Is there a proof of the Krull Intersection Theorem without using the Artin-Rees lemma?

Question

Textbooks on commutative algebra always prove the Krull Intersection Theorem using the Artin-Rees lemma. But is it really necessary to do so? Can one prove the Krull Intersection Theorem without using the Artin-Rees lemma? Is it even possible? I would like to see such a proof.

Answer 1

I reproduce the proof from the CRing project, which in turn cites ''Hervé Perdry. An elementary proof of Krull’s intersection theorem. The American Mathematical Monthly, 111(4):356–357, 2004''.

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Theorem. Suppose $R$ is a noetherian ring, $I \subset R$ an ideal. Suppose $b \in \bigcap I^n$. Then, $(b) = (b) I$, as ideals.

Note that the theorem implies the intersection $\bigcap I^n$ to be zero under either of the following condition:

1. $I$ is contained in the Jacobson radical of $R$. (In particular, $R$ is local, and $I$ is the maximal ideal.)
2. $R$ is a domain and $I \neq R$.

Proof of the Theorem. Let $a_1, \ldots, a_k \in I$ be generators of $I$. Note that $I^n$ is generated by all monomials of degree $n$ in the $a_i$. Thus, if $b \in \bigcap I^n$, then for each $n$, there exists a homogeneous polynomial $P_n \in R[x_1, \ldots, x_k]$ of degree $n$ such that $$b = P_n(a_1, \ldots, a_k).$$ Now, note that the ideal $(P_1, P_2, \ldots)$ is itself finitely generated since $R[x_1, \ldots, x_k]$ is noetherian.
Thus, there exist $N > 0$ and polynomials $Q_i$ such that $$P_{N + 1} = Q_1 P_1 + \cdots + Q_N P_N.$$ Note that since the $P_i$ are all homogeneous of degree $i$, we may pick $Q_i$ to be homogeneous of degree $N + 1 - i > 0$. (We only need that the $Q_i$s have no constant term.)
Now, evaluating the above expression at $a_1, \ldots, a_k$ gives $$b = b\underbrace{(Q_1(a_1, \ldots, a_k) + \cdots + Q_N(a_1, \ldots, a_k))}_{\in I}.$$
Thus, $b \in (b)I$. $\hspace{1cm} \Box$