Cohen-Macaulay is polynomial-closed

This article gives the statement, and possibly proof, of a commutative unital ring property satisfying a commutative unital ring metaproperty
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Statement

Property-theoretic statement

The property of commutative unital rings of being Cohen-Macaulay satisfies the metaproperty of commutative unital rings of being polynomial-closed.

Verbal statement

The polynomial ring in one variable over a Cohen-Macaulay ring, is again Cohen-Macaulay.

Definitions used

Cohen-Macaulay ring

Further information: Cohen-Macaulay ring

A Noetherian ring is termed Cohen-Macaulay if for every ideal, the depth equals the codimension.

Facts used

• Cohen-Macaulay is strongly local: A ring is Cohen-Macaulay iff its localization at every maximal ideal is Cohen-Macaulay.
• Krull's principal ideal theorem

Proof

Given: A Cohen-Macaulay ring $R$, and the polynomial ring $R[x]$

To prove: $R[x]$ is a Cohen-Macaulay ring

Reduction to a local case

By the strongly local nature of the Cohen-Macaulay property, it suffices to show that for every maximal ideal $P$ of $R[x]$, the localization $R[x]_{P}$ is a Cohen-Macaulay ring. Let $Q$ be the contraction of $P$ to $R$ (in other words, $Q=P\cap R$). Clearly $Q$ is a prime ideal in $R$ (because primeness is contraction-closed).

A little thought reveals that:

$R[x]_{P}=R_{Q}[x]_{P_{Q}}$

where $R_{Q}$ denotes localization at the prime ideal $Q$, and $P_{Q}$ denotes the ideal generated by $P$ in $R_{Q}[x]$. Thus, by passing to $R_{P}$, we may without loss of generality assume that $R$ is a local ring with maximal ideal $Q$. In particular, we may asume that $R/Q$ is a field.

Note that for this step of the reduction, we're using the fact that since $R$ is Cohen-Macaulay, so is $R_{Q}$.

The depth increases by at least one

Notation as before: $R$ is a local Cohen-Macaulay ring, $Q$ is its unique maximal ideal, and $P$ is an ideal of $R[x]$ that contracts to $Q$.

Now, consider $R[x]/QR[x]$. This is the same as $(R/Q)[x]$, which is a polynomial ring over a field, hence a principal ideal. The image $P/QR[x]$ is thus a principal ideal in this field, and since it is also prime, it must be generated by a monic polynomial, say $f$. Pulling back, we see that $P$ is generated by $Q$ and a monic polynomial (any pullback of $f$).

Now, any regular sequence in $Q$ in $R$ continues to remain a regular sequence inside the ring $R[x]$. Augmenting with $f$ at the end, we get a regular sequence for $P$ in $R$ (we use the fact that by our construction $f$ is not a zero divisor in the quotient). Thus, the depth of $P$ in $R[x]$ is at least 1 more than the depth of $Q$ in $R$.

The codimension increases by at most one

This follows from Krull's principal ideal theorem.

Putting the pieces together

• Since $R$ is Cohen-Macaulay, the depth and codimension of $Q$ are equal
• The codimension of $P$ is at most one more than this, and the depth of $P$ is at least one more than this
• Hence the depth of $P$ is at least equal to the codimension of $P$
• On the other hand, we know that the depth of $P$ is at most equal to the codimension of $P$, for any $P$
• Hence, the depth and codimension of $P$ are equal

References

Textbook references

• Book:Eisenbud, Page 456-457