Abstract Nonsense

Crushing one theorem at a time

Prime Ideals (Pt. IV)


Point of Post: This is a continuation of this post.

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Miscellaneous Facts

We now record some of the basic miscellaneous facts concerning prime ideals, things like what the spectrum of a product looks like and that morphisms pull back prime ideals:

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Theorem: Let R_1,\times,R_n be a collection of commutative unital rings. Then,

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\displaystyle \text{Spec}\left(R_1\times\cdots\times R_n\right)=\left\{\prod_{\alpha\in[n]}A_\alpha:\text{There exi}\text{sts }\alpha_0\in[n]\text{ such that }A_{\alpha_0}\in\text{Spec}(R_{\alpha_0})\text{ and }A_\alpha=R_\alpha\text{ for }\alpha\ne\alpha_0\right\}

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Proof: This follows immediately from the characterization of prime ideals as being those whose quotient ring is an integral domain, the fact that the ideals of a product of commutative unital rings are of the form \mathfrak{a}_1\times\cdots\times\mathfrak{a}_n, and the observation that (R_1\times\cdots\times R_n)/(\mathfrak{a}_1\times\cdots\times\mathfrak{a}_n)\cong (R_1/\mathfrak{a}_1)\times\cdots\times (R_n/\mathfrak{a}_n). \blacksquare

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Theorem: Let R and S be commutative rings and f:R\to S a morphism. Then, if \mathfrak{p}\in\text{Spec}(S) one has that f^{-1}(\mathfrak{p})=R or f^{-1}(\mathfrak{p})\in\text{Spec}(R).

Proof: Suppose that a,b\notin f^{-1}(\mathfrak{p}) then f(a),f(b)\notin\mathfrak{p} and so f(a)f(b)=f(ab)\notin\mathfrak{p} and thus ab\notin\mathfrak{p}. The conclusion readily follows. \blacksquare

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By considering the inclusion map from a subring S into a ring R we get the following corollary:

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Corollary: Let R be a commutative ring, S a subring, and \mathfrak{p}\in\text{Spec}(R). Then, \mathfrak{p}\cap S\in\text{Spec}(S).

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We last make an obvious note about how a prime ideal can end up in a product of ideals which mirrors precisely the case for the primes in \mathbb{Z}:

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Theorem: Let R be a commutative ring, \mathfrak{a}_1,\cdots,\mathfrak{a}_n ideals of R. If \mathfrak{p}\in\text{Spec}(R) and \mathfrak{p}\supseteq\mathfrak{a}_1\cdots\mathfrak{a}_n then \mathfrak{p}\supseteq\mathfrak{a}_j for some j.

Proof: By way of contradiction assume that there exists some a_j\in\mathfrak{a}_j-\mathfrak{p} fore ach j\in[n], then by assumption that \mathfrak{p} is prime we find that a_1\times a_n\notin\mathfrak{p}\supseteq\mathfrak{a}_1\cdots\mathfrak{a}_n which is an evident contradiction. \blacksquare

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

3.  Reid, Miles. Undergraduate Commutative Algebra. Cambridge: Cambridge UP, 1995. Print.

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August 17, 2011 - Posted by | Uncategorized | , , , , , ,

1 Comment »

  1. […] be a commutative unital ring and a unitally mulitplicative subset.  We define an equivalence relation on given by if and only if there exists with . This is […]

    Pingback by Localization (Pt. I) « Abstract Nonsense | September 29, 2011 | Reply


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