I am reading Wikipedia, proof of the Krull's height theorem ( https://en.wikipedia.org/wiki/Krull%27s_principal_ideal_theorem ) and some question arises : Let $A$ be a noetherian ring.
I am now trying to understand the underlined statements.
Q.1. In this proof, where the property $\mathfrak{p}= \sqrt{(x_1, \dots, x_n)}$ is used?
Q.2. Why "If $\bar{\tau}$ is minimal prime over $\bar{x_1}$, then $\tau$ contains $x_1, x_2^{r_2}, \dots, x_n^{r_n}$" ?
And why from this $\tau = \mathfrak{p}$ ? Since $\mathfrak{p} = \sqrt{(x_1, \dots, x_n)}$ ? I guess that it is true since from $x_1, x_2^{r_2}, \dots , x_n^{r_n} \in \tau$, we have $x_1, x_2, \dots x_n \in \tau$ ( $\tau$ is prime ) so that $(x_1, \dots, x_n) \subseteq \tau$. Since $\mathfrak{p}$ is minimal over $(x_1, \dots , x_n)$, $\tau = \mathfrak{p}$. Correct?
EDIT (Other way?) : Note that $ (x_1 , x_2^{r_2}, \dots , x_n^{r_n}) = (x_1, y_2,\dots , y_n)$. My interlude question is, $\mathfrak{p}$ is minimal over $ (x_1 , x_2^{r_2}, \dots , x_n^{r_n})$? If so, then $\mathfrak{p}$ is minimal over $(x_1, y_2,\dots , y_n)$, and $\mathfrak{p}/ (y_2, \dots, y_n)$ is minimal over $(x_1,y_2,\dots, y_n)/(y_2,\dots, y_n)$. But this last ideal is a principal ideal in $A/(y_2,\dots, y_n)$. So
$$\operatorname{ht}(\mathfrak{p}/(y_2, \dots, y_n)) \le 1 $$
by the principal ideal theorem. This implies $\operatorname{ht}(\mathfrak{q}/(y_2, \dots, y_n))=0$. so $\mathfrak{q}$ is a prime ideal that is minimal over $(y_2, \dots, y_n)$. By induction, this implies $\operatorname{ht}(\mathfrak{q}) \le n-1$, which completes the proof. So, the above interlude question is true?
Q.3. I'm now struggling with showing $\operatorname{ht}({\mathfrak{p}}) \le n$ from $\operatorname{ht}(\mathfrak{q}) \le n-1$. If $\mathfrak{q} \subsetneq \mathfrak{p}$ are prime ideals such that there is no prime strictly between them, then $\operatorname{ht}(\mathfrak{p}) \ \le \operatorname{ht}(\mathfrak{q})+1 ( \le n)$ ? Let $L_{\mathfrak{p}}$ be the set of length of all chains of prime ideals contained in $\mathfrak{p}$ so that $\operatorname{ht}(\mathfrak{p}) = \operatorname{sup}L_{\mathfrak{p}}$ Then can we show that $\operatorname{ht}(\mathfrak{q}) +1$ is an upper bound of $L_{\mathfrak{p}}$? Can anyone helps?
Best Answer
Q.2. If $\bar{\mathfrak{r}}\subset \bar{A}$ (Fraktur r) is minimal prime over $\bar{x_1}$, its preimage $\mathfrak{r}\subset A$ contains $x_1$ and $y_i$ and consequently $x_i^{r_i}=y_i+a_ix_1$ (since $\mathfrak{r}$ is an ideal). I think you have deduced $\mathfrak{r}=\mathfrak{p}$ correctly.
Another way is possible: we have inclusuions $$\mathfrak{p}\supset (x_1, x_2, \dots ,x_n)\supset (x_1, x_2^{r_2}, \dots,x_n^{r_n})\supset (x_1,\dots,x_n)^N$$ for sufficiently large $N$. Taking radicals, $$\mathfrak{p}\supset \sqrt{(x_1, x_2 ,\dots, x_n)}\supset \sqrt{(x_1, x_2^{r_2}, \dots,x_n^{r_n})}\supset \mathfrak{p}.$$
This yields $\mathfrak{p}= \sqrt{(x_1, x_2^{r_2}, \dots,x_n^{r_n})}$. The latter equation answers your interlude question affirmatively, since a prime $P$ is minimal over $Q$ if $P=\sqrt{Q}$.
Q.3. Note that the above argument applies to any $\mathfrak{q}$ maximal among primes strictly contained in $\mathfrak{p}$. By definition the height of $\mathfrak{p}$ is the supremum of length of prime chains in $\mathfrak{p}$. For a maximal chain, $\operatorname{ht}\mathfrak{q}\le n-1$ implies its length $\le n$. Taking their supremum we still have $\operatorname{ht}\mathfrak{p}\le n$.