Abstract Nonsense

Crushing one theorem at a time

Product of Rings (Pt. II)


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

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Unit Group of a Product of Rings

The next question we’d like to answer is whether or not we can describe the group of units \displaystyle U\left(\prod_{\alpha\in\mathcal{A}}R_\alpha\right) in terms of the groups of units U\left(R_\alpha\right). The first obvious fact is that, as sets we have:

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Theorem: Let \left\{R_\alpha\right\}_{\alpha\in\mathcal{A}} be a set of unital rings. Then, as sets

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\displaystyle U\left(\prod_{\alpha\in\mathcal{A}}R_\alpha\right)=\prod_{\alpha\in\mathcal{A}}U(R_\alpha)

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Proof: Let f be in \displaystyle \prod_{\alpha\in\mathcal{A}}U(R_\alpha) and define g to be such that g(\alpha)=f(\alpha)^{-1}. We see then that f(\alpha)g(\alpha)=1_\alpha and g(\alpha)f(\alpha)=1_\alpha for every \alpha\in\mathcal{A} and so fg=gf=1 and so f is a unit in the product with f^{-1}=g.

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Conversely, if f is in \displaystyle U\left(\prod_{\alpha\in\mathcal{A}}R_\alpha\right) then we know that f(\alpha)\in U(R_\alpha) since we know that the canonical projection is a unital homomorphism and so preserves units and f(\alpha)=\pi_\alpha(f). \blacksquare

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In particular we see that the product of rings always has a non-zero non-unitRecalling our recasting of the definition of the product when \mathcal{A} is finite we see in particular that U(R_1\times\cdots\times R_n) is setwise equal to U(R_1)\times\cdots\times U(R_n). The interesting thing though is that comparing the definition of the product of groups we get the following theorem:

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Theorem: Let \left\{R_\alpha\right\}_{\alpha\in\mathcal{A}} be a set of maps. Then, as groups

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\displaystyle U\left(\prod_{\alpha\in\mathcal{A}}R_\alpha\right)=\prod_{\alpha\in\mathcal{A}}U(R_\alpha)

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Isomorphism Issues

We would now like to discuss certain isomorphism issues which come up naturally when discussing products. Things like whether (R\times S)/(\mathfrak{a}\times\mathfrak{b})\cong (R/\mathfrak{a})\times(R/\mathfrak{b}), whether R_1\cong S_1 and R_2\cong S_2 implies R_1\times R_2\cong S_1\times S_2, etc.

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We’ve actually secretly done the first of these. Namely:

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Theorem: Let \left\{R_\alpha\right\}_{\alpha\in\mathcal{A}} be a set of rings with a set of ideals \left\{\mathfrak{a}_\alpha\right\}_{\alpha\in\mathcal{A}} with \mathfrak{a}_\alpha\in\mathfrak{L}(R_\alpha). Then,

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\displaystyle \frac{\displaystyle \prod_{\alpha\in\mathcal{A}}R_\alpha}{\displaystyle \prod_{\alpha\in\mathcal{A}}\mathfrak{a}_\alpha}\cong\prod_{\alpha\in\mathcal{A}}\frac{R_\alpha}{\mathfrak{a}_\alpha}

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Proof: We’ve already discussed that the product of the canonical epimorphism R_\alpha\to R_\alpha/\mathfrak{a}_\alpha is an epimorphism \displaystyle \prod_{\alpha\in\mathcal{A}}R_\alpha\to \prod_{\alpha\in\mathcal{A}}R_\alpha/\mathfrak{a}_\alpha which has kernel \displaystyle \prod_{\alpha\in\mathcal{A}}\mathfrak{a}_\alpha. The conclusion then follows from the first isomorphism theorem. \blacksquare

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As a corollary we find things such as (R\times S)/(R\times\{0\})\cong S.

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We finish with an obvious but important theorem:

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Theorem: Let \left\{R_\alpha\right\}_{\alpha\in\mathcal{A}} and \left\{S_\alpha\right\}_{\alpha\in\mathcal{A}} be two sets of rings with R_\alpha\cong S_\alpha then \displaystyle \prod_{\alpha\in\mathcal{A}}R_\alpha\cong\prod_{\alpha\in\mathcal{A}}S_\alpha.

Proof:  Let f_\alpha:R_\alpha\to S_\alpha be one of the guaranteed isomorphism then we now prove that

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\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha:\prod_{\alpha\in\mathcal{A}}R_\alpha\to\prod_{\alpha\in\mathcal{A}}S_\alpha

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is an isomorphism. We already know that it’s a morphism, and so it suffices to show that it’s injective and surjective. Surjectivity is obvious since if g is in then we know for each \alpha\in\mathcal{A} there exists some r_\alpha\in R_\alpha such that g(\alpha)=f_\alpha(r_\alpha) and so if we define f in the domain by f(\alpha)=r_\alpha then evidently \displaystyle \left(\prod_{\alpha\in\mathcal{A}}f_\alpha\right)(f)=g and so our map is surjective. It’s evidently injective since if f\ne g for f,g in the domain then f(\alpha)\ne g(\alpha) for some \alpha and so we know f_\alpha(f(\alpha))\ne f_\alpha(g(\alpha)) and so evidently \displaystyle \left(\prod_{\alpha\in\mathcal{A}}f_\alpha\right)(f)\ne \left(\prod_{\alpha\in\mathcal{A}}f_\alpha\right)(g) and so our map is injective. The conclusion follows. \blacksquare

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We thus have from previous theorem that:

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Theorem: If \left\{R_\alpha\right\}_{\alpha\in\mathcal{A}} and S_\alpha be two sets of rings. Then,

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\displaystyle U\left(\prod_{\alpha\in\mathcal{A}}R_\alpha\right)\cong U\left(\prod_{\alpha\in\mathcal{A}}S_\alpha\right)

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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.

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July 11, 2011 - Posted by | Algebra, Ring Theory | , , , , , , , , , ,

2 Comments »

  1. […] 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 , and the observation […]

    Pingback by Prime Ideals (Pt. IV) « Abstract Nonsense | August 17, 2011 | Reply

  2. […] ring isomorphism induces a group isomorphism on the group of units so that as groups we know . Recalling that the unit group of a product is the product group of the units we may finally conclude that . […]

    Pingback by Chinese Remainder Theorem (Pt. III) « Abstract Nonsense | September 6, 2011 | Reply


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