《交换代数》习题研讨 (I)

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  1. Units
    , nilpotents, and idempotents lift from A/\mathfrak{N} to A.Proof: Units and nilpotents are obvious. In fact they lift to any of their representatives.
    For idempotents, if x^2=x\in A/\mathfrak{N}, then (1-x)x=0 \in A/\mathfrak{N}, so (1-x)^kx^k=0\in A for sufficiently large k. And (1-x)^k+x^k=1-x+x=1\in A/\mathfrak{N}, so lifts to a unit (1-x)^k+x^k. Moreover, its inverse u=1\in A/\mathfrak{N}. So (ux)^k(u(1-x))^k=0,ux^k+u(1-x)^k=1\in A and ux=x,u(1-x)=1-x\in A/\mathfrak{N}.
    This can be interpreted by sheaf theory, which is to be discussed in later posts.
  2. Prime ideals of A_1\times...\times A_n is of the form A_1\times...\times p_i\times ... \times A_n, where p_i is a prime ideal of A_i. What about countable products? (Profinite exists. Boolean Ring)

    Proof:
     Multiplying by (0,...,1,...,0) we see I=I_1\times...\times I_n. Then (A_1\times...\times A_n)/I=A_1/I_1\times...\times A_n/I_n. It is a domain iff n-1 factors are 0 and the other is a domain. Actually the index set does not matter, as this is a product. Direct sums are of interest, and we will discuss it later.
    The projection onto each factor corresponds geometrically to inclusion into the disjoint union. Multiplication by (0,...,1,...,0) means restrict the function to i-th component. The above demonstrates that ideals of a product works independently on factors, and so the subset is irreducible, iff it is restricted in one part, and irreducible there.
    1. Let f:A\rightarrow B be surjective. Then f(\mathfrak{R}(A))\subseteq \mathfrak{R}(B). The inclusion may be strict. What about \mathfrak{N}?
    2. If A is semilocal then the above is an equality.

    Proof:

    1. Since 1+f^{-1}(b) a is invertible, so is 1+b f(a) for all b\in B. Let f be the quotient map from a domain A by some principle ideal generated by a power. Then \mathfrak{R}\supseteq \mathfrak{N}\supsetneq (0)=f(\mathfrak{R}(A)).
      For non-surjective morphisms, the two thing may have no relation at all. For example, let A be a local domain and f the embedding into B, its field of fractions. Then f(\mathfrak{R}(A))=\mathfrak{R}(A) is very large but \mathfrak{R}(B)=0.
      Since prime ideals always pull back, we always have f(\mathfrak{N}(A))\subseteq \mathfrak{N}(B). For Jacobson radicals, the reason actually is the same since when f 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}, this reduces to trivial intuition.
    2. Denote the kernel by I and the collection of maximal ideals \mathcal{M}. It is equivalent to \cap_{\mathfrak{m} \in\mathcal{M}}\mathfrak{m} + I=\cap_{\mathfrak{m}\supseteq I}\mathfrak{m}. Passing to A/\cap_{\mathfrak{m} \in\mathcal{M}} \mathfrak{m}\cong \prod_{\mathfrak{m} \in\mathcal{M}}  A/\mathfrak{m}, it is equivalent to I=\cap_{\mathfrak{m}\supseteq I}\mathfrak{m}. This is a product of fields, so by 2. above, all ideals are products of the whole field or 0. I has 0 in the components of \mathfrak{m}\supseteq I while k_i otherwise, which is exactly equal to \cap_{\mathfrak{m}\supseteq I}\mathfrak{m}. This does not work when |\mathcal{M}| is infinite, because then Chinese remainder theorem does not hold.
      Continuing the discussion of a., 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}, p vanishes on the single point \mathrm{Spec}(\mathbb{Z}/p^2\mathbb{Z}), but cannot be induced by some elements vanishing on all of \mathrm{Spec}\mathbb{Z}: such elements must be 0. 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}), this holds, as we can always pull back to (2,3,5,...).
  3. An integral domain A is a UFD iff both of the following are satisfied:
    1. Every irreducible element is prime.
    2. Principle ideals satisfy A.C.C.


    Proof:
     For UFDs, it is crystal clear that these are satisfied. Conversely, we can easily split a into a finite product of irreducible elements, by A.C.C.. The product is unique because irreducibles are prime. We should care about the cases when irreducible element is not prime.

  4. Let \{P_{\lambda}\}_{\lambda\in\Lambda} be a non-empty totally ordered (by inclusion) family of prime ideals. Then \cap P_{\lambda} is prime. Thus for any ideal I, there is some minimal prime ideal containing I.
    Proof: If ab\in\cap P_{\lambda}, then for all \lambda, either a,b is in P_{\lambda}. So the one of the collections of primes containing a and b respectively is not bounded below. Thus either of a,b is in the intersection. The corollary then follows from Zorn’s lemma.
  5. Let A be a ring, P_1,...,P_r ideals. Suppose r-2 of them are prime. Then if I\subseteq \cup_i P_i, then \exists i:I\subseteq P_i.
    Proof: This is mysterious. Proof is not hard, but I do not know why. I will write when I know its meaning or usage.
  6. In a ring A, if every ideal I\subsetneq \mathfrak{N} contains a nonzero idempotent, then \mathfrak{N}=\mathfrak{R}.
    Proof: Notice when A is reduced, this amounts to say if every ideal contains a nonzero idempotent, then \mathfrak{R}=0: If a\ne 0, then (a) contains a nonzero idempotent e=ka, with e(1-e)=0, so 1-ka is not a unit, and a\notin R. The general case follows by passing to A/\mathfrak{N}. But this is more like an awkward exercise.
  7. A local ring contains no idempotents \ne 0,1.
    Proof: Otherwise it would split as a direct product. By 2. above, it has at least two maximal ideals. Geometrically, a local picture cannot be a disjoint union.
  8. The ideal \mathfrak{Z} of zero-divisors is a union of prime ideals.
    Proof: Non-zero-divisors form a multiplicative set: If a,b are not zero-divisors, and a b x=0, we have b x=0 and x =0. 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.

The topics are from Matsumura, 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 Atiyah, 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.

Friday, August 2, 2013

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