Abstract Nonsense

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.