Since I wrote a few posts on Krull dimension and defined Krull dimension a few times, I thought it might be time to have a post dedicated to Krull dimension. Along the way, I have included some exercises for which there are solutions that you can click to reveal.

N.B. The notation $X\subset Y$ always means that $X$ is a *proper* subset of $Y$. That is, $X\subset Y$ means for each $x\in X$, we also have $x\in Y$ *and* there exists a $y\in Y$ such that $y\not\in X$.

## The definition

Let $R$ be a commutative ring. A prime ideal of $R$ is a proper ideal $P\subset R$ such that $xy\in P$ implies either $x\in P$ or $y\in P$. We really mean proper: even though $R$ trivially satisfies this property, it is not considered to be prime.

**Exercise.**An proper ideal $M\subset R$ is called maximal if there does not exist a proper ideal $N\subset R$ with $M\subset N$. Prove that a maximal ideal is a prime ideal.

## Click for the solution

Let $x\in R$ and $x\not\in M$. Then the ideal generated by $M$ and $x$ is the whole ring $R$ because $M$ is maximal. Therefore, there exists an $r\in R$ such that $m + rx = 1$. Now suppose $xy\in M$.

Assume for the sake of contradiction that neither $x$ nor $y$ is in $M$. Then there exists $r,s\in R$ and $m_1,m_2\in M$ such that $m_1 + rx = 1$ and $m_2 + sy = 1$. Taking the product we get that

$$1 = (m_1 + rx)(m_2 + sy) = m_1m_2 + m_1sy + rxm_2 + rsxy.$$ Each of these four elements is in $M$, showing that $1\in M$, a contradiction. Hence, either $x\in M$ or $y\in M$.

A chain of prime ideals $P_0\subset P_1\subset \cdots P_{n}$ of $n+1$ primes is said to have length $n$ (think of the inclusion symbols as links in a chain). If $R$ is a commutative ring, we define the **Krull dimension** of $R$ to be the supremum over the lengths of all chains of prime ideals in $R$. We denote this quantity by $\dim(R)$.

**Exercise.**If $F$ is a field, prove that the Krull dimension of the polynomial ring $F[x]$ is one.

## Click for the solution

We use the fact that every nonzero element in $F[x]$ can be factored uniquely as a product of irreducibles up to order and multiplication by a unit and that $F[x]$ is a principal ideal domain.

We see that thus that every ideal $I$ is of the form $I = (f)$ for some polynomial $f\in F[x]$. If $f$ is not irreducible, then $I$ cannot be prime, for then $f = gh$ where $\deg(g) < \deg(f)$ and $\deg(h) < \deg(f)$. Since all multiple of $f$ have degree greater than or equal to $f$, we see that $gh\in (f)$ but $g\not\in (f)$ and $h\not\in (f)$. Thus $f$ cannot be prime.

On the other hand, if $f$ is irreducible and $gh\in (f)$, then we see by unique factorization that either $g$ or $h$ must have $f$ as one of its factors, so $(f)$ is prime. Therefore, an ideal $I = (f)$ in $F[x]$ is prime if and only if $f$ is irreducible.

If $(f) \subseteq (g)$ are ideals with $f$ and $g$ irreducible, then $f = gh$ for some $h$, which must be a unit by the irreducibility of $f$. Therefore, $(f)$ is a maximal ideal and the only prime ideal properly included in it is $(0)$. So $(0)\subset (f)$ is a maximal chain of primes.

## Basic examples

Here are some basic examples

- If $F$ is a field then $\dim(F) = 0$
- If $F$ is a field then $\dim(F[x_1,\dots,x_n]) = n$
- $\dim(\Z) = 1$
- If $R$ is Noetherian then $\dim(R[x_1,\dots,x_n]) = \dim(R) + n$ (note, the Noetherian hypothesis here is crucial).
- The ring $\R[x,y]/(x – y^2)$ has Krull dimension one.

## Bounds in the non Noetherian case

We already remarked that if $R$ is a Noetherian ring, then the Krull dimension of $R[x]$ is $\dim(R) + 1$. This is a little boring to tell you the truth. It’s far more interesting to look at the cases when $R$ is *not* Noetherian. In this case we can only expect bounds:

**Theorem.**(Seidenberg [1]) If $R$ is an $n$-dimensional ring then $R[x]$ is at least $n+1$ dimensional and at most $(2n+1)$-dimensional.

There is a more general criteria for the dimension of polynomial extensions to behave like the Noetherian case in the one-dimensional case:

**Theorem.**(Seidenberg [1]) Let $R$ be a 1-dimensional ring. The ring $R[x]$ is 2-dimensionl if and only if every nontrivial quotient of the integral closure of $R$ is a valuation ring.

Recall that a valuation ring is a ring $R$ such that for any two $r,s\in R$, either $r$ divides $s$ or $s$ divides $r$. For example, although this theorem is overkill for the integers $\Z$, we can see it applies because the integral closure of $\Z$ is $\Z$ itself, and each quotient of $\Z$ is a product of rings of the form $\Z/p^k$. Any two nonunits in such a ring are just units times a power of $p$, so it is a valuation ring.

There are also similar oddities in power series rings. Jimmy T. Arnold discovered that there is an interesting property that is weaker than Noetherian that in some sense governs the behaviour of Krull dimension on the passage to power series rings. If $R$ is a commutative ring, we say than an ideal $I\subseteq R$ is of *strong finite type* if there exists a finitely generated ideal $J\subseteq I$ and a fixed natural number $k$ such that $x^k\in J$ for each $x\in I$. In other words, the power map $y\mapsto y^k$ sends $I$ into a finitely-generated ideal $J$. We say that $R$ is of *strong finite type* if each ideal of $R$ is of strong finite type. Clearly, every Noetherian ring is of strong finite type. Then we have:

**Theorem.**(Arnold [2]) Let $R$ be a commutative ring. If $R$ is not of strong finite type then $R[[x]]$ has

*infinite*Krull dimension.

**Theorem.**(Arnold [2]) Let $R$ be a 0-dimensional commutative ring. Then $R[[x]]$ is infinite dimensional if and only if $\dim R[[x]] \not= 1$.

Therefore, in the case that $R$ is of strong finite type, it is guaranteed that $\dim R[[x]] = 1$ if $R$ is zero dimensional. Arnold gives a class of examples in his paper of zero dimensional rings $R$ whose power series ring $R[[x]]$ is infinite-dimensional. One such ring is the ring $R = \Q[x_0,x_1,x_2,\dots]/(x_0^2,x_1^2,x_2^2,\dots)$.

Incidentally, in general, we can have finite-dimensional rings $R$ that are of strong finite type but such that $R[[x]]$ is still infinite dimensional. An example of such a ring is given by Jim Coykendall in [4].

Arnold in another paper also consider discrete valuation rings. These are principal ideal domains with exactly one nonzero maximal ideal. In fact, a prominent example of a discrete valuation ring is a power series ring $R[[t]]$ when $R$ is a *field*.

**Exercise.**Prove that if $R$ is a field then $R[[t]]$ is a discrete valuation ring.

## Click for the solution

Any power series of the form $a_0 + a_1t + a_2t^2 + \cdots$ with $a_0\not=0$ is invertible in $R[[t]]$. Therefore, a power series of the form $a_0t^k + a_1t^{k+1} + \cdots$ is a unit multiple of $t^k$. So, if $I$ is an ideal of $R$ generated by an arbitrary nonempty set $S$ of power series, then $I$ is generated by $t^k$ where $k$ is the least integer such that there exists an $f\in S$ which is a unit multiple of $t^k$.

Therefore, we have shown that every proper ideal in $R[[t]]$ is principal and of the form $(t^k)$ for $k=1,2,3,\dots$. These form a chain of ideals, and hence there is exactly one maximal ideal: $(t)$.

**Theorem.**(Arnold [3]) If $R$ is a discrete valuation ring with $\dim(R) = m$ then $\dim R[[x_1,\dots,x_n]] = mn + 1$.

The proof is elementary but tricky.

## References

- Seidenberg, A. A note on the dimension theory of rings. Pacific J. Math. 3 (1953), 505-512
- Arnold, Jimmy T. Krull dimension in power series rings. Trans. Amer. Math. Soc. 177 (1973), 299-304.
- Arnold, Jimmy T. Power series rings over discrete valuation rings. Pacific J. Math. 93 (1981), no. 1, 31-33.
- Coykendall, Jim. The SFT property does not imply finite dimension for power series rings. J. Algebra 256 (2002), no. 1, 85-96.