Theorem 14.47. Existence of SOPs.
Let \((R,\fm)\) be a noetherian local ring of dimension \(d\text{.}\) There there exist \(x_1,\dots,x_d\in \fm\) such that \(\fm= \sqrt{(x_1,\dots,x_d)}\text{.}\)
Section 14.4 Systems of Parameters
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Proof.
Let \(d := \dim(R)\text{.}\) If \(d=0\text{,}\) then \(\fm=\sqrt{(0)}\text{,}\) so the statement holds.
In general, we will show that we can choose \(x_1,\dots, x_i\in \fm\) inductively such that every minimal prime of \(J_i= (x_1,\dots,x_i)\) has height \(i\text{.}\) The case \(i=0\) is clear from the comment above. Say that we have chosen the first \(i\) elements. If \(i<d\text{,}\) note that \(\fm\) is not a minimal prime of \(J_i\text{,}\) as this would contradict. Note that \(R\) is noetherian and thus \(J_i\) has finitely many minimal primes, by . Thus, by , we can choose \(x_{i+1} \in \fm\) not in any minimal prime of \(J_i\text{.}\) Then by every minimal prime of \(J_{i+1} := J_i +(x_{i+1})\) has height at most \(i+1\text{.}\) On the other hand, every \(Q \in \Min(J_{i+1})\) contains some \(P \in \Min(J_i)\text{,}\) and in fact we must have \(Q \supsetneq P\text{,}\) since \(x_{i+1}\in Q \smallsetminus P\text{.}\) Since \(P\) has height \(i\) and \(P \supsetneq Q\text{,}\) then \(Q\) must have height at least \(i+1\text{.}\) Therefore, \(Q\) must have height exactly \(i+1\text{,}\) completing the induction.
Since every minimal prime of \(J_d = (x_1,\dots,x_d)\) has height \(d\text{,}\) its unique minimal prime must be \(\fm\text{.}\) It follows that \(\sqrt{J_d}=\fm\text{.}\)
Corollary 14.48. Dim \(<\)= EmbDim.
Let \((R,\fm,k)\) be a noetherian local ring. Then
\begin{equation*}
\dim(R)=\min\{ n \ | \ \sqrt{(f_1,\dots,f_n)}=\fm \textrm{ for some } f_1,\dots,f_n \} \leqslant \mu(\fm).
\end{equation*}
In particular, a noetherian local ring has finite dimension.
Proof.
Suppose that \(\sqrt{(f_1,\dots,f_n)}=\fm\text{.}\) Then \(\fm\) is a minimal prime of \((f_1, \ldots, f_n)\text{,}\) and thus \(\fm\) has height at most \(n\) by. The dimension of a local ring is the height of its maximal ideal, so \(\dim(R) \leqslant n\text{,}\) and thus \(\dim(R)\) is bounded above by the minimum in the middle. On the other hand, if \(d = \dim(R)\) then by there exist \(f_1, \ldots, f_d\) such that \(\fm = (f_1, \ldots, f_d)\text{,}\) which gives the other inequality, showing
\begin{equation*}
\dim(R)=\min\{ n \ | \ \sqrt{(f_1,\dots,f_n)}=\fm \textrm{ for some } f_1,\dots,f_n \}.
\end{equation*}
Finally, since \(\fm\) is generated by \(\mu(\fm)\) elements, there are in particular \(\mu(\fm)\) elements whose radical is \(\fm\text{.}\)
Definition 14.49. System of Parameters.
A sequence of \(d\) elements \(x_1,\dots,x_d\) in a \(d\)-dimensional noetherian local ring \((R,\fm)\) is a system of parameters or SOP if \(\sqrt{(x_1,\dots,x_d)}=\fm\text{.}\) If \(k\) is a field, a sequence of \(d\) homogeneous elements \(x_1,\dots,x_d\) in a \(d\)-dimensional \(\N\)-graded finitely generated \(k\)-algebra \(R\text{,}\) with \(R_0=k\text{,}\) is a homogeneous system of parameters if \(\sqrt{(x_1,\dots,x_d)}=R_+\text{.}\)
We say that elements \(x_1,\dots,x_t\) are parameters if they are part of a system of parameters; this is a property of the set, not just the elements.
Definition 14.50. Absolutely Minimal Prime.
Let \(R\) be a noetherian ring. A prime \(P\) of \(R\) is an absolutely minimal prime of \(R\) if \(\dim(R)=\dim(R/P)\text{.}\)
Theorem 14.51. SOP Stuff Theorem.
Let \((R,\fm)\) be a noetherian local ring, and \(x_1,\dots,x_t\in \fm\text{.}\) (a) \({\dim(R/(x_1,\dots,x_t))\geqslant \dim(R)-t}.\) (b) \(x_1,\dots,x_t\) are parameters if and only if \(\dim(R/(x_1,\dots,x_t))=\dim(R)-t\text{.}\) (c) \(x_1,\dots,x_t\) are parameters if and only if \(x_1\) is not in any absolutely minimal prime of \(R\) and \(x_i\) is not contained in any absolutely minimal prime of \(R/(x_1,\dots,x_{i-1})\) for each \(i=2,\dots,t\text{.}\)
Proof.
If \(\dim(R/(x_1,\dots,x_t)) = s\text{,}\) then take a system of parameters \(y_1,\dots,y_s\) for \(R/(x_1,\dots,x_t)\text{,}\) and pull back to \(R\) to get \(x_1,\dots,x_t,y'_1,\dots,y'_s\) in \(R\) such that the quotient of \(R\) modulo the ideal generated by these elements has dimension zero. By Krull height, we get that \(t+s\geqslant \dim(R)\text{.}\)
Let \(d=\dim(R)\text{.}\) Suppose first that \(\dim(R/(x_1,\dots,x_t))=d-t\text{.}\) Then, there is a SOP \(y_1,\dots,y_{d-t}\) for \(R/(x_1,\dots,x_t)\text{;}\) lift back to \(R\) to get a sequence of \(d\) elements \(x_1,\dots,x_t,y_1,\dots,y_{d-t}\) that generate an \(\fm\)-primary ideal. This is a SOP, so \(x_1,\dots,x_t\) are parameters.
On the other hand, if \(x_1,\dots,x_t\text{,}\) are parameters, extend to a SOP \(x_1,\dots,x_d\text{.}\) If \(I\) is the image of \((x_{t+1},\dots,x_d)\) in \(R'=R/(x_1,\dots,x_t)\text{,}\) we have \(R'/I\) is zero-dimensional, hence has finite length, so \(\Ass_{R'}(R'/I)=\{\fm\}\text{,}\) and \(I\) is \(\fm\)-primary in \(R'\text{.}\) Thus, \(\dim(R')\) is equal to the height of \(I\text{,}\) which is then \(\leqslant d-t\) by Krull height. That is, \(\dim(R') \leqslant d-t\text{,}\) and using the first statement, we have equality.
This follows from the previous statement and the observation that \(\dim(S/(f)) < \dim(S)\) if and only if \(f\) is not in any absolutely minimal prime of \(S\text{.}\)
Let \(R=\Z_{(2)}[x]\text{.}\) This ring has dimension at least two, but \(R/(2x-1)\cong \mathbb{Q}\) has dimension zero.
For an example that is a finitely generated algebra over a field, consider \(R = S/I\) with \(S = k[x,y,z]\) and \(I =(xy,xz,x^2-x)\text{.}\) This ring has dimension two: \(A=k[y,z]\) is a noether normalization, as \(x^2-x=0\) makes \(x\) integral over \(A\text{,}\) and no nonzero polynomial in \(y,z\) in \(k[x,y,z]\) can belong to the ideal \((x) \supseteq (xy,xz,x^2-x)\text{.}\) However,
\begin{equation*}
\displaystyle R/(x-1) \cong \frac{k[x,y,z]}{(y,z,x-1)}\cong k
\end{equation*}
has dimension zero.
