By Julio R. Bastida

Initially released in 1984, the primary target of this booklet is to make the final conception of box extensions available to any reader with a modest heritage in teams, earrings and vector areas. Galois idea is mostly considered as one of many significant and most lovely components of algebra and its production marked the end result of investigations by means of generations of mathematicians on one of many oldest difficulties in algebra, the solvability of polynomial equations through radicals.

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**Example text**

Proposition. Let K be afield, and let A be a subdomain ofK. Then the subfield of K generated by A is the only subfield of K that is a field of fractions of A; it consists of all elements of K of the form a/fi with a,fi^A and ft * 0. Proof. Let F denote the set of all elements of K of the form a /ft with a , | 8 E i and /? =£ 0. It is clear that A c F. Moreover, in view of the equalities (a/P)±(y/S) = («8±Py)/P8 and (a/P)(y/8) = ay/P8, which hold when a , j 8 , y , S e ^ and /? =£ 0 =£ 8, we see that F is a subdomain of K.

For this reason, we shall omit the tedious details required in the verification of some assertions. Consider a domain A, and let E = A X (A - {0}). y. It is easily verified that this defines an equivalence relation on E. ). 1. Fields of Fractions Now suppose that (a, / J ) , ( Y, 8),(a, /3),(y, 8) e E, and that ( « , / ? ) (y, 6) and (a, P) ~ (y, 8). £) ~( Y Y,88~). > , < « , £ » - < « « , / ? )e £. , a, /? e A and /? # 0 ¥= /?. A tedious, but completely elementary, computation now shows that this addition and this multiplication define a field structure on F with respect to which the zero and unit elements are, respectively, (0,1) and (1,1).

Thus, (a) and (b) are equivalent. To prove that (a) implies (d), suppose that C\\ax(A) = p. Z, and hence Z/Ker(w) = Z/p. Since Z/Ker(w) and Im(w) are isomorphic, it follows that Im(w) and Z/p are isomorphic. It is obvious that (d) implies (c). For if Z/p is isomorphic to the subring Im(w) of A, then it is embeddable in A. To conclude, we now show that (c) implies (a). Suppose that Z/p is embeddable in A, and choose a monomorphism v from Z/p to A. If w denotes the natural projection from Z to Z/p, it is clear that u = v ° w.