Basic Ring Theory - 6

Groups, Rings, Domains, Modules, etc, Galois theory
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Tsakanikas Nickos
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Joined: Tue Nov 10, 2015 8:25 pm

Basic Ring Theory - 6

#1

Post by Tsakanikas Nickos »

We follow the conventions of the previous posts regarding the ring \( \displaystyle A \).

Let \( M \) be an \( A \)-module and let \( N \) and \( P \) be submodules of \( M \). Show that \[ \displaystyle ( N \, \colon P ) \; \; \colon = \left\{ a \in A \; \Big | \; aP \subseteq N \right\} \]is an ideal of \( A \). If we set \( N=0 \) and \( P=M \), then we get the annihilator \( Ann(M) \) of \( M \). If \( I \) is an ideal of \( A \) such that \( I \subseteq Ann(M) \), then show that \( M \) has an \( A / I \)-module structure.


Additionally, prove the following:
  1. \( Ann(M+N)=Ann(M) \cap Ann(N) \)
  2. \( ( N \, \colon P ) = Ann( (N+P)/P ) \)

The post was edited, because it contained a mistake (Instead of \( M \) I had written \( Ann(M) \) in the second part - now the correct statement is marked in red)
Last edited by Tsakanikas Nickos on Sat Dec 12, 2015 3:09 pm, edited 2 times in total.
Papapetros Vaggelis
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Re: Basic Ring Theory - 6

#2

Post by Papapetros Vaggelis »

Obviously, \(\displaystyle{0\in (N:P)}\) . Let \(\displaystyle{x\,,y\in (N:P)}\) and \(\displaystyle{a\in A}\) .

We have that \(\displaystyle{x\,P\subseteq N\,,y\,P\subseteq N}\) .

If \(\displaystyle{z\in (x-y)\,P}\) and \(\displaystyle{w\in (a\,x)\,P}\), then :

\(\displaystyle{z=(x-y)\,p\,,p\in P\,\,,w=(a\,x)\,q\,,q\in P}\), or :

\(\displaystyle{z=x\,p-y\,p\,\,,w=a\,(x\,q)}\), where \(\displaystyle{x\,p\in x\,P\,,y\,p\in y\,P\,,x\,q\in x\,P}\)

that is \(\displaystyle{x\,p\,,y\,p\in N\,,x\,q\in N}\) and since \(\displaystyle{N}\) is submodule, we get :

\(\displaystyle{z\in N\,,w\in N}\). So, \(\displaystyle{(x-y)\,P\subseteq N\,,(a\,x)\,P\subseteq N}\)

which means that \(\displaystyle{x-y\in (N:P)\,,a\,x\in (N:p)}\) and \(\displaystyle{(N:P)}\) is an

ideal of \(\displaystyle{A}\) .

Some part of this post was deleted due to erroneous statement as asked by Vaggelis and Nickos.

T:
Tsakanikas Nickos
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Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Re: Basic Ring Theory - 6

#3

Post by Tsakanikas Nickos »

My apologies, Vaggelis, for the mistake in the statement of the second part of the exercise!

Define a map
\[ \displaystyle \mu \ \colon A/I \times M \longrightarrow M \; , \; \left( a + I , m \right) \mapsto \left(a+I \right) \star m := am \]
Then \( \star \) is well-defined, because \( I \subseteq Ann(M) \). Specifically:
\[ \displaystyle (a+I,m)=(b+I,n) \implies a-b \in I \, , \, m=n \implies (a-b)m=0 \implies am=bm=bn \implies (a+I) \star m = (b+I) \star n \]
It is now easy to verify that \( ( M , \mu ) \) is an \( A/I \)-module.
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