Basic Ring Theory - 3

Groups, Rings, Domains, Modules, etc, Galois theory
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Tsakanikas Nickos
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Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Basic Ring Theory - 3

#1

Post by Tsakanikas Nickos »

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

Let \( \displaystyle \mathfrak{a} \) and \( \displaystyle \mathfrak{b} \) be two ideals of a ring \( \displaystyle A \). Prove the following:
  • If \( \displaystyle \mathfrak{a} \) and \( \displaystyle \mathfrak{b} \) are coprime, then \( \displaystyle \mathfrak{a} \mathfrak{b} = \mathfrak{a} \cap \mathfrak{b} \).
  • \( \displaystyle \mathfrak{a} \) and \( \displaystyle \mathfrak{b} \) are coprime if and only if there exist \( \displaystyle x \in \mathfrak{a} \) and \( \displaystyle y \in \mathfrak{b} \) such that \( \displaystyle x + y = 1_{A} \).
Can you generalise the second result (that is, find a characterisation of "coprime") for a finite family \( \displaystyle \{ \mathfrak{a}_{i} \} _{i=1}^{n} \) of ideals of \( A \)?
Papapetros Vaggelis
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Joined: Mon Nov 09, 2015 1:52 pm

Re: Basic Ring Theory - 3

#2

Post by Papapetros Vaggelis »

Hi Nickos.

1st part

If \(\displaystyle{a}\) and \(\displaystyle{b}\) are coprime, then \(\displaystyle{A=a+b}\) .

So, \(\displaystyle{1\in A\implies \left(\exists\,x\in a\right)\,\left(\exists\,y\in b\right): x+y=1}\) .

If \(\displaystyle{r\in a\cap b}\), then :

\(\displaystyle{r=r\,1=r\,x+r\,y=x\,r+r\,y\in a\,b}\) since \(\displaystyle{x\in a\,,r\in a\cap b\,,y\in b}\) .

Therefore, \(\displaystyle{a\cap b\subseteq a\,b}\) .

On the other hand, if \(\displaystyle{r\in a\,b}\) , then \(\displaystyle{r=\sum_{i=1}^{n}x_{i}\,y_{i}}\) for some \(\displaystyle{n\in\mathbb{N}}\), where :

\(\displaystyle{x_{i}\in a\,,y_{i}\in b\,,1\leq i\leq n}\) .

Since \(\displaystyle{a\,,b}\) are ideals of \(\displaystyle{\left(A,+,\cdot\right)}\), we get :

\(\displaystyle{\forall\,i\in\left\{1,...,n\right\}: x_{i}\,y_{i}\in a\implies r=\sum_{i=1}^{n}x_{i}\,y_{i}\in a}\)


\(\displaystyle{\forall\,i\in\left\{1,...,n\right\}: x_{i}\,y_{i}\in b\implies r=\sum_{i=1}^{n}x_{i}\,y_{i}\in b}\)

which means that \(\displaystyle{r\in a\cap b}\), so: \(\displaystyle{a\,b\subseteq a\cap b}\) .

Finally, \(\displaystyle{a\,b=a\cap b}\) .

2nd part

If \(\displaystyle{a\,,b}\) are coprime, then, as I said above, \(\displaystyle{1=x+y}\) for some \(\displaystyle{x\in a\,,y\in b}\) .

Now, if \(\displaystyle{x+y=1}\), where \(\displaystyle{x\in a\,,y\in b}\), then, using the fact that \(\displaystyle{a+b\subseteq A}\),

\(\displaystyle{a\,,b}\) are ideals of \(\displaystyle{(A,+,\cdot)}\) and the following relation

\(\displaystyle{\forall\,r\in A: r=r\,1=r\,x+r\,y\in a+b}\),

we have that \(\displaystyle{A=a+b}\), that is, \(\displaystyle{a\,,b}\) are coprime.

Generalization

(I am not so sure about my answer, so, please help me if something goes wrong).

If \(\displaystyle{\left\{a_{i}\right\}_{i=1}^{n}}\) is a finite family of ideals of the commutative ring \(\displaystyle{(A,+,\cdot)}\),

then,

\(\displaystyle{a_{i}}\) and \(\displaystyle{\sum_{k=1\,,k\neq i}^{n}a_{k}}\) are coprime if, and only if,

\(\displaystyle{1=\sum_{k=1}^{n}x_{k}}\) for some \(\displaystyle{x_{k}\in a_{k}\,,1\leq k\leq n}\)

and the above is true for every \(\displaystyle{i\in\left\{1,...,n\right\}}\) .
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