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{ }

and

\text{ }

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

\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

\text{ }

\text{ }

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.

Advertisements

June 27, 2011 - Posted by | Algebra, Ring Theory | , , , , , , , ,

3 Comments »

  1. […] now use the universal characterization of quotient rings to prove the first ring isomorphism […]

    Pingback by First Ring Isomorphism Theorem « Abstract Nonsense | June 27, 2011 | Reply

  2. […] to form the ‘product ring’ of the set of rings . We will then show that similarly to quotient rings there is a universal mapping property which characterizes product rings. We will also discuss the […]

    Pingback by Product of Rings (Pt. I) « Abstract Nonsense | July 11, 2011 | Reply

  3. […] knowing what we do about rings it’s not shocking that there is a universal characterization of quotient rings (i.e. a set […]

    Pingback by Quotient Modules « Abstract Nonsense | November 7, 2011 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: