Completeness of the ring of polynomials (Q479320)
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| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Completeness of the ring of polynomials |
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Completeness of the ring of polynomials (English)
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5 December 2014
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Let \(k\) be a field. Let \(R = k[x_1, \ldots x_n]\) denotes the polynomial ring in \(n\) variables. Set \(d = 0\) if \(k\) is finite, otherwise \(d\) is defined by the cardinality equation \(|k| = \aleph_d\). A recent conjecture of L. Gruson (priv. comm., 2013) states: {Conjecture:} \(\mathrm{Ext}^i_R(K,R) \neq 0 \iff i = \text{inf}\{d+1,n\}\). This conjecture generalizes for all \(i\) the following two 40 years old conjectures, {Conjecture}. With same notation, let \(m \subset R\) be a maximal ideal. Then the following bi-implications hold: {a} \(\mathrm{Ext}^1_R(K,R) \neq 0 \iff n = 1 \text{ or } |k| \leq \aleph_0\) [\textit{C. U. Jensen}, Bull. Am. Math. Soc. 78, 831--834 (1972; Zbl 0256.20069)]. {b} \(\mathrm{Ext}^1_{R_m}(K,R_m) \neq 0 \iff n = 1 \text{ or } |k| \leq \aleph_0\) [\textit{L. Gruson}, in: Sympos. math. 11, Algebra commut., Geometria, Convegni 1971/1972, 243--254 (1973; Zbl 0291.13007)]. The work under review proves these latter two conjectures. In fact, as discussed in the introduction to the paper, both these conjectures are implied by Theorem 4 of the work: {Theorem}. Assume \(|k| \geq \aleph_1\), that \(n \geq 2\), and that \(R = U^{-1}R_0\) is a localization of \(R_0 = k[X_1, \ldots X_n]\) with a multiplicative subset \(U \subset R_0\). In addition, assume that every maximal ideal of \(R\) contracts to a maximal ideal of \(R_0\). Then \(R\) is complete. The proof of this theorem is presented in a series of lemma and part of it is true in a more general setup where \(R\) is an integral domain. The proof depends crucially on the following result of \textit{L. Gruson} [in: Sympos. math. 11, Algebra commut., Geometria, Convegni 1971/1972, 243--254 (1973; Zbl 0291.13007)]: {Proposition}. Assume \(n = 2\). The localization \(R_m\) of \(R\) at any maximal ideal \(m \subset R\) is complete.
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