Why isn't $\mathbb{C}[x,y,z]/(xz-y)$ a flat $\mathbb{C}[x,y]$-module

Why isn't $M = \mathbb{C}[x,y,z]/(xz-y)$ a flat $R = \mathbb{C}[x,y]$-module?

The reason given on the book is "the surface defined by $y-xz$ doesn't lie flat on the $(x,y)$-plane". But I don't understand why this can be a reason.

Since

An $R$-module $M$ is flat if and only if for any ideal $S$ of $R$, the map $1_M \otimes i: M \otimes_RS \rightarrow M \otimes_RR$ is injective. Here, $i: S \rightarrow R$ is the inclusion map.

I tried to find an ideal $S$ in $R$ such that $M \otimes_RS \rightarrow M \otimes_RR = M$ is not injective, but I didn't succeed. Please give me some help. Thank you.

Solution 1:

This answer is similar to the others; perhaps it will help to see the same points made by yet another person.

First of all, it might help to note that $\mathbb C[x,y,z]/(xz-y)$ is isomorphic to $\mathbb C[x,z]$. So you are looking at the map $\mathbb C[x,y] \to \mathbb C[x,z]$ defined by $x \mapsto z, y \mapsto x z$, and asking why it is not flat.

Geometrically, this is the map $\mathbb A^2 \to \mathbb A^2$ defined by $(x,z) \mapsto (x,xz)$. Note that a whole line in the first copy of $\mathbb A^2$ in the source (the line where $x = 0$) is mapped to a single point of the target (the point $(0,0)$), whereas the map is an open immersion on the complement of this line. Since open immersions are flat, this says that the point $(0,0)$ in the target is where we should focus our attention when looking for non-flatness.

(Here is a translation of my remark about open immersions in algebraic terms: if $f$ is any polynomial in $\mathbb C[x,y]$ with zero constant term, then the map on localizations $\mathbb C[x,y]_f \to \mathbb C[x,z]_f$ is flat --- check this!)

There is one ideal that is particularly "sensitive" to the point $(0,0)$, namely its corresponding maximal ideal $(x,y) \subset \mathbb C[x,y]$. So let's try this ideal.

We have to look at the induced map $(x,y)\otimes \mathbb C[x,z]\to \mathbb C[x,z]$. The very equation $y = x z$ defining the map $\mathbb C[x,y] \to \mathbb C[x,z]$ suggests an element in the kernel, namely the element $y\otimes 1 - x\otimes z$. I leave it as an exercise to check that this element is non-zero in $(x,y)\otimes\mathbb C[x,z]$. (If you don't know how to make this sort of computation, then you should probably ask it as a separate question --- but try first!)

One lesson to draw from this is that the geometry of the situation informs the algebra. A more specific remark is that the map $\mathbb A^2 \to \mathbb A^2$ from your question is an affine patch of the blow up of $\mathbb A^2$ at the origin, and this example illustrates the general fact that non-trivial blow-ups are never flat.

Added: Looking over the other answers, it seems that one of the points of the question is to really check that $y \otimes 1 - x \otimes z$ is non-zero in $(x,y)\otimes \mathbb C[x,z]$.

There is a standard way to compute tensor products: by generators and relations. While there can be other tricks in particular cases (see e.g. Michael Joyce's answer), it might be worth explaining this standard approach, since it doesn't require any cleverness; you can always just do it.

We have to begin with a presentation of the ideal $(x,y)$ as a $\mathbb C[x,y]$-module. This is easy: it has two generators, $x$ and $y$, which satisfy the relation $y x - x y = 0$. So we have the presentation $$ 0 \to \mathbb C[x,y] \cdot e \to \mathbb C[x,y]\cdot f_1 \oplus \mathbb C[x,y]\cdot f_2 \to (x,y) \to 0,$$ where $e$, $f_1$, and $f_2$ are just names for basis elements of free modules, and the maps are given by $e \mapsto (y f_1, -x f_2)$, and $f_1\mapsto x, f_2 \mapsto y$.

Now we tensor with $\mathbb C[x,z]$, to obtain the presentation $$ \mathbb C[x,z] \cdot e \to \mathbb C[x,z] \cdot f_1 \oplus \mathbb C[x,z] \cdot f_2 \to (x,y)\otimes \mathbb C[x,z] \to 0,$$ where again the maps are given by $e \mapsto (y f_1, -x f_2) = (x z f_1, - x f_2) = x(z f_1,-f_2),$ and $f_1 \mapsto x, f_2 \mapsto y = x z$. (Note that in this particular case this exact sequence is also exact on the left, but that is not a general feature of this approach to computing tensor products, since generally tensoring is right-exact, but not exact.)

From this presentation of $(x,y)\otimes \mathbb C[x,z]$, we see that $x\otimes z - y$ (which is the image of $(z f_1, -f_2)$) is non-zero, since $(z f_1, -f 2)$ is not in the image of the map from $\mathbb C[x,z]\cdot e.$

On the other hand, it is a torsion element --- it is killed by multiplication by $x$ (since $x(z f_1, -f_2)$ is in the image of $\mathbb C[x,z] \cdot e$; indeed it is the image of $e$). This reflects the fact that if we localize away from $x = 0$ (i.e. invert $x$), the original map becomes flat, and so the map $(x,y)\otimes\mathbb C[x,z] \to \mathbb C[x,z]$ must become injective after inverting $x$; hence its kernel must consist of $x$-torsion elements.

Solution 2:

Here is a

New version of the answer

I'll leave the old version below so that the comments remain understandable.

Let $K$ be a commutative ring and $x,y,z$ be indeterminates. Put $$ M:=\frac{K[x,y,z]}{(xz-y)}\quad. $$ In particular, $M$ is an $K[x,y]$-module.

We claim that $M$ is not $K[x,y]$-flat.

Set $$ t:=1\otimes y-z\otimes x\in K[x,y,z]\underset{K}{\otimes}(x,y). $$ In view of the flatness criterion mentioned in the question, it suffices to check that

$(*)$ the image of $t$ in $$ M\underset{K[x,y]}{\otimes}(x,y) $$ is nonzero.

The $K$-bilinear map $$ \phi:K[x,y,z]\times(x,y)\to K $$ defined by $$ \phi(f,g)=f(0,0,0)\ \frac{\partial g}{\partial y}(0,0) $$ induces a $K[x,y]$-bilinear map $$ \overline\phi:M\times(x,y)\to K=\frac{K[x,y]}{(x,y)} $$ satisfying $$ \overline\phi(1,y)=1,\quad\overline\phi(z,x)=0. $$ This proves $(*)$.

Old version of the answer

Let $K$ be a commutative ring and $x,y,z$ be indeterminates. Put $$ A=K[x,y],\quad B:=K[x,y,z],\quad M:=B/(xz-y). $$ In particular, $M$ is an $A$-module.

We claim that $M$ is not $A$-flat.

Let denote the image in $M$ of the element $b\in B$ by $b_M$.

In view of the flatness criterion mentioned in the question, it suffices to check $$ t:=z_M\otimes x-1_M\otimes y\neq0\in M\otimes_A(x,y). $$ Let $N$ be the quotient of $B$ by the sub-$A$-module generated by $$ xz-y,\quad x^2,\quad z^2,\quad xy,\quad yz. $$ Put $P:=(x,y)/(x,y)^2$.

It suffices to prove that the image of $t$ in $N\otimes_AP$ is nonzero.

But $N$ admits the $K$-basis $$ 1_N,\quad x_N,\quad y_N,\quad z_N $$ (obvious notation), whereas $P$ admits the $K$-basis $x_P,y_P$ (obvious notation).

The claim follows easily from these observations.

EDIT. Thanks to Michael Joyce and Georges Elencwajg for their comments. I'll try to salvage the argument.

Let $n_1,n_2,n_3,n_4$ be the $K$-basis of $N$ mentioned above, and $p_1,p_2$ be the $K$-basis of $N$ mentioned above.

Consider the $K$-bilinear map $f$ from $N\times P$ to $K$ mapping $(1_N,y_P)$ to $1$ and the other $(n_i,p_j)$ to $0$.

Let $x$ and $y$ act by $0$ on $K$.

Then it suffices to check that $f$ is $A$-bilinear, which (I think) is clear.