Get Algebra DeMYSTiFieD (2nd Edition) PDF

By Rhonda Huettenmueller

ISBN-10: 0071743626

ISBN-13: 9780071743624

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Extra resources for Algebra DeMYSTiFieD (2nd Edition)

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Any homomorphic image of a Noetherian ring is Noetherian. Furthermore, if Ro is a Noetherian ring, and R is a finitely generated algebra over Ro, then R is Noetherian. Proof. Given an ideal I in R/ J, with R Noetherian, the preimage of I in R is finitely generated, and the images of its generators generate I. Since R is a finitely generated algebra over Ro, R is a homomorphic image of S := R O[Xl, ... ,xrJ for some r. 2 and induction on r, we see that S is Noetherian. Since a homomorphic image of a Noetherian ring is Noetherian, we are done.

The map P takes p to q. The beauty of this description of the morphisms from X to Y is that it is independent of the description of A( X) and A(Y) as quotients of particular polynomial rings~that is, it is independent of the particular embeddings of X c k n and Y c km ! In particular, we see that X and Yare isomorphic by polynomial maps iff A(X) and A(Y) are isomorphic as k-algebras. 10. The category of affine algebraic sets and morphisms (over an algebraically closed field k) is equivalent to the category of affine kalgebras with the arrows reversed.

Let k be an algebraically closed field and let X cAn be an algebraic set. Every maximal ideal of A(X) is of the form mp := (Xl - al, ... ,Xn - an)/I(X) for some p = (al, ... ,an) EX. In particular, the points of X are in one-to-one correspondence with the maximal ideals of the ring A(X). Proof. The maximal ideals of A(X) correspond to the maximal ideals of k[XI, ... ,xnl containing I(X), so it suffices to treat the case X = An, A(X) = k[XI"'" xnl. To prove the first statement, note that any maximal ideal m--even any prime ideal -is a radical ideal, and thus I (Z (m)) = m by the Nullstellensatz.

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Algebra DeMYSTiFieD (2nd Edition) by Rhonda Huettenmueller


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