Abstract Nonsense

Crushing one theorem at a time

Quotient Rings (Pt. II)

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

\text{ }

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

\text{ }

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
\text{ }

 \begin{array}{ccc} R & \xrightarrow{\;\phi\;} & Q\\ {^f}\big\downarrow & \swarrow _\eta & \\ S & &\end{array}

\text{ }

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,

\text{ }

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

\text{ }


\text{ }


\text{ }

 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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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 - Posted by | Algebra, Ring Theory | , , , , , , , ,


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