By Falko Lorenz
From Math reports: "This is a captivating textbook, introducing the reader to the classical components of algebra. The exposition is admirably transparent and lucidly written with merely minimum necessities from linear algebra. the recent thoughts are, at the least within the first a part of the publication, outlined within the framework of the advance of conscientiously chosen difficulties. hence, for example, the transformation of the classical geometrical difficulties on buildings with ruler and compass of their algebraic surroundings within the first bankruptcy introduces the reader spontaneously to such basic algebraic notions as box extension, the measure of an extension, etc... The publication ends with an appendix containing workouts and notes at the past components of the ebook. despite the fact that, short ancient reviews and recommendations for extra interpreting also are scattered in the course of the text."
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Extra resources for Algebra: Fields and Galois Theory
But if n is prime we will see in F11 that f is indeed irreducible. This is not the place for a comprehensive study of the problem mentioned in the ﬁrst paragraph. Nonetheless, we would like to shed some light on certain theoretical aspects of the issue. Two very natural questions will guide us: (i) Is the ring ޚŒX a UFD? (ii) Is every polynomial f 2 ޚŒX that is irreducible in ޚŒX also irreducible in ޑŒX ? These questions are intimately connected and were both answered by Gauss in the afﬁrmative.
F with the following property: If Ä W R ! E is any injective ring homomorphism from R into a ﬁeld E, there is a unique ring homomorphism W F ! E such that ıÃ D Ä — in other words, making the following diagram commutative: ✲ E F ✻ ✒ (23) Ã Ä R Such a ﬁeld F is called a fraction ﬁeld of R. It is uniquely determined up to isomorphism: more precisely, if F 0 is another fraction ﬁeld and Ã0 W R ! F 0 the corresponding map, there exists a unique isomorphism W F ! F 0 such that (24) F ✻ Ã ✲ F0 ✻0 Ã id✲ R R commutes.
Proof. Assume the opposite. Then f D gh, where g; h are nonconstant polynomials in RŒX (nonconstant because f is primitive). Taking the image in RŒX gives f D gh. Since an is nonzero, we must have deg g D deg g 1 and deg h D deg h 1. Since R is an integral domain, we get a contradiction with the assumption that f is irreducible. ˜ In applying F9, one is usually dealing with a unique factorization domain R, because apart from the fact that otherwise one has hardly any control over the primitivity of f , it is also not permissible in the general case to deduce that f is irreducible over K D Frac R.