Is the ring $A[x_1, \ldots, x_n]$ Cohen-Macaulay? Does the dimension formula hold?

Question

Let $A[x_1, \ldots, x_n]$ be a polynomial ring over a Noetherian, commutative ring, $A$. Is the polynomial ring Cohen-Macaulay?

If not, does it follow the dimension formula, $ \mathrm{dim} (A[x_1, \ldots, x_n]) = \mathrm{dim} (A[x_1, \ldots, x_n]/\mathfrak{p}) + \mathrm{ht} (\mathfrak{p} ), $ for a prime ideal, $\mathfrak{p}$ in $A[x_1, \ldots, x_n]$.

It depends on A, of course: take, for example, $n=0$. – Mariano Suárez-Álvarez Jul 22 '15 at 05:11 — Mariano Suárez-Álvarez, Jul 22 '15 at 05:11
If A satisfies the dimension formula is assumed, then? – Zoey Jul 22 '15 at 05:40 — Zoey, Jul 22 '15 at 05:40

score 3 · Accepted Answer · answered Jul 22 '15 at 06:16

3

It's well known that $A[x_1, \ldots, x_n]$ is CM iff $A$ is CM.
Let $A=\mathbb Z_{(2)}$. Then $\dim\mathbb Z_{(2)}[X]=2$, $\dim\mathbb Z_{(2)}[X]/(2X-1)=0$ and $\operatorname{ht}(2X-1)=1$.

answered Jul 22 '15 at 06:16

user26857

53,190

related question: http://math.stackexchange.com/questions/1369700/when-will-ax-1-ldots-x-n-satisfy-the-dimension-formula – Zoey Jul 22 '15 at 06:58

Is the ring $A[x_1, \ldots, x_n]$ Cohen-Macaulay? Does the dimension formula hold?

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