Abstract Nonsense

Subrings

Point of Post: In this post we define subrings and prove some basic properties (transitivity, intersections, etc.)

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Motivation

As always, soon after one defines a new structure one is morally bound to define the correlated substructure. I doubt that anyone reading this (if there is anyone) would even need to be told what the definition is. That said, there is an interesting phenomenon concerning subrings. Namely, while it turned out that normal subgroups were nicer than general subgroups they weren’t the only thing that was important. For ring theory, while a slight overstatement, the analogous statement is true. Namely, it turns out that subrings are, in many ways, unimportant and the notion of ‘normal subring’ (kernels of ring morphisms–called ideals) are what really matters. That said, it is still useful to define them for a later date.

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Subrings

Let $R$ be a ring and $S\subseteq R$. We say that $S$ is a subring of $R$ if $S$ is a subgroup of the group structure of $R$ and it’s closed under multiplication. Said more concretely, $S$ is a subring if the addition and multiplication of $R$, when restricted to $S$, make into a ring. Natural examples are $2\mathbb{Z}$ sitting inside $\mathbb{Z}$, the ring $C^1(\mathbb{R})$ of continuously differentiable functions sitting inside $C(\mathbb{R})$, etc.

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To get an idea of these subrings let’s define the analogous idea of the center of a group for a ring and show it’s a subring. Namely, if $R$ is a ring then we define the center of $R$ to be the set $Z(R)=\left\{r\in R:rs=sr\text{ for all }s\in R\right\}$.

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Theorem: Let $R$ be any ring. Then, $Z(R)$ is a subring of $R$.

Proof: Evidently if $a,b\in Z(R)$ then $(a-b)s=as-bs=sa-sb=s(a-b)$ for all $a\in R$ and so $a-b\in Z(R)$ and so $Z(R)\leqslant R$. To see it’s closed under multiplication we merely note that if $a,b\in Z(R)$ then for any $s\in R$ we have that $(ab)s=a(bs)=a(sb)=(as)b=(sa)b=sab$ and so $ab\in Z(R)$. The conclusion follows. $\blacksquare$

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We finish off this post by proving that the intersection of subrings is a subring. Indeed:

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Theorem: Let $\left\{S_\alpha\right\}$ be a collection of subrings of a ring $R$. Then, $\displaystyle \bigcap_{\alpha\in\mathcal{A}}S_\alpha$ is a subring of $R$.

Proof: Since each $S_\alpha\leqslant R$ we know from group theory that $\displaystyle \bigcap_{\alpha\in\mathcal{A}}S_\alpha\leqslant R$. Thus, it suffices to prove that $\displaystyle \bigcap_{\alpha\in\mathcal{A}}S_\alpha$ is closed under multiplication. That said, this is clear since if $a,b$ is in the intersection then $a,b\in S_\alpha$ for each $\alpha\in\mathcal{A}$ and since $S_\alpha$ is a subring we have that $ab\in S_\alpha$ for every $\alpha\in\mathcal{A}$ and so $ab$ is in the intersection. $\blacksquare$

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References:

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

June 15, 2011 -

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