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Thread: Subring of R generated by X

  1. #1
    Super Member Bernhard's Avatar
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    Subring of R generated by X

    Given any ring R AND a subset X of R, we can consider the set of all subrings of R containing X and form their intersection, T, say.

    Show that T is the smallest subring of R containing X.
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  2. #2
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    Re: Subring of R generated by X

    Quote Originally Posted by Bernhard View Post
    Given any ring R AND a subset X of R, we can consider the set of all subrings of R containing X and form their intersection, T, say.

    Show that T is the smallest subring of R containing X.
    This is actually a straightforward manipulation of definitions. You'll need to prove two things: $\displaystyle T$ is a subring, then $\displaystyle T$ is the smallest subring containing $\displaystyle X$. Suppose your set of all subrings of $\displaystyle R$ containing $\displaystyle X$ is given by $\displaystyle \mathcal{S}=\{R_\alpha |\alpha \in \Lambda\}$.


    For the first one, you really just need to check that $\displaystyle r,s\in T=\displaystyle{\bigcap_{\alpha\in \Lambda} R_{\alpha}}\implies r-s,rs\in T=\displaystyle{\bigcap_{\alpha\in \Lambda} R_{\alpha}}$. The notation is a bit cumbersome, but this fact is actually quite trivial if you think about it. We also have $\displaystyle X\subseteq T$ by definition of $\displaystyle T$.



    The second bit is even easier: Let $\displaystyle R_\beta$ be a subring of $\displaystyle R$ containing $\displaystyle X$. Since this is a subring of $\displaystyle R$ containing $\displaystyle X$, we have $\displaystyle R_\beta \in \mathcal{S}$ by definition of $\displaystyle \mathcal{S}$. Using the basic property $\displaystyle A\cap B \Rightarrow A\cap B \subseteq A$ of intersections, this says that $\displaystyle T\subseteq R_\beta$
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