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2 Fundamental perfect fields

In this section, we will build fundamental perfect fields of positive characteristic and show the structure of perfect fields. Recall that a perfect closure F of a field E is a smallest perfect extension of E in an algebraic closure E of E; i.e., F F ′ for all F ′ perfect and E F ′ E. Let p be a ⊆ ⊆ ⊆ positive prime. Denote by Zp the field of congruence classes modulo p. And throughout this paper, we always consider Zp as a subfield of fields of characteristic p. Remark 2.1. Let H be a field of characteristic p and a H. Then for any n 0, the n ∈ ≥ n polynomial xp a has a unique root in an algebraic closure of H and denote by p√a − n this unique root. And if H is perfect then p√a H. ∈ Let X = xi i I be a set of independent variables and assume that Zp(X) is the field { } ∈ pn of all rational functions in X over Zp. For any u Zp(X) and any n 0, denote by √u n n p ∈ p ≥ pn the root of y u in an algebraic closure Zp(X) of Zp(X). Set √X = √xi i I for all − n n+1 n { } ∈ n+1 p p p pn+1 p p p n 0. It is clear that √X =( √X) = ( √xi) i I and Zp( √X) Zp( √X) ≥ { n } ∈ ⊆ Z p√ Z for all n 0. Consequently, Fp(I) = Sn 0 p( X) is a subfield of p(X). Now, we ≥ ≥ assign Fp(I)= Zp to the case that I = . Note that the structure of Fp(I) only depends ∅ on the cardinality of I; i.e., if I = J then Fp(I) ∼= Fp(J). Then we have the following| result.| | |

Proposition 2.2. Fp(I) is a perfect closure of Zp(X).

pn Proof. Let a be an element of Fp(I). There exists n such that a Zp( √X). Hence ∈

pn pn g( √x1,..., √xd) a = pn pn h( √x1,..., √xd) for g(y1,...,yd); h(y1,...,yd) are two polynomials of the polynomial ring Zp[y1,...,yd] in y1,...,yd over Zp. Since Zp is perfect, it follows that

p p p p p p g(y1,...,yd)= u(y1,...,yd) and h(y1,...,yd)= v(y1,...,yd)

2 for u(y ,...,yd); v(y ,...,yd) Zp[y ,...,yd]. Consequently, we get 1 1 ∈ 1

pn pn pn+1 p pn+1 p pn+1 pn+1 p g( √x1,..., √xd) g(( √x1) ,..., ( √xd) ) u( √x1,..., √xd) a = = = . pn pn pn+1 p pn+1 p pn+1 pn+1 p h( √x1,..., √xd) h(( √x1) ,..., ( √xd) ) v( √x1,..., √xd)

pn+1 pn+1 n+1 p u( √x1,..., √xd) Z p √ Hence a = b for b = pn+1 pn+1 p( X) Fp(I). So Fp(I) is perfect. v( √x1,..., √xd) ∈ ⊆ Now, assume that H is a perfect field and Zp(X) H Zp(X). Since X H, it pn ⊆ ⊆ ⊆ follows that √X H for all n 0 by Remark 2.1. Hence Fp(I) H. Thus, Fp(I) is ⊆ ≥ ⊆ a perfect closure of Zp(X).

We call a field k of characteristic p > 0 is a fundamental perfect field if k ∼= Fp(I) for some I. Then we obtain the following interesting result. Theorem 2.3. A field of positive characteristic is a perfect field if and only if it is an algebraic extension of a fundamental perfect field.

Proof. Let E be a perfect field of positive characteristic p. Assume that S = si i I is { } ∈ a transcendence base of E over Zp. Then it is easily seen that E is algebraic over Zp(S). Let X = xi i I be a set of independent variables. Since S = si i I is algebraically { } ∈ Z Z Z { } ∈ independent over p, it follows that p(S) ∼= p(X). Let F be a perfect closure of Zp(S). Since E is a perfect field, we can consider F E. Remember that Fp(I) is a Z Z ⊆ Z perfect closure of p(X) by Proposition 2.2 and p(S) ∼= p(X), F ∼= Fp(I) by [1]. So F is a fundamental perfect field. And since E is algebraic over Zp(S), E is algebraic over F. Conversely, if K is an algebraic extension of a fundamental perfect field k then K is perfect. Indeed, for any algebraic extension H of K, H is an algebraic extension of k. Since k is perfect, H is separable over k. Hence H is separable over K. Thus, K is perfect. The proof is complete. A subfield F of a perfect field P is called a maximal fundamental perfect subfield of P if F is a fundamental perfect field and P is an algebraic extension of F. Then from the proof of Theorem 2.3, we find the following corollary. Corollary 2.4. Let P be a perfect field of positive characteristic p. Then the following statements hold.

(i) Q is a maximal fundamental perfect subfield of P if and only if there exists is a transcendence base S = si i I of P over Zp such that Q is the perfect closure of Z { } ∈ p(S) in P. In this case, Q ∼= Fp(I). (ii) The maximal fundamental perfect subfields of P are isomorphic.

(iii) The structure of maximal fundamental perfect subfields of P only depends on the cardinality of transcendence bases of P over Zp.

3 Proof. Q is a maximal fundamental perfect subfield of P if and only if Q ∼= Fp(I) for some I and P is an algebraic extension of Q. On the one hand, Q ∼= Fp(I) if and only if there exists is a transcendence base S = si i I of Q over Zp such that Q is the { } ∈ perfect closure of Zp(S) in P by the proof of Theorem 2.3. On the other hand, P is an algebraic extension of Q if and only if S = si i I is a transcendence base of P over { } ∈ Zp. So we get (i). Since if I = J then Fp(I) = Fp(J), hence from the proof of (i) | | | | ∼ and remember that the cardinalities of transcendence bases of P over Zp are the same, we have (ii). From (i) and the proof of (ii) we get (iii). Finally, we would like to give the following proposition for perfect closures of fields.

Proposition 2.5. Let E be a field of positive characteristic p and S = si i I a { } ∈ transcendence base of E over Zp. Let V = vj j J be a base for the vector space E over { } ∈ Zp(S) and Q a perfect closure of Zp(S) in an algebraic closure E of E. Then Q(V ) is a perfect closure of E, and E is perfect if and only if Q E. ⊆

Proof. Since V is algebraic over Zp(S), Q(V ) algebraic over Q. Hence Q(V ) is perfect. Now, assume that E F E is perfect. Then Q F and hence Q(V ) F. So Q(V ) is a perfect closure of⊆E. From⊆ this it follows that E⊆is perfect if and only⊆ if E = Q(V ). This is equivalent to Q E. ⊆ References

[1] N. Bourbaki, Algebra II, Springer, ISBN 978-3-540-00706-7 (2003).

[2] H. Matsumura, Commutative ring theory, Translated from the Japanese by M. Reid. Cambridge Studies in Advanced Mathematics, 8 (2003).

[3] B. Olivier and C. Brian, CMI Summer School notes on p-adic Hodge theory, re- trieved 2010-02-05 (2009).

[4] J. P. Serre, Local fields, Graduate Texts in Mathematics, 67, Springer-Verlag, ISBN 978-0-387-90424-5, MR 554237 (1979).

Department of Mathematics Hanoi National University of Education 136 Xuan Thuy street, Hanoi, Vietnam [email protected] and [email protected]

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