Show this ideal in polynomial ring of four variable is finitely generated free module over polynomial in two variables Let $I \subset K[X, Y, Z,W]$ be the ideal given by $I = (Y W -Z^{2}, XW - YZ,XZ-
Y^{2}),$
where $K$ is a field.
Show that
$A = K[X, Y, Z,W]/I$
is
finitely generated free module over $B = K[X,W]$.
I tried looking at the general element in $K[X, Y, Z,W]$, But IDK how to proceed.
 A: There is probably a nice solution to this problem, but I’ve decided to go with a tried and true method instead.
The polynomial ring $k[X, Y, Z, W] ≅ B[Y, Z]$ has the monomials $Y^n Z^m$ with $n, m ≥ 0$ as a $B$-basis.
In $A$, we can rewrite these monomials via the rewriting rules
$$
 Y^2 \longrightarrow XZ \,,
 \quad
 YZ \longrightarrow XW \,,
 \quad
 Z^2 \longrightarrow YW \,.
$$
This allows us to express those monomials with $n + m ≥ 2$ in terms of those monomials for which $n + m ≤ 1$.
(Each of the three rewriting rules reduces the sum $n + m$, whence the repeated use of these rewriting rules always terminates.)
As an example, we have
$$
 [Y^3 Z^2]
 =
 [X Y Z^3]
 =
 [X Y^2 W Z]
 =
 [X^2 W Z^2]
 =
 [X^2 Y W^2]
 =
 X^2 W^2 ⋅ [Y] \,.
$$
(We denote here, and in the following, the action of a ring on a module by “$⋅$”.)
The condition $n + m ≤ 1$ is only satisfied in three cases: either $n = 0$ and $m = 0$, or $n = 1$ and $m = 0$, or $n = 0$ and $m = 1$.
We therefore conjecture that $A$ has the three elements
$$
 [1] = [Y^0 Z^0] \,,
 \quad
 [Y] = [Y^1 Z^0] \,,
 \quad
 [Z] = [Y^0 Z^1]
$$
as a $B$-basis.
Our above discussion already explains why these three elements generate $A$ as a $B$-module.
But we also need to show that they are linearly independent over $A$.
We use for this a standard method from representation theory.
Step 1.
The $B$-algebra $B[Y, Z]$ is generated by the two elements $Y$ and $Z$.
We compute the action of these two algebra generators on the proposed basis elements $[1]$, $[Y]$, $[Z]$ of $A$, and express the results as $B$-linear combination of these three proposed basis elements.
We find that
\begin{gather*}
 Y ⋅ [1] = [Y] \,,
  \quad
 Y ⋅ [Y] = X ⋅ [Z] \,,
  \quad
 Y ⋅ [Z] = XW ⋅ [1] \,,
 \\
 Z ⋅ [1] = [Z] \,,
  \quad
 Z ⋅ [Y] = XW ⋅ [1] \,,
  \quad
 Z ⋅ [Z] = W ⋅ [Y] \,.
\end{gather*}
Step 2.
Let now $M$ be the free $B$-module with basis $e_1$, $e_Y$, $e_Z$.
There exist unique $B$-linear endomorphisms $f_Y$ and $f_Z$ on $M$ such that
\begin{gather*}
  f_Y(e_1) = e_Y \,,
  \quad
  f_Y(e_Y) = X ⋅ e_Z \,,
  \quad
  f_Y(e_Z) = XW ⋅ e_1 \,,
  \\
  f_Z(e_1) = e_Z \,,
  \quad
  f_Z(e_Y) = XW ⋅ e_1 \,,
  \quad
  f_Z(e_Z) = W ⋅ e_Y \,.
\end{gather*}
These two endomorphisms $f_Y$ and $f_Z$ commute.
This can be checked on the basis elements $e_1$, $e_Y$ and $e_Z$ of $M$, where we find that
\begin{gather}
 f_Z( f_Y( e_1 ) ) = f_Z( e_Y ) = XW ⋅ e_1 = f_Y( e_Z ) = f_Y( f_Z( e_1 ) ) \,,
 \\
 f_Z( f_Y( e_Y ) ) = f_Z( X ⋅ e_Z ) = X ⋅ f_Z( e_Z ) = XW ⋅ e_Y  = XW ⋅ f_Y(e_1) = f_Y(XW ⋅ e_1) = f_Y( f_Z( e_Y ) )
 \\
 f_Z( f_Y( e_Z ) ) = f_Z( XW ⋅ e_1 ) = XW ⋅ f_Z(e_1) = XW ⋅ e_Z = W ⋅ f_Y(e_Y) = f_Y( W ⋅ e_Y ) = f_Y( f_Z( e_Z ) ) \,.
\end{gather}
It follows that the $B$-module structure of $M$ extends to a $B[Y, Z]$-module structure, such that $Y$ acts via $f_Y$ and $Z$ acts via $f_Z$.
Step 3.
We claim that this $B[Y, Z]$-module structure descends to an $A$-module structure.
For this, we need to check that the action of the three elements $YW - Z^2$, $XW - YZ$ and $XZ - Y^2$ on $M$ is zero.
This can again be checked on the $B$-basis elements $e_1$, $e_Y$ and $e_Z$ of $M$.
For the element $YW - Z^2$ we find
\begin{gather*}
 YW ⋅ e_1 = W ⋅ e_Y = Z ⋅ e_Z = Z^2 ⋅ e_1 \,,
 \\
 YW ⋅ e_Y = XW ⋅ e_Z = XZW ⋅ e_1 = Z^2 ⋅ e_Y \,,
 \\
 YW ⋅ e_Z = XW^2 ⋅ e_1 = ZW ⋅ e_Y = Z^2 ⋅ e_Z \,,
\end{gather*}
for the element $XW - YZ$ we find
\begin{gather*}
 XW ⋅ e_1 = Y ⋅ e_Z = YZ ⋅ e_1 \,,
 \\
 XW ⋅ e_Y = XYW ⋅ e_1 = YZ ⋅ e_Y \,,
 \\
 XW ⋅ e_Z = YW ⋅ e_Y = YZ ⋅ e_Z \,,
\end{gather*}
and for the element $XZ - Y^2$ we find
\begin{gather*}
 XZ ⋅ e_1 = X ⋅ e_Z = Y ⋅ e_Y = Y^2 ⋅ e_1 \,,
 \\
 XZ ⋅ e_Y = X^2 W ⋅ e_1 = XY ⋅ e_Z = Y^2 ⋅ e_Y \,,
 \\
 XZ ⋅ e_Z = XW ⋅ e_Y = XYW e_1 = Y^2 ⋅ e_Z \,.
\end{gather*}
Step 4.
We have now arrived at the $A$-module structure on $M$.
We observe that
$$
 ( b_1 [1] + b_2 [Y] + b_3 [Z] ) ⋅ e_1
 =
 b_1 [1] ⋅ e_1 + b_2 [Y] ⋅ e_1 + b_3 [Z] ⋅ e_1
 =
 b_1 e_1 + b_2 e_Y + b_3 e_Z
$$
for all coefficients $b_1, b_2, b_3 ∈ B$.
The basis elements $e_1$, $e_Y$ and $e_Z$ are by construction linear independent over $B$, so it follows that the elements $[1]$, $[Y]$ and $[Z]$ are also linear independent over $B$.
