Theorem 14.34. Krull’s Principal Ideal Theorem.
Let \(R\) be a noetherian ring, and \(f\in R\text{.}\) Then, every minimal prime of \((f)\) has height at most one.
Section 14.3 Height and Number of Generators
“You are not judged by the height you have risen, but from the deoth you have climbed.”―Fredrick Douglass
Proof.
Suppose the theorem is false, so that there is some ring \(R\text{,}\) a prime \(P\text{,}\) and an element \(f\) such that \(P\) is minimal over \((f)\) and \(\ht(P)>1\text{.}\) If we localize at \(P\) and then mod out by an appropriate minimal prime, we obtain a noetherian local domain \((R,\fm)\) of dimension at least two in which \(\fm\) is the unique minimal prime of \((f)\text{.}\) Let’s work over that noetherian local domain \((R, \fm)\text{.}\) Note that \(\overline{R} = R/(f)\) is zero-dimensional, since \(\fm\) is the only minimal prime over \((f)\text{.}\) Back in \(R\text{,}\) let \(Q\) be a prime strictly in between \((0)\) and \(\fm\text{,}\) and notice that we necessarily have \(f \notin Q\text{.}\)
Consider the symbolic powers \(Q^{(n)}\) of \(Q\text{.}\) We will show that these stabilize in \(R\text{.}\) Since \(\overline{R} = R/(f)\) is Artinian, the descending chain of ideals
\begin{equation*}
Q \overline{R} \supseteq Q^{(2)}\overline{R} \supseteq Q^{(3)}\overline{R} \supseteq \cdots
\end{equation*}
stabilizes. We then have some \(n\) such that \(Q^{(n)} \overline{R} = Q^{(m)} \overline{R}\) for all \(m \geqslant n\text{,}\) and in particular, \(Q^{(n)} \overline{R} = Q^{(n+1)} \overline{R}\text{.}\) Pulling back to \(R\text{,}\) we get \(Q^{(n)} \subseteq Q^{(n+1)} + (f)\text{.}\) Then any element \(a \in Q^{(n)}\) can be written as \(a = b + fr\text{,}\) where \(b \in Q^{(n+1)} \subseteq Q^{(n)}\) and \(r \in R\text{.}\) Notice that this implies that \(fr \in Q^{(n)}\text{.}\) Since \(f\notin Q\text{,}\) we must have \(r\in Q^{(n)}\text{.}\) This yields \(Q^{(n)} = \fq^{(n+1)} + f \fq^{(n)}\text{.}\) Thus, \(Q^{(n)} / Q^{(n+1)} = f(Q^{(n)} / \fq^{(n+1)})\text{,}\) so \(Q^{(n)} / Q^{(n+1)} = \fm(Q^{(n)} / Q^{(n+1)})\text{.}\) By , \(Q^{(n)} = Q^{(n+1)}\) in \(R\text{.}\) Similarly, we obtain \(Q^{(n)} = Q^{(m)}\) for all \(m \geqslant n\text{.}\)
Now, if \(a \in Q\) is nonzero, we have \(a^n\in Q^n \subseteq Q^{(n)} = Q^{(m)}\) for all \(m\text{,}\) so
\begin{equation*}
\bigcap_{m \geqslant 1} Q^{(m)} = \bigcap_{m \geqslant n} Q^{(m)} = Q^{(n)}.
\end{equation*}
Notice that \(Q^n \neq 0\) because \(R\) is a domain, and so \(Q^{(n)} \supseteq Q^n\) is also nonzero. So
\begin{equation*}
\bigcap_{m \geqslant 1} Q^{(m)} = Q^{(n)} \neq 0.
\end{equation*}
On the other hand, \(Q^{(m)} = Q^m R_Q \cap R\) for all \(m\text{,}\) and
\begin{equation*}
\bigcap_{m \geqslant 1} Q^{(m)} R_Q \subseteq \bigcap_{m \geqslant 1} Q^m R_{Q} = \bigcap_{m \geqslant 1} (Q R_{Q})^m = 0
\end{equation*}
by . Since \(R\) is a domain, the contraction of \((0)\) in \(R_Q\) back in \(R\) is \((0)\text{.}\) This is the contradiction we seek. So no such \(Q\) exists, so that \(R\) has dimension \(1\text{,}\) and in the original ring, all the minimal primes over \(f\) must have height at most \(1\text{.}\)
Remark 14.35.
Note that this is stronger than the statement that the height of \((f)\) is at most one: that would only mean that some minimal prime of \((f)\) has height at most one.
Noetherianity is necessary, as the next example shows.
Example 14.36. Noetherianity is Necessary.
\(R = k[x,xy,xy^2,\ldots] \subseteq k[x,y]\text{.}\) Note that \((x)\) is not prime: for \(a>0\text{,}\) \(xy^a \notin (x)\text{,}\) since \(y^a\notin R\text{,}\) but \((xy^a)^2 = x \cdot x y^{2a} \in (x)\text{.}\) Thus, \(\fm = (x,xy,xy^2,\dots) \subseteq \sqrt{(x)}\text{,}\) and since \(\fm\) is a maximal ideal, we have equality, so \(\Min{(x)} = \{ \fm \}\text{.}\) However, the ideal \(P = (xy,xy^2, xy^3,\dots) = (y) k[x,y] \cap R\) is prime, and the chain \((0) \subsetneq P \subsetneq \fm\) shows that \(\ht(\fm)>1\text{.}\)
Lemma 14.37. Proper Containment of Ideals in Noetherian Ring.
Let \(R\) be noetherian, \(P \subsetneq Q \subsetneq \fa\) be primes, and \(f \in \fa\text{.}\) Then there is some \(Q'\) with \(P \subsetneq Q'\subsetneq \fa\) and \(f \in Q'\text{.}\)
Proof.
If \(f \in P\text{,}\) there is nothing to prove, since we can simply take \(Q' = Q\text{.}\) Suppose \(f \notin P\text{.}\) After we quotient out by \(P\) and localize at \(\fa\text{,}\) we may assume that \(\fa\) is the unique maximal ideal and that our ring is a domain. Thus all we need is to find a nonzero prime \(Q'\) which is not maximal. Note that by assumption \(\ht(\fa) \geqslant 2\text{.}\) Our assumption implies that \(f \neq 0\text{,}\) and then by the principal ideal theorem
[provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Theorem - Krull’s Principal Ideal Theorem|Krull’s Principal Ideal Theorem]], minimal primes of \((f)\) have height one, hence are not maximal nor \(0\text{.}\) We can take \(Q'\) to be one of the minimal primes of \(f\text{.}\)
Theorem 14.38. Krull’s Height Theorem.
Let \(R\) be a noetherian ring. If \(I\) is an ideal generated by \(n\) elements, then every minimal prime of \(I\) has height at most \(n\text{.}\)
Proof.
By induction on \(n\text{.}\) The case \(n=1\) is .
Let \(I=(f_1,\dots,f_n)\) be an ideal, \(P\) a minimal prime of \(I\text{,}\) and \(P_0 \subsetneq P_1 \subsetneq \cdots \subsetneq P_{h} = P\) be a saturated chain of length \(h\) ending at \(P\text{.}\) If \(f_1 \in P_1\text{,}\) then we can apply the induction hypothesis to the ring \(\overline{R} = R/((f_1) + P_0)\) and the ideal \((f_2,\dots,f_n)\overline{R}\text{.}\) Then by induction hypothesis, the chain \(P_1 \overline{R} \subsetneq \cdots \subsetneq P_{h} \overline{R}\) has length at most \(n-1\text{,}\) so \(h-1 \leqslant n-1\) and \(P\) has height at most \(n.\)
If \(f_1 \notin P_1\text{,}\) we use the previous lemma to replace our given chain with a chain of the same length but such that \(f_1 \in P_1\text{.}\) To do this, note that \(f_1 \in P_i\) for some \(i\text{;}\) after all, \(f_1 \in I \subseteq P\text{.}\) So in the given chain, suppose that \(f_1 \in P_{i+1}\) but \(f_1 \notin P_i\text{.}\) If \(i>0\text{,}\) apply the previous lemma with \(\fa = P_{i+1}\text{,}\) \(Q = P_i\text{,}\) and \(P = P_{i-1}\) to find \(Q_i\) such that \(f_1 \in Q_i\text{.}\) Replace the chain with
\begin{equation*}
P_0 \subsetneq P_1 \subsetneq \cdots \subsetneq P_{i-1} \subsetneq Q_i \subsetneq P_i \subsetneq \cdots \subsetneq P_{h} = P.
\end{equation*}
Repeat until \(f_1 \in P_1\text{.}\)
Example 14.39. .
If \(k\) is a field and \(R = k[x_1, \ldots, xn]\text{,}\) the ideal \((x_1,x_2,\dots,x_n)\) generated by \(n\) variables has height~\(n\text{.}\) There are many other ideals that attain the bound given by . For example, \((u^3-xyz,x^2+2xz-6y^5,vx+7vy)\in k [u,v,w,x,y,z]\text{.}\) An ideal of height \(n\) generated by \(n\) elements is called a complete intersection.
Example 14.40. .
The ideal \((xy,xz)\) in \(k[x,y,z]\) has minimal primes of heights 1 and 2.
Example 14.41. .
In \(R=k[x,y]/(x^2,xy)\text{,}\) the ideal generated by zero elements (the zero ideal) has an associated prime of height two, namely \((x,y)\text{.}\) The same phenomenon can happen even in a nice polynomial ring. For example, consider the ideal \(I=(x^3, y^3, x^2 u + xy v + y^2 w) \subseteq R= k[u,v,w,x,y]\text{.}\) Note that \((u,v,w,x,y) = (I : x^2 y^2)\text{,}\) so \(I\) has an associated prime of height 5.
Example 14.42. .
Let \(R=k[x,xy,xy^2,\dots] \subseteq k[x,y]\text{.}\) For all \(a \geqslant 1\text{,}\) \(xy^a \notin (x)\text{,}\) since \(y^a\notin R\text{,}\) but \((xy^a)^2 = x \cdot x y^{2a}\in (x)\text{.}\) Then \((x)\) is not prime in \(R\text{,}\) and moreover \(\fm=(x,xy,xy^2,\dots) \subseteq \sqrt{(x)}\text{.}\) Since \(\fm\) is a maximal ideal, we have equality, so \(\Min{(x)}=\{\fm\}\text{.}\) However, \(\p=(xy,xy^2, xy^3,\dots) = (y) k[x,y] \cap R\) is prime, and the chain \((0) \subsetneq \p \subsetneq \fm\) shows that \(\ht(\fm)>1\text{.}\)
Corollary 14.43. Ideals in Noeth Rings have Finite Height.
If \(R\) is a noetherian ring, then any ideal has finite height. If \((R,\fm,k)\) is also local, then \(\dim(R) \leqslant \dim_k(\fm/\fm^2) <\infty\text{.}\)
Proof.
If \(R\) is noetherian, then every ideal is finitely generated, by , and thus by every ideal has finite height.
Now suppose that \((R,\fm,k)\) is a noetherian local ring. By , \(\fm\) is generated by \(\dim_k(\fm/\fm^2)\) elements, so
\begin{equation*}
\dim(R) = \ht(m) \leqslant \dim_k(\fm/\fm^2).
\end{equation*}
Definition 14.44. Embedding Dimension.
The embedding dimension of a local ring \((R, \fm)\) is the minimal number of generators of \(\fm\text{,}\) \(\mu(\fm)\text{.}\) We write \(\textrm{embdim}(R) := \mu(\fm)\) for the embedding dimension of \(R\text{.}\)
Definition 14.45. Regular Local Ring.
A noetherian local ring \((R, \fm)\) is regular if \(\dim(R) = \edim(R)\text{.}\)
Remark 14.46.
Power series rings \(k [[ x_1, \ldots, x_d ]]\) are regular local rings. In general, a ring is regular if all its localizations are regular local rings. In order for this definition to make sense, we need to first make sure that regularity localizes, meaning that if \((R, \fm)\) is a regular local ring, then \(R_P\) is also regular for all primes \(P\text{.}\) But to do that, we need some homological algebra. However (spoiler alert!), things do work out alright, and as you might expect, polynomial rings over fields are also regular.
