Abstract Algebra by Prabhat Choudhary

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By Prabhat Choudhary

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Ak) for the subfield of E generated by F and the ai' Thus F(a l , .. ,' ak) is the smallest subfield of E containing all elements of F along with the ai' ("Smallest" means that F(a l , .. " ak) is the intersection of all such subfields,) Explicitly, F(a l , .. (X - a l ) .. , (X - a k) for some aI' .. " a k E E and>.. E F, (There is a subtle point that should be mentioned. We would like to refer to the a i as "the" roots off, but in doing so we are implicitly assuming that if f3 is an element of some extension E of E andj(f3) = 0, then f3 must be one of the a i .

The desired isomorphism is given by aD + ala + ... + an_Ian- 1 ~ aD + alf3 + ... + a n_ I f3n- l . If/is a polynomial in FIX] and F is isomorphic to the field F' via the isomorphism i, we may regard/as a polynomial over F'. We simply use i to transfer! Thus if/= aD +alX + ... ajf', then/' = i(f) = i(ao) + i(al)X + ... + i(an)Xn. There is only a notational difference between/and/', and we expect that splitting fields for/and/, should also be essentially the same. We prove this after the following definition.

Ak) is the intersection of all such subfields,) Explicitly, F(a l , .. (X - a l ) .. , (X - a k) for some aI' .. " a k E E and>.. E F, (There is a subtle point that should be mentioned. We would like to refer to the a i as "the" roots off, but in doing so we are implicitly assuming that if f3 is an element of some extension E of E andj(f3) = 0, then f3 must be one of the a i . This follows upon substituting f3 into the equationj(X) = >.. (X - 0. 1) ... ) If K is an extension of F and/ E F1x], 'We say that K is a splittingfield for/over F if / splits over K but not over any proper subfield of K containing F.

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