Basic Ring Theory - 14

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 - 14

#1

Post by Tsakanikas Nickos »

Let \( A \) be a commutative ring with unity \( 1_{A} \). Let \( \left\{ A_{i} = (a_{i}) \right\}_{i=1}^{n} \), where \( a_{i} \in A , \, 1 \leq i \leq n \), be principal ideals of \( A \). Show that \[ \displaystyle \prod_{i=1}^{n} A_{i} = (a_{1}) \dots (a_{n}) = (a_{1} \dots a_{n} ) \]
Papapetros Vaggelis
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Joined: Mon Nov 09, 2015 1:52 pm

Re: Basic Ring Theory - 14

#2

Post by Papapetros Vaggelis »

Obviously, \(\displaystyle{\left(a_1\cdot ...\cdot a_n\right)\subseteq (a_1)\cdot ...\cdot (a_n)}\).

Let \(\displaystyle{x\in (a_1)\cdot ...\cdot (a_n)}\). Then,

\(\displaystyle{x=\sum_{i=1}^{m}\,\prod_{j=1}^{n}x_{ij}}\), where \(\displaystyle{x_{ij}\in (a_{j})}\).

So, \(\displaystyle{x_{ij}=r_{ij}a_{j}\,,r_{ij}\in A}\). Using the commutativity, we get

\(\displaystyle{x=\sum_{k=1}^{m\,,n}r_{k}\,(a_1\cdot...\cdot a_n)\in \left(a_1\cdot ...a_{n}\right)\,\,(\star)}\).

Finally, \(\displaystyle{(a_1)\cdot ...\cdot (a_n)=(a_1\cdot...\cdot a_n)}\).

In order to explain the relation \(\displaystyle{(\star)}\) better, let us take \(\displaystyle{n=2}\).

Then, if \(\displaystyle{x\in (a_1)\cdot (a_2)}\), then,

\(\displaystyle{x=x_1\,y_1+x_2\,y_2+...+x_m\,y_m\,,x_i\in (a_1)\,,y_j\in (a_2)\,,1\leq i\,,j\leq m}\).

But, \(\displaystyle{x_i=r_i\,a_1\,,y_j=s_j\,a_2}\) and now,

\(\displaystyle{\begin{aligned} x&=r_1\,a_1\,s_1\,a_2+r_2\,a_1\,s_2\,a_2+...+r_m\,a_1\,s_m\,a_2\\&=(r_1\,s_1)\,(a_1\,a_2)+(r_2\,s_2)\,(a_1\,a_2)+...+(r_m\,s_m)\,(a_1\,a_2)\\&=\sum_{i=1}^{m}(r_i\,s_i)\,(a_1\,a_2)\end{aligned}}\).
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