I have a problem with showing that the $k$-algebra morphism $$k[x,y,z]/(y^2-xz) \to k[u,v]$$ defined by $x\mapsto u^2$, $y\mapsto uv$, $z\mapsto v^2$ is not flat. Here $k$ is a field.
I usually checked flatness by checking torsion-freeness or checking the equidimensionality of a fibre (or the dimension formula between noetherian schemes), but I have no idea why this morphism is not flat. What other techniques are there to determine flatness of a ring/algebra morphism?
On
1st step: $\varphi$ is injective.
Indeed \begin{equation} \ker(\varphi)=\left\{\sum_{(i,j,k)\in\left(\mathbb{N}_{\geq0}\right)^3}a_{i,j,k}x^iy^jz^k\in\mathbb{K}[x,y,z]_{\displaystyle/(y^2-xz)}\mid\sum_{(i,j,k)\in\left(\mathbb{N}_{\geq0}\right)^3}a_{i,j,k}u^{2i+j}v^{j+2k}=0\right\}=\dotsc=\{0\}. \end{equation} So $R_1=\mathbb{K}[u,v]$ is a module over $R_0=\mathbb{K}[x,y,z]_{\displaystyle/(y^2-xz)}$ via $\varphi$.
2nd step: $\_\otimes_{R_0}R_1$ is not an exact functor. Equivalentely $\varphi$ is not flat.
Considering the short exact sequence \begin{equation} 0\to R_0\xrightarrow{\_\times y}R_0\to\mathbb{K}[x,z]_{\displaystyle/(xz)}\to0, \end{equation} applying $\_\otimes_{R_0}R_1$ one has the right exact sequence \begin{equation} R_1\xrightarrow{\_\times uv}R_1\to\mathbb{K}[u,v]_{\displaystyle/(u^2v^2)}\to0, \end{equation} which is not exact on the left.
Using tag 00HD (3), it suffices to find an ideal $I$ of $R=k[x,y,z]/(y^2-xz)$ s.t. the map $I\otimes_RM\to M$ is not injective where $M=k[u,v]$.
A potential candidate is that $x\otimes v-y\otimes u \mapsto u^2v-uv\cdot u=0$. Hence let $I$ be the ideal generated by $x$ and $y$. It remains to show that $x\otimes v-y\otimes u \neq0$ in $I\otimes_R M$.
The strategy is to find triples $(N,\phi:I\to N,\varphi:M\to N)$ where $N$ is an $R$-module and $\phi,\varphi$ are $R$-linear maps s.t. the induced map $(\phi,\varphi):I\otimes_R M\to N$ is non-zero at $x\otimes v-y\otimes u$.
Fix a triple $(N,\phi,\varphi)$, we want that $\alpha:=\phi(x)\varphi(v)-\phi(y)\varphi(u)\neq0$. Since $I$ is generated by $x$ and $y$, there exists an exact sequence of $R$-modules: $$0\to \ker q\to x_0R\oplus y_0R\stackrel{q}{\to}I\to 0$$ where $q(x_0)=x$ and $q(y_0)=y$. Then $\phi:I\cong\frac{x_0R\oplus y_0R}{\ker q}\to N$ can be viewed as an $R$-map $\psi:x_0R\oplus y_0R\to N$ s.t. $\psi(\ker q)=0$.
Claim: Every element of $\ker q$ is the image of $(x_0z-y_0y)f(x,z)+(x_0y-y_0x)g(x,z)$ in $x_0R\oplus y_0R$ for some $f,g$. The reverse is also true.
Proof: The reverse is trivially true. Clearly $R$ is a free module over $k[x,z]$ with basis $\{1,y\}$. So every element of $x_0R\oplus y_0R$ is of the form $t=x_0(f_1+yf_2)+y_0(g_1+yg_2)$ where $f_i,g_i\in k[x,z]$. Assume $t\in \ker q$, then $q(t)=0=x(f_1+zg_2)+y(xf_2+g_1)$. So $f_1=-zg_2$ and $g_1=-xf_2$.
Then $t=x_0(-zg_2+yf_2)+y_0(-xf_2+yg_2)=(x_0y-xy_0)f_2+(y_0y-x_0z)g_2$.$\square$
So $\psi(\ker q)=0 \Leftrightarrow y\cdot\psi(x_0)-x\cdot \psi(y_0)=0=z\cdot \psi(x_0)-y\cdot \psi(y_0).$
For any $n\in N$, $\varphi(u^2)=x\cdot \varphi(1),\varphi(uv)=y\cdot \varphi(1),\varphi(v^2)=z\cdot \varphi(1)$. So we may suppose $N$ is a commutative ring and $\varphi$ is a ring map as well.
Now $\psi(\ker q)=0 \Leftrightarrow \varphi(u)[\varphi(v)\psi(x_0)-\varphi(u)\psi(y_0)]=0=\varphi(v)[\varphi(v)\psi(x_0)-\varphi(u)\psi(y_0)]$. Since $\phi(x)=\psi(x_0)$ and $\phi(y)=\psi(y_0)$. It boils down to that we want $\varphi(u)\alpha=\varphi(v)\alpha=0$ and $\alpha=\phi(x)\varphi(v)-\phi(y)\varphi(u)\neq0$. And it can be done by simply set $N=\frac{k[u,v]}{(u^2-uv,v^2-uv)}$ with $\varphi:M\to N,u\mapsto u,v\mapsto v$ and $\psi:x_0 R\oplus y_0R\to N,x_0\mapsto 1,y_0\mapsto 1$. The result follows.