Cohen-Macaulay is polynomial-closed

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.

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

Textbook references

 * , Page 456-457