy \leadsto xyw- wxy = 0.$$

As they both resolve (i.e. rewrite to zero using the three relations defining your ideal) you know by Buchberger's criterion that you have a Gröbner basis. Even more so, you know that the normal monomials, those that give a $\mathbb k$-linear basis for your ring, are those monomials not divisible by the leading terms $z^2,yz,y^2$, which are then just those polynomials obtained from $\mathbb k[x,w]$ multiplying by $1,y$ or $z$.

The fact that that the big ring is free over the smaller one is just a consequence, then, of the fact that you have a Gröbner basis: linear independence follows from Buchberger's criterion. The idea is that left multiplication by $x$ and $w$ preserves the subspace of normal monomials, and so one can extract a basis (in this case, those normal monomials that do not contain $x$ or $z$, that is $1,y,z$). This coincides with what was proved, with other methods, in the other answer.

The idea above is a version of what one can consider of as a 'freeness theorem via Gröbner bases', in the spirit of this paper.

Write $\mathbb k[X]$ for the free commutative algebra on the set of variables $X$ and $R$ for a set of relations (polynomials in $X$), $\mathbb k[X\mid R]$ is the quotient $\mathbb k[X] / (R)$. With this at hand, one can state a version of Theorem 4 in the article above.

Theorem. Suppose that $f:\mathbb k[X\mid R] =A \to \mathbb k [X\sqcup Y , R\sqcup S]=B$ is an injection of commutative rings, and suppose that $R$ is a Gröbner basis for $A$, and that $R\sqcup S$ is a Gröbner basis for $B$ obtained from $R$ by adding relations whose leading terms do not contain any variable from $A$. Then $B$ is a free $A$-module.

Proof. Let $V$ be the space of normal forms that do not contain a variable from $A$. There is a map $A\otimes K\to B$ that is surjective, for if $m$ is a monomial that is irreducible for $R\sqcup S$, then it clearly can be written as $m'm''$ where $m'$ is a monomial in $A$ and $m''$ is a monomial in $K$. The map is also injective, since the condition we imposed on the set $S$ implies that we cannot reduce the image $ak$ of an element $a\otimes k\in A\otimes K$: if $k$ is normal, then $ak$ is normal too, as multiplying by elements of $A$ cannot make a leading term appear, by our condition.

Your situation happens with $R = \varnothing$ (!) and $S = \{z^2-yw,yz-xw,y^2-xz\}$. The intuition is that the behaviour here is exactly the same as the one you would expect if your relations were instead just $\{z^2,yz,y^2\}$, that yield the obvious free module $\mathbb k[x,w]\langle 1,y,z \rangle$.

$1$: The order I had chosen did not give the desired leading terms. I'll change it later.

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