# Abstract Nonsense

## 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.