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 September 16th, 2012, 05:02 PM #1 Newbie   Joined: Oct 2009 Posts: 14 Thanks: 0 ideal hey lets say that (a, b) is an ideal in R. What exactly is that set? How can i describe an element of (a, b)? I would like to know this, because i have to show that some kind of set is equivalent to an ideal (generated by 2 elements) and i haven't got a clue how to do that :-s help is appreciated very very very much!
 September 16th, 2012, 10:04 PM #2 Senior Member   Joined: Aug 2010 Posts: 195 Thanks: 5 Re: ideal The ideal $(a,b)$ is the same as the set $aR + bR$. It is the set of all sums of elements $a \cdot r_1 + b \cdot r_2$ for any $r_1,r_2 \in R$. Note that $0 \in R$, so elements of the form $a \cdot r_1 + b \cdot 0= a \cdot r_1$ and $a \cdot 0 + b \cdot r_2= b \cdot r_2$ are in $(a,b)$, hence $(a) \subseteq (a,b)$ and $(b) \subseteq (a,b)$.
 September 17th, 2012, 08:03 PM #3 Newbie   Joined: Oct 2009 Posts: 14 Thanks: 0 Re: ideal Thank you! But what is then the difference between the ideal (a, b) and the sum of the ideals (a) and (b)? And if we multiply the ideal with itself? Is it then the same ideal?
 September 20th, 2012, 07:47 PM #4 Senior Member   Joined: Aug 2010 Posts: 195 Thanks: 5 Re: ideal There is no difference at all between $(a,b)$ and $(a) + (b)$, they are the same sets. The product $(a)(b)$ is the set of finite sums $\sum_{i=1}^n x_iy_i$ where $x_i=a \cdot x#39;_i \in (a)$ and $y_i=b \cdot y#39;_i \in (b)$, so the product is actually given by finite sums $\sum_{i=1}^n ab \cdot x'_iy'_i = ab \sum_{i=1}^n x'_iy#39;_i$, which is to say that $(a)(b)=(ab)$. So in order that $(a)(a)= (a)$, it must be that $(a^2)=(a)$ which is unlikely, but not impossible depending on the ring.

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