How to prove a specific quotient of polynomial ring is a free module? 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:

*

*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 $w^a x^b z^{\{0, 1, 2\}}$.

*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.

*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.

*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?
 A: 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 $p=q=r=0$.
A: 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.
