Finite ring extensions doesn't preserve non-zero-divisors Can you give an example of a finite ring extension $A \hookrightarrow B$ of noetherian rings $A,B$ which maps a non-zero-divisor to a zero-divisor.
Background: In this case $A \hookrightarrow B$ does not induce a homomorphism $Q(A) \to Q(B)$ of total quotient rings.
 A: Yes: take $\mathbb{Z} \to \mathbb{Z}[x]/(2x, x^2)$. Then 2 becomes a zero divisor in the latter. They are noetherian, and it is finite since the latter ring is generated by $1,x$ over $\mathbb{Z}$.
How did I find this? If you try a proof of the fact that non zero divisors are sent to non zero divisors, you stuck at some point. Let $y$ be a non zero divisor, and suppose there exist $x$ in the extended ring such that $\phi(y)x =0$. If x was in the image of phi, then yes, by injectivity you could lift to an element in the original ring. If $y$ was invertible, then the image would be invertible; in particular it would be non zero divisor.
But what about if $x$ is not in the image and $y$ is not invertible? It can happen. Indeed, for any ring $A$, consider a non zero divisor $y$ which is not invertible. Let's add an element $x$ that makes the work: it is not in the original ring, and annihilates $y$. Formally, set $B=A[x]/(yx,x^2)$. The second $x^2$ is put so to get a finite ring extension. And you are done!
