<\/p>\n
Proof:<\/b><\/p>\n The topics are from\u00a0Matsumura, H. (June 30, 1989). “Chapter 1: Commutative Rings and Modules”. Commutative Ring Theory. Cambridge University Press. p. 6. ISBN 978-0-521-36764-6. and\u00a0Atiyah, M. F.; MacDonald, I. G. (February 21, 1994). “Chapter 1: Rings and Ideals”. Introduction to Commutative Algebra. Westview Press. p. 11. ISBN 978-0-201-40751-8.<\/u><\/p>\n Friday, August 2, 2013<\/p>\n <\/p>","protected":false},"excerpt":{"rendered":"\n
\nFor non-surjective<\/span><\/b> morphisms, the two thing may have no relation at all. For example, let A<\/span> be a local domain<\/span><\/b> and f<\/span> the embedding into B<\/span>, its field of fractions. Then f(\\mathfrak{R}(A))=\\mathfrak{R}(A)<\/span> is very large but \\mathfrak{R}(B)=0<\/span>.
\nSince prime ideals always pull back, we always have\u00a0f(\\mathfrak{N}(A))\\subseteq\u00a0\\mathfrak{N}(B)<\/span>. For Jacobson radicals, the reason actually is the same since when f<\/span> is surjective, maximal ideals pull back. This is like saying, if a function vanishes on every closed point, then it vanishes on every closed point of a closed subset. If it vanishes on every point, then its pullback vanishes on every point. In the polynomial case, since \\mathfrak{N}=\\mathfrak{R}<\/span>, this reduces to trivial intuition.<\/li>\n
\nContinuing the discussion of a.<\/i><\/b>, this is saying if in addition closed points are finite, then a function vanishing on a subset of them must be induced by some function vanishing on all of them. Taking the example of \\mathbb{Z}<\/span>, p<\/span> vanishes on the single point \\mathrm{Spec}(\\mathbb{Z}\/p^2\\mathbb{Z})<\/span>, but cannot be induced by some elements vanishing on all of \\mathrm{Spec}\\mathbb{Z}<\/span>: such elements must be 0<\/span>. This happens because we fail to let it vanish at all other primes simultaneously: infinite product does not make sense. However in \\mathrm{Spec}(\\prod_{p=2,3,5,...}\\mathbb{Z}\/p^2\\mathbb{Z})<\/span>, this holds, as we can always pull back to (2,3,5,...)<\/span>.<\/li>\n<\/ol>\n<\/li>\n\n
\nProof:<\/b>\u00a0For UFDs, it is crystal clear that these are satisfied. Conversely, we can easily split a<\/span> into a finite product of irreducible elements, by A.C.C.. The product is unique because irreducibles are prime.\u00a0We should care about the cases when irreducible element is not prime.<\/b><\/li>\n
\nProof:<\/b>\u00a0If ab\\in\\cap P_{\\lambda}<\/span>, then for all \\lambda<\/span>, either a,b<\/span> is in P_{\\lambda}<\/span>. So the one of the collections of primes containing a<\/span> and b<\/span> respectively is not bounded below<\/b><\/span>. Thus either of a,b<\/span> is in the intersection. The corollary then follows from Zorn’s lemma<\/i><\/b>.<\/li>\n
\nProof:<\/b>\u00a0This is mysterious. Proof is not hard, but I do not know why. I will write when I know its meaning or usage.<\/li>\n
\nProof:<\/b>\u00a0Notice when A<\/span> is reduced<\/b><\/span>, this amounts to say if every ideal contains a nonzero idempotent, then \\mathfrak{R}=0<\/span>: If a\\ne 0<\/span>, then (a)<\/span> contains a nonzero idempotent e=ka<\/span>, with e(1-e)=0<\/span>, so 1-ka<\/span> is not a unit, and a\\notin R<\/span>. The general case follows by passing to A\/\\mathfrak{N}<\/span><\/span><\/b>. But this is more like an awkward exercise.<\/li>\n
\nProof: Otherwise it would split<\/span><\/b> as a direct product. By 2.<\/i><\/b> above, it has at least two maximal ideals. Geometrically, a local picture cannot be a disjoint union.<\/li>\n
\nProof: Non-zero-divisors form a multiplicative set<\/span>: If a,b<\/span> are not zero-divisors, and a b x=0<\/span>, we have b x=0<\/span> and x =0<\/span>. The primes in the localization with respect to this set corresponds exactly to primes consisting of zero-divisors. Everything is clear. This is similar to the case of non-nilpotent elements is out of some prime ideals, or that localization with respect to a prime ideal is local.<\/li>\n<\/ol>\n