TITLE: Flat families of schemes (Exercize of Vakil's notes)
QUESTION [5 upvotes]: I'm studying Vakil's notes "Fundation of algebraic geometry" (http://math.stanford.edu/~vakil/216blog/FOAGmar2313public.pdf) and I have a problem understanding exercize 24.4.O (page 638). In fact if we consider the coordinate ring
$$ A= k[x,y,m,t]/(m^2-t(x+1),ty-mx,y^2-x^3-x^2) $$
professor Vakil says that $t$ isn't a zero divisor on $A$, but if we consider in the quotient the relations $ty=mx$ and $m^2x=tx(x+1)$ we obtain that $t(my-x(x+1))=0$. What is wrong? Thank you!

REPLY [4 votes]: Let $B= k[x,y]/(y^2-x^3-x^2)$
There is a canonical $k$-algebra morphism $\phi: A\to B$ sending $m,t$ to $0$ and $x,y$ to $x,y$.
That morphism sends  $my-x(x+1)\in A$ to  $-x(x+1)\in B$, which is clearly $\neq0$.
This proves that:  
1) $my-x(x+1)\neq 0 $ in $A$   since its image in $B$ under $\phi$ is $-x(x+1)\neq0$
2) $t$ is a zero divisor in $A$ since $t(my-x(x+1))=0$ although $my-x(x+1)\neq 0\in A$.
3) Professor Vakil may have made a mistake.