Basic Ring Theory - 14
-
- Community Team
- Posts: 314
- Joined: Tue Nov 10, 2015 8:25 pm
Basic Ring Theory - 14
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 \[ \displaystyle \prod_{i=1}^{n} A_{i} = (a_{1}) \dots (a_{n}) = (a_{1} \dots a_{n} ) \]
-
- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Re: Basic Ring Theory - 14
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,
\(\displaystyle{\begin{aligned} x&=r_1\,a_1\,s_1\,a_2+r_2\,a_1\,s_2\,a_2+...+r_m\,a_1\,s_m\,a_2\\&=(r_1\,s_1)\,(a_1\,a_2)+(r_2\,s_2)\,(a_1\,a_2)+...+(r_m\,s_m)\,(a_1\,a_2)\\&=\sum_{i=1}^{m}(r_i\,s_i)\,(a_1\,a_2)\end{aligned}}\).
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,
\(\displaystyle{\begin{aligned} x&=r_1\,a_1\,s_1\,a_2+r_2\,a_1\,s_2\,a_2+...+r_m\,a_1\,s_m\,a_2\\&=(r_1\,s_1)\,(a_1\,a_2)+(r_2\,s_2)\,(a_1\,a_2)+...+(r_m\,s_m)\,(a_1\,a_2)\\&=\sum_{i=1}^{m}(r_i\,s_i)\,(a_1\,a_2)\end{aligned}}\).
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 9 guests