An example of a (necessarily non-Noetherian) ring $R$ such that $\dim R[T]>\dim R+1$ What is an example of a non-Noetherian ring $R$ such that the Krull dimension of $R[T]$ is greater than dim$R+1$? 
 A: For a commutative ring $R$ of finite Krull dimension $d$, we have $1+d\leq \dim(R[X])\leq 2d+1$. In fact, for every $n$ and $m$ with $1+n\leq m\leq 2n+1$ there is an integral domain $D$ such that $\dim(D)=n$ and $\dim(D[X])=m$. A proof of this can be found in the paper "On the dimension theory of rings (II)" by A. Seidenberg.
Here is an example of W. Krull of a ring $R$ with $\dim(R)=1$ and $\dim(R[X])=3$. $R$ consists of the rational functions (over a field $K$) $r(x,y)$ which, when written in lowest terms, have denominators not divisible by $x$, and such that $r(0,y) \in K$. Every element of $R$ is of the form $$r(x,y)=\frac {xp(x,y)+kf(y)} {xq(x,y)+f(y)}$$ where $p$,$q \in K[X,Y]$, $f \in K[Y]\setminus\{0\}$ and $k \in K$. I claim that the only nonzero prime ideal of $R$ consists of the rational functions whose numerator is of the form $xg$ with $g \in K[X,Y]$. It is straightforward to prove that $P$ is an ideal. To show that $P$ is prime, let $r \in R$, $r \notin P$. Then $r^{-1} \in R$ and so $R-P$ is the set of units of $R$. This shows $P$ is prime (and, in fact, maximal).
To show that $P$ is the only prime ideal of $R$, let $s$ be an arbitrary element of $P$. $$s=\frac {x^tp_1(x,y)} {xq_1(x,y)+f_1(y)}$$ where $x$ does not divide $p_1(x,y)$. Let $r \in P$, $$r=\frac {xp_2(x,y)} {xq_2(x,y)+f_2(y)}.$$ Then $$r^{t+1}=x\left(\frac {p_2(x,y)} {xq_2(x,y)+f_2(y)}\right)^{t+1}\cdot\frac {xq_1(x,y)+f_1(y)} {p_1(x,y)}\cdot s=a\cdot s,$$ where the denominator of $a$ is not divisible by $x$ and $a(0,y)=0$. Thus $a \in R$ and $r^{t+1} \in (s)$.
Suppose $q$ is a prime ideal contained in $P$. Let $p \in P$ and $u \neq 0 \in q$. Then $p^n\in (u) \subseteq q$ for some $n>0$. Since $q$ is prime, $p \in q$. Thus $q=P$ and $P$ is the only nonzero proper prime ideal of $R$. It follows that $R$ is $1$-dimensional.
The polynomials which vanish for $Z=y$ form a prime ideal $Q$ of $R[Z]$, since $R[Z]$ is an integral domain. $x$ and $xy$ are elements of $R$, and $g(Z)=xZ-xy \in Q$. Thus $Q$ is nonzero. I claim that $Q \subseteq P[Z]$. Let $h(Z)=\sum r_iZ^i\in Q$. Then letting $x=0$ and $Z=y$ in $h$, we have $\sum r_i(0,y)y^i=0$ and $r_i(0,y) \in K$. Since $y$ is an indeterminate over $K$, $r_i(0,y)=0$ for all $i$.  Thus $r_i \in P$ for all $i$ and so $Q \subseteq P[Z]$. Since $f(Z)=xZ \in P[Z]$ and $f(y) \neq 0$, $Q \subset P[Z]$. Thus $R[Z]$ is $3$-dimensional.
