# Abstract Nonsense

## Quotient Rings (Pt. II)

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

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What shall now see though is that this property characterizes $R/\mathfrak{a}$ in a sense. In particular:

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Theorem (Universal Characterization of Quotient Rings): Let $R$ be a ring and $\mathfrak{a}\in\mathfrak{L}(R)$. Suppose that $Q$ is a ring with a fixed epimorphism $\phi:R\twoheadrightarrow Q$ with $\ker\phi=\mathfrak{a}$. Moreover, suppose that $Q$ is such that whenever there exists a morphism $f:R\to S$, for some ring $S$, with $\mathfrak{a}\subseteq\ker f$ there exists a unique morphism $\eta:Q\to S$ such that the following diagram commutes
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$\begin{array}{ccc} R & \xrightarrow{\;\phi\;} & Q\\ {^f}\big\downarrow & \swarrow _\eta & \\ S & &\end{array}$

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Then, $Q\cong R/\mathfrak{a}$.

Proof: Consider that  since $\pi:R\to R/\mathfrak{a}$ and $\ker \pi=\mathfrak{a}$ we have that there exists a unique morphism $\eta:Q\to R/\mathfrak{a}$ with $\pi=\eta\circ \phi$. Now, since $\phi:R\to Q$ and $\ker \phi=\mathfrak{a}$ we know there exists a unique morphism $\xi:R/\mathfrak{a}\to Q$ such that $\phi=\xi\circ\pi$. Thus,

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$\text{id}_Q\circ\phi=\phi=\xi\circ \pi=\xi\circ\eta\circ\phi\quad\mathbf{(1)}$

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and

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$\text{id}_{R/\mathfrak{a}}\circ\pi=\eta\circ\phi=\eta\circ\xi\circ\pi\quad\mathbf{(2)}$

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Now, since $\phi$ is an epimorphism we may conclude from $\mathbf{(1)}$ that $\text{id}_Q=\xi\circ\eta$ and since $\pi$ is an epimorphism we may conclude from $\mathbf{(2)}$ that $\text{id}_{R/\mathfrak{a}}=\eta\circ\xi$. Thus, $\eta$ and $\xi$ are isomorphisms with $\eta^{-1}=\xi$. The conclusion follows. $\blacksquare$

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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.

June 27, 2011 -