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A Note on the Algebraic Closure of a Author(s): Robert Gilmer Source: The American Mathematical Monthly, Vol. 75, No. 10 (Dec., 1968), pp. 1101-1102 Published by: Taylor & Francis, Ltd. on behalf of the Mathematical Association of America Stable URL: https://www.jstor.org/stable/2315743 Accessed: 28-09-2019 03:55 UTC

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This content downloaded from 128.205.114.91 on Sat, 28 Sep 2019 03:55:54 UTC All use subject to https://about.jstor.org/terms This content downloaded from 128.205.114.91 on Sat, 28 Sep 2019 03:55:54 UTC All use subject to https://about.jstor.org/terms 1102 CLASSROOM NOTES [December constant polynomial over Ei has a root in Ei41. Then if L U 1 Ei, it is easy to show that each nonconstant polynomial over E splits into linear factors over L so that L contains an algebraic closure of E. (See [2] pp. 169-170.) The purpose of this note is to observe that in the above process, the field E1 already contains an algebraic closure of E. This we do by establishing the following result:

THEOREM. If K is a subfield of afield L and if each nonconstant polynomial with coefficients in K has a root in L, then each nonconstant polynomial with coefficients in K splits into linear factors in L [X].

Proof. It suffices to prove that for f a nonconstant irreducible monic poly- nomial with coefficients in K, f is a product of linear factors in L [X]. If K has 0, we define p 1; otherwise p denotes the characteristic of K. (In Bourbaki's terminology [i, p. 71], p is the characteristic exponent of K.) In a F of f over K, the roots of f all have the same multiplicity pe for some nonneg,ative e:f= H' , (X- ai)PeI =H1(XPe -a). Further, if f = XnP+f_X(n-l)p d- ? +f1XP +fo, and if f3= apoe for each i, { * * s } is the set of roots of the irreducible, g Xn +f,_-Xn-1 + +fo over K [3, p. 120]. The field K(01, * *, ,Bn) is a finite, normal, of K. It is therefore a simple extension of K: K(1, 0, ) =K(,y). If h is the minimal polynomial for y over K, then K(y) is a splitting field for h over K and there is, by hypothesis, a root 0 of h in L. The fields K(0) and K(y) are each K-isomorphic to K [X]/(h), and hence are K-isomorphic to each other. It follows that k(0) is a splitting field of g over K. Therefore, if pe = 1-that is, if f is separable over K, then L contains a splitting field of f over K. Hence we have proved our theorem in case the algebraic closure of K in L is separable over K. In particular, our proof is complete if K is a . In case K is not perfect, we let Ko be the subfield of L consisting of those elements which are purely inseparable over K. We show that the field Ko is perfect. Hence if sCKo and if sOeCK, theen by hypotlhesis, XPeO _Spe has a root t in L. Therefore, tP1 CK, tC-Ko, and (tP-s)Pe=0, implying that tP=s so that Ko is perfect [3, p. 124]. Further, if q(x) = '0 qiXi is any nonconstant polynomial over Ko, then for some positive integer e, [q(X) ]P =v(X) - qp C [X so that v (X) has a root a in L. We have 0= v(a)= [q(a)]P', implying thlat q (a) =0. We conclude that Ko is a perfect field with the property that each nonconstant polynomial over Ko has a root in the extension field L of Ko. By our previous proof, each nonconstant polynomial over Ko splits into linear factors in L [X]. This property holds then, in particular, for nonconstant polynomials over K, and this completes the proof of our theorem.

References

1. N. Bourbaki, Elements de Mathematique, Alghbre, Book II, Chap. 5, Hermann, Paris, 1950. 2. S. Lang, Algebra, Addison-Wesley, Reading, 1965. 3. B. L. van der Waerden, Modern Algebra, vol. I, English ed., Ungar, New York, 1948.

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