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Course: MATH 7, Fall 2008
School: UPenn
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Notes Lecture Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 19, 2006 1 TALK SLOWLY AND WRITE NEATLY!! 0.1 Examples of Fields We are now ready to begin the study of elds which will eventually lead us to Galois theory. Back when I took this class this was by far the coolest part of it. In the study of elds we will often consider pairs of eld F K. However unlike the case of groups where we...

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Notes Lecture Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 19, 2006 1 TALK SLOWLY AND WRITE NEATLY!! 0.1 Examples of Fields We are now ready to begin the study of elds which will eventually lead us to Galois theory. Back when I took this class this was by far the coolest part of it. In the study of elds we will often consider pairs of eld F K. However unlike the case of groups where we often start with a group G and look at subgroups, in the case of elds we will be more interested in starting with a eld F and looking as dierent extensions of F to elds K. Extension Field Denition 0.1.0.1. Let F be a eld. If F K then we say K is a extension eld of F . The three most important classes of elds we will encounter are the following. 2 (1) Number Fields: A number eld K is any subeld of C. As any subeld of C contains 1 it must also contain Q. The number elds most commonly studied are those were every element is algebraic. (2) Finite Fields: A eld having only nitely many elements. If K is nite then the kernel of the unique homomorphism Z K is a prime ideal since Z is innite. So in particular K contains a subeld isomorphic to Z/(p) = Fp for some prime p. And hence we can consider K as an extension of Fp. (3) Function Fields: Certain extensions of the eld C(x). 3 Specically, suppose we have an irreducible polynomial in f C[x, y] which is not a polynomial in x alone. Since it is irreducible in C[x, y] we know that it is irreducible in C[x][y] and hence, by the generalized Guasss Lemma we have that we have that f is also irreducible in F = C(x) the fraction eld of C[x]. In particular this means that the idea (f ) F [y] is maximal. and hence F [y]/(f ) is a eld. 0.2 Algebraic and Transcendental Elements Similarly to the case of the complex numbers we dene Algebraic/Transcendental Extensions Denition 0.2.0.2. Let F be a eld. And let K 4 such that K F . We say that is algebraic over F if there is a polynomial over F which is satised by . I.e. if n + an1n1 + + a1 + = 0 for some a0, . . . , an1 F We say that is transcendental over F if it is not algebraic over F . We can think of the two possibilities in terms of the evaluation homomorphism. Specically Lemma 0.2.0.3. Let F K be elds with K. Then if : F [x] K f (x) f () we have is transcendental if and only if is injective. Or more specically if the kernel of is 0. Proof. Immediate 5 Similarly we have Irreducible Polynomial Denition 0.2.0.4. Let F K be elds with K. Further let : F [x] K f (x) f () We then know that ker( ) is principle as F [x] is a principle ideal domain. So in particular it is generated by a single element f (x) F [x]. But because K is a eld we must have f (x) is irreducible (because otherwise K would have a zero divisor) Hence f (x) is the only irreducible polynomial in (f (x)) (because every element of the ideal is a multiple of f (x)) and we call f the Irreducible Polynomial for over F . It is important to note that the base eld we are working over is crucial when determining the irreducible polyno- 6 mial of an element. Or for that matter even if a polynomial is irreducible. For example Let = i. Then the irreducible polynomial for over Q is x4 + 1. However, over Q[i] the irreducible polynomial is x2 1. And what is more, over Q[i] x4 + 1 is not irreducible as x4 + 1 = (x2 + i)(x2 i) F () Denition 0.2.0.5. Let F () be smallest the eld containing both and F . Similarly let F (1, . . . , n) be the smallest eld containing 1, . . . , n and F . Connection between F [] and F () Lemma 0.2.0.6. Recall that F [] is the ring {ann : an F } 7 and is the smallest ring containing both F and . We then have F () is isomorphic to the eld of fractions of F [] In particular we have that if is transcendental then F [x] F [] is an isomorphism and hence F () is isomorphic to the eld F (x) of rational functions. Proof. Immediate Notice that this means that if and are both transcendental over F then F () F (). For example this = means that Q() Q(e) which is not at all obvious at = rst glance. However, if and are algebraic the case is very dierent. Isomorphism of Equivalent Extensions Theorem 0.2.0.7. (a) Suppose that is algebraic over 8 F and let f (x) be its irreducible polynomial over F . The map F [x]/(f ) F [] is an isomorphism and F [] is a eld. Thus F [] = F () (b) More generally let 1, . . . , n be algebraic elements of a eld extension K of F . Then F [1, . . . , n] = F (1, . . . , n). Proof. Let be the map which takes f (x) to f (). Since f (x) generates ker(), we know that F [x]/(f ) is isomorphic to the image of which is F [x]. Since f is irreducible we know that (f ) is a maximal ideal and hence F [x]/(f ) is a eld. Since F () is the smallest eld containing F [] we must have F [x]/(f ) = F (). The second part follows from the rst and is let as an exercise. Field Extensions as Vector Spaces 9 Theorem 0.2.0.8. Let be an algebraic over F and let f (x) be its irreducible polynomial. Suppose f (x) has degree n. Then (1, , . . . , n1) is a basis for F [] as a vector space over F Proof. This is a special case of the same theorem for rings which we have already seen. The following is one of the most important results concerning eld extensions Isomorphism of Equivalent Extensions Theorem 0.2.0.9. Let K and L be algebraic elements of two extensions of F . There is an isomorphism of elds : F () F () which is the ...

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