2011
DOI: 10.1007/s11856-011-0104-7
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Artin-Schreier extensions in NIP and simple fields

Abstract: We show that NIP fields have no Artin-Schreier extension, and that simple fields have only a finite number of them.

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Cited by 51 publications
(95 citation statements)
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“…Proof. By [23,Corollary 4.5] every infinite NIP field of characteristic p > 0 contains F alg p . But then M cannot satisfy the strong Erdős-Hajnal property by the proposition above, contradicting distality.…”
Section: 3mentioning
confidence: 99%
“…Proof. By [23,Corollary 4.5] every infinite NIP field of characteristic p > 0 contains F alg p . But then M cannot satisfy the strong Erdős-Hajnal property by the proposition above, contradicting distality.…”
Section: 3mentioning
confidence: 99%
“…Since char(K) = p it is perfect and so vK is p-divisible. Moreover, as K is dependent it follows from the proof of [24,Proposition 5.3] that Kv is Artin-Schreier closed, and therefore infinite.…”
Section: (Claim)mentioning
confidence: 99%
“…It will suffice, therefore, to find a definable non-trivial valuation on some finite extension L ≥ K (since if O is a non-trivial valuation ring on L then O ∩ K is a non-trivial valuation ring in K). It is therefore, harmless to assume that √ −1 ∈ K. By [22,Theorem 4.4] K is Artin-Schreier closed. So the same is true of any finite extension L ≥ K. This implies (e.g., [27, Lemma 2.4] 6 ) that either K is separably closed, or there exists some finite separable extension L ≥ K and q = char(K) such that (L × ) q = L × (in fact, by [22,Corollary 4.5] K has no finite separable extensions of degree divisible by p).…”
Section: N ′mentioning
confidence: 99%
“…It is therefore, harmless to assume that √ −1 ∈ K. By [22,Theorem 4.4] K is Artin-Schreier closed. So the same is true of any finite extension L ≥ K. This implies (e.g., [27, Lemma 2.4] 6 ) that either K is separably closed, or there exists some finite separable extension L ≥ K and q = char(K) such that (L × ) q = L × (in fact, by [22,Corollary 4.5] K has no finite separable extensions of degree divisible by p). Since √ −1 ∈ L it follows that, letting L(q) denote the q-closure of L, we have [L(q) : L] = ∞ ([8, Theorem 4.3.5]).…”
Section: N ′mentioning
confidence: 99%