There are two kinds of primes $\mathfrak q \subset R[T]$. The first possibility is that $\mathfrak q = \mathfrak p[T]$ with $\mathfrak p = R \cap \mathfrak q$. Let us call such a prime "small". The second possibility is that the inclusion $\mathfrak p[T] \subset \mathfrak q$ is strict. Let us call such a prime "big".

Suppose we have a sequence of primes $\mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset\mathfrak q_r$ in $R[T]$. Then we get a sequence of big/small. If the sequence has only one switch from small to big, then of course $r \leq \dim(R) + 1$. The problem comes from a sequence with multiple switches. But thinking about it for a moment we see that it suffices to prove the following.

**Scholium:** If $\mathfrak q_0 \subset \mathfrak q_1$ in $R[T]$ lies over $\mathfrak p_0 \subset \mathfrak p_1$ in $R$ and if $\mathfrak q_0$ is big and $\mathfrak q_1$ is small, then there is a prime strictly in between $\mathfrak p_0$ and $\mathfrak p_1$.

To prove this we argue by contradiction and assume there is no prime strictly in between. Observe that in any case $\mathfrak p_0 \not = \mathfrak p_1$ by our definition of big and small. After replacing $R$ by $(R/\mathfrak p_0)_{\mathfrak p_1}$ we reach the situation where $R$ is a local Noetherian domain of dimension $1$. Then $\mathfrak p_0 = (0)$ and $\mathfrak p_1 = \mathfrak m$ is the maximal ideal.

Translating we have to derive a contradiction from the following: we have a nonzero prime $\mathfrak q \subset \mathfrak m[T]$ with $\mathfrak q \not = \mathfrak m[T]$.

Let $K$ be the fraction field of $R$. Let $\mathfrak q_K \subset K[T]$ be the ideal generated by $\mathfrak q$ in $K[T]$. Then $\mathfrak q = \mathfrak q_K \cap R[T]$.

For every $n \geq 0$ let $R[T]_{\leq n}$ be the polynomials of degree $\leq n$. Let $M_n = \mathfrak q \cap R[T]_{\leq n}$ and $Q_n = R[T]_{\leq n}/M_n$ so that we have a short exact sequence
$$
0 \to M_n \to R[T]_{\leq n} \to Q_n \to 0
$$
Now observe that $Q_n$ is a finite $R$-module, is torsion free, and has rank bounded independently of $n$. Namely, over $K$ we know that $\mathfrak q_K$ is generated by a polynomial of degree $d$ and we see that $Q_n \otimes_R K$ has dimension over $K$ at most $d$.

Pick $a \in \mathfrak m$ nonzero. Then (1) $R/aR$ has finite length $c$, (2) for any finite torsion free module $Q$ of rank $r$ the length of $Q/aQ$ is $rc$, and (3) a module $Q$ with length $Q/aQ$ bounded by $rc$ is generated by $\leq rc$ elements. [Hints for elementary proofs: To prove (1) you show for any $b \in \mathfrak m$ some power of $b$ is in $aR$ otherwise $R/aR$ would have a second prime. To prove (2) you choose $R^{\oplus r} \subset Q$ and you use the snake lemma for multiplication by a on the corresponding ses. To prove (3) use Nakayama and that a finite length module is generated by at most its length number of elements.]

Take $n > dc$ where $d$ is the upper bound for the ranks of all $Q_n$ found above. Then we conclude that there exists an element in $M_n$ which is not in $\mathfrak m(R[T]_{\leq n})$ because we have seen above that $Q_n$ can be generated by $\leq dc$ elements. Small standard argument omitted.

This is the desired contradiction because we assumed $\mathfrak q \subset \mathfrak m[T]$. QED

This answer shows that with usual commutative algebra there is a very short proof. Enjoy!

Examples of Commutative Rings, Example 27: Let $k$ be a field and let $R = k(y)[[x]] \times_{k(y)} k$ the ring of those power series in $x$ with coefficients in $k(y)$ resp. $k$ for the constant term. Then $\dim(R)=1$ and $\dim(R[T])=3$. I've also read that actually every number between $\dim(R)+1$ and $2 \dim(R)+1$ may appear as $\dim(R[T])$. $\endgroup$ – Martin Brandenburg Aug 4 '14 at 14:13On the dimension theory of rings (II),Pacific J. Math. 4 (1954), 603-614. $\endgroup$ – Fred Rohrer Aug 25 '14 at 21:2113more comments