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.
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 $B$.