Question

If $A$ is a Cohen Macaulay local ring, and $B$ is a quotient ring of $A$ and $B$ is also Cohen Macaulay, Then is $B$ always a quotient by a regular sequence of $A$?

Answer 1

Here is an even better example:

$A=(k[x,y,z]/(xy-z^2))_{\mathfrak m}$ with ${\mathfrak m}=(x,y,z)$, $B=k[x]_{(x)}$. Here $A$ is Gorenstein and $B$ is regular. The map is given by $x\mapsto x$ and $y,z\mapsto 0$.

Geometrically, $B$ corresponds to a line going through the vertex of a quadric cone corresponding to $A$. It is not a quotient by a regular sequence, because that would have to be just a regular element for dimension reasons, but the line is not a Cartier divisor, so it cannot be defined by a single equation. To see that it is not a Cartier divisor, for example one can compute its self-intersection number which is $\frac 12$.

Answer 2

A more general series of examples (when this fails) is given by choosing $B$ to be Cohen-Macaulay, but not Gorenstein and $A$ to be Gorenstein.

An explicit example is when $B=(k[x,y,z]/(xy,yz,zx))_{\mathfrak m}$ is the affine coordinate ring of three lines in $\mathbb A^3$ meeting in one point, but not contained in a plane and $A=k[x,y,z]_{\mathfrak m}$ where ${\mathfrak m}=(x,y,z)$.

Answer 3

As another example, say $\mathfrak{m}$ is the maximal ideal of $A$. Then $B:=A/\mathfrak{m}$ is Cohen-Macaulay (field), but $\mathfrak{m}$ is not generated by a regular sequence, unless $A$ is regular.

Answer 4

The answer is no. Let $k$ be a field. Let $A=k[x]/x^2$, and let $B=k$.

$A$ is Artinian, so it is Cohen Macaulay. $B$ is clearly Cohen Macaulay. But the only non-trivial ideal in $A$ is $(x)$ which is not generated by a regular sequence, because every element of it is a zero-divisor.