4 deleted 9 characters in body

OOPS: As Sándor points out, I missed the assumption that the ring is local, in the following , and will attempt a separate answer.example of the wrong thing:

"Inside 3-space, glue together a plane $y=0$ transversely with a parabola $z=0, x=y^2$ and a line $z=1, x=0$ meeting it in a separate point. This is reduced, and I'm pretty sure its depth is $1$. Because of the plane, its dimension is $2$.

Now cut it with $x=0$, which cuts the plane to a line and the parabola to a double point leaving the line alone. Hence $x$ is a zero divisor in $k[x,y,z]/(\langle y\rangle \cap \langle z,x-y^2\rangle \cap \langle z-1,x\rangle)$, and cutting with it drops the dimension from $2$ to $1$.

The resulting space is generically reduced, but not reduced, so I'm pretty sure its depth is $0$.

3 added 37 characters in body

OOPS: As Sándor points out, I missed the assumption that the ring is local, in the following:, and will attempt a separate answer.

"Inside 3-space, glue together a plane $y=0$ transversely with a parabola $z=0, x=y^2$ and a line $z=1, x=0$ meeting it in a separate point. This is reduced, and I'm pretty sure its depth is $1$. Because of the plane, its dimension is $2$.

Now cut it with $x=0$, which cuts the plane to a line and the parabola to a double point leaving the line alone. Hence $x$ is a zero divisor in $k[x,y,z]/(\langle y\rangle \cap \langle z,x-y^2\rangle \cap \langle z-1,x\rangle)$, and cutting with it drops the dimension from $2$ to $1$.

The resulting space is generically reduced, but not reduced, so I'm pretty sure its depth is $0$.

2 added 95 characters in body

OOPS: As Sándor points out, I missed the assumption that the ring is local, in the following:

Inside 3-space, glue together a plane $y=0$ transversely with a parabola $z=0, x=y^2$ and a line $z=1, x=0$ meeting it in a separate point. This is reduced, and I'm pretty sure its depth is $1$. Because of the plane, its dimension is $2$.

Now cut it with $x=0$, which cuts the plane to a line and the parabola to a double point leaving the line alone. Hence $x$ is a zero divisor in $k[x,y,z]/(\langle y\rangle \cap \langle z,x-y^2\rangle \cap \langle z-1,x\rangle)$, and cutting with it drops the dimension from $2$ to $1$.

The resulting space is generically reduced, but not reduced, so I'm pretty sure its depth is $0$.

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