# 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.

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【Sincere thanks to everyone. The answers and comments after them explain this problem deeply in different aspects. The answers are all enlightening and splendid. It's a pity that I can accept only one of them.】 – ShinyaSakai Feb 21 '12 at 17:05

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.

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It's possibly worth mentioning that $N \otimes_A P$ is a four-dimensional $K$-vector space with basis $1_N \otimes x_P, 1_N \otimes y_P, z_N \otimes x_P, z_N \otimes y_P$. – Michael Joyce Feb 18 '12 at 14:08
Dear Pierre-Yves, I am not quite convinced. I agree that the point of your nice construction (and very clear notation) is that $\bar t=z_N\otimes x_N-1_N\otimes y_N\in N\otimes _A P$ is nonzero. And I also agree that $\bar{\bar t}=z_N\otimes x_N-1_N\otimes y_N\in N_K\otimes _K P_K$ is non zero, where $N_K, P_K$ mean $P,N$ seen as $K$-modules $via$ restriction of scalars $K\to A$. (to be continued) – Georges Elencwajg Feb 18 '12 at 14:15
But I fail to see why this implies that $\bar t$ is non zero, since I cannot see a map $N \otimes_A P \to N_K \otimes_ K P_K$ that would send $\bar t\mapsto \bar{\bar t}$ . – Georges Elencwajg Feb 18 '12 at 14:20
Dear Pierre-Yves: touché! You are absolutely right and paradoxically I think I am more relieved than you, because my role as a censor here was becoming very unpleasant in my own eyes. So kudos , and I strongly encourage everybody else to also upvote this highly non-trivial answer. – Georges Elencwajg Feb 18 '12 at 15:10
Dear @Georges: Once more, I don't know how to thank you!!! - Of course I see no censorship when you try (successfully) to help all of us. I can't tell you how happy I am that other users (and especially you) try help me write things which are not too incorrect and/or stupid! – Pierre-Yves Gaillard Feb 18 '12 at 15:16

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.

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The question seems to be about gaining geometric intuition for flat morphisms of (affine) schemes. Let $A = \mathbb{C}[x, y, z]/(x z - y)$, $X = \operatorname{Spec} A$ and $\mathbb{A}^2_\mathbb{C} = \operatorname{Spec} \mathbb{C}[x, y]$. The claim is that the projection $\pi : X \to \mathbb{A}^2_\mathbb{C}$ is not a flat morphism of schemes. Let's look at a picture.

$X$ is quite evidently twisted. More formally, if one looks at the dimensions of the fibres over the points of $\mathbb{A}^2_\mathbb{C}$, one sees that at $(0, 0)$ there is a 1-dimensional fibre, but everywhere else the fibre is 0-dimensional. It is a fact that a finitely generated module over a noetherian domain is locally free if and only if the dimensions of the fibres over maximal ideals is constant, so this confirms our intuition that $X$ is not flat over $\mathbb{A}^2_\mathbb{C}$

Of course, we could take a more direct approach. From the above discussion we know that there should be a problem at $(0, 0)$, so let $\mathfrak{m} = (x, y) \subset \mathbb{C}[x, y]$. We have a short exact sequence $$0 \to \mathfrak{m} \to \mathbb{C}[x, y] \to \mathbb{C} \to 0$$ and upon tensoring with $A$, we get a right exact sequence $$A \otimes \mathfrak{m} \to A \to \mathbb{C}[z] \to 0$$ I claim that the first map is not injective. Indeed, in $A$, $$z \cdot x = 1 \cdot y$$ but in $A \otimes \mathfrak{m}$ $$z \otimes x \ne 1 \otimes y$$ since there are no $f \in \mathbb{C}[x, y]$, $g \in A$, and $h \in \mathfrak{m}$ such that \begin{align*} g f & = z & h & = x \\ g & = 1 & f h & = y \end{align*} Hence $A$ is not flat over $\mathbb{C}[x, y]$.

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+1 for great picture and great explanation. – Michael Joyce Feb 18 '12 at 11:48
Dear Zhen, I don't understand your argument that $z \otimes x \ne 1 \otimes y$. What criterion are you using to show that two elementary tensors of the form $u \otimes v$ and $u' \otimes v'$ are not equal? Please explain that in the abstract context of two $R$-modules $U,V$ over a ring $R$. – Georges Elencwajg Feb 18 '12 at 12:29
@Georges: I'm assuming that the linearity relations for $\otimes$ do not play a role here – this is somewhat unjustified, I agree. Perhaps the fact that the modules in question are all $\mathbb{C}$-vector spaces justifies my assumption. The relations I am thinking of are the ones of the form $g f \otimes h = g \otimes f h$. – Zhen Lin Feb 18 '12 at 13:11
@ZhenLin This may seem like an elementary question but can you explain how you plotted $\operatorname{Spec}(A)$? – user38268 Feb 19 '12 at 3:47
@BenjaminLim: You're thinking far too high-tech. I plotted the equation $x z - y$. – Zhen Lin Feb 19 '12 at 10:11

How much algebraic geometry have you had? Their explanation is based on the fact that the fibers of flat morphism have constant dimension. In your case, the ring homomorphism $R \rightarrow M$ corresponds to the map of schemes $X := \text{Spec } M \rightarrow Y := \text{Spec } R$. The fibers have dimension $0$ ($= \dim X - \dim Y$) if $x \neq 0$ but have dimension $1$ if $x = y = 0$. Thus, $f$ cannot be flat.

Algebraically, this means you should consider $I := (x,y) \subseteq R$ and the corresponding map $I \otimes_R M \rightarrow R \otimes_R M = M$. This map is induced by the bilinear map $I \times M \rightarrow M$ given by $(f,m) \mapsto fm$. It suffices to find a non-zero element of $I \otimes_R M$ that maps to $0$, and I leave it as an exercise for you to check that $x \otimes z - y \otimes 1$ does the trick.

EDIT: As per Georges' request, I am adding the relevant algebra to give a full proof. It is clear that $x \otimes z - y \otimes 1$ maps to $0$ in $M$, but what needs proof is that $x \otimes z - y \otimes 1$ is not already $0$ in $I \otimes_R M$.

Note that $M$ is isomorphic to the polynomial ring $k[x,z]$. We have $x \otimes P(x,z) =y \otimes Q(x,z)$ if and only if $x \mid Q(x,z)$ and $xz \mid P(x,z).^*$ Of course, this can happen, e.g. $y \otimes x = x \otimes xz$. If $x \otimes z = y \otimes 1$, then we must have $x \mid 1$ and $y = xz \mid z$ in $M$, which is not the case. (This is easy to check only because $M$ is a polynomial ring, for which we know the units are the constant functions.)

In fact, you can similarly show that a basis for $I \otimes_R M$ as a $k$-vector space is given by $x \otimes x^a z^b$, as $a,b$ ranges over pairs of non-negative integers, and $y \otimes z^b$, as $b$ ranges over non-negative integers.

${}^*$ This is the key statement. One can see it directly by invoking the explicit construction of the tensor product in this case. But there must be better ways to see this. Pierre-Yves Gaillard has given one such alternative approach -- quotienting out by appropriate submodules of $I$ and $M$ so that you reduce to finite dimensional $k$-vector spaces and compute there.

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Dear Michael, I think you should completely solve your "exercise" yourself in an edit to your answer: what you write in your second section is completely obvious apart from that. – Georges Elencwajg Feb 18 '12 at 12:08
Sure, it's certainly a fair point that all I did was use the geometry to isolate the algebraic problem that needs to be solved. (But that's not always obvious'' to someone studying commutative algebra.) Showing $x \otimes z - y \otimes 1$ is not zero is the non-trivial algebra required to finish the proof. – Michael Joyce Feb 18 '12 at 12:46
Dear Michael, we agree! – Georges Elencwajg Feb 18 '12 at 13:05
@Georges: I attempted to add the details that you suggested. Ultimately, I have not found what you are really seeking ... an abstract "functorial" argument. I'll think about it some more during my morning run and will post an edit if I come up with something along the lines of what I take it you are asking for. – Michael Joyce Feb 18 '12 at 14:22
Thanks Michael, but please don't ruin your running because of me: it would make me very sorry ! – Georges Elencwajg Feb 18 '12 at 15:17