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July 10th, 2014, 11:58 AM  #1 
Member Joined: Mar 2013 Posts: 71 Thanks: 4  Demonstration involving ideals of a ring
Please, check my solution to the following exercise: "Let $I \subset A$ and $J \subset A$ be ideals of $A$. Show that: a) $I \cap J$ is an ideal of A. b) $I+J=\left\{x+y;x \in I, y \in J \right \}$ is an ideal of $A$. a) since $I$ and $J$ are ideals, then $I \neq \varnothing$ and $J \neq \varnothing$. It implies that $I \cap J \neq \varnothing$. $a,b \in I \cap J \rightarrow a,b \in I$ and $a,b \in J \rightarrow a+b \in I \cap J$ and $ab \in I \cap J$. Hence $I \cap J$ is an ideal. b) $I \neq \varnothing$ and $J \neq \varnothing \rightarrow I+J \neq \varnothing$. $a,b \in I+J \rightarrow a= x+y; b= v+w; x,v \in I; y,w \in J$. But then $a+b =(x+v)+(y+w) \in I + J$, because $(x+v) \in I$ and $(y+w) \in J$. Additionaly, if $a,b \in I+J$ then $ab= xv+xw+yv+yw$ with $xv \in I; yw \in J; xw \in I \cap J; yv \in I \cap J$. Then $ab=(xv+xw)+(yv+yw) \in I+J$. Hence $I+J$ is an ideal. Thanks! 
July 10th, 2014, 12:55 PM  #2 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 
The intersection of two nonnull sets is not necessarily nonnull. For example, if $A = \{1\}$ and $B = \{2\}$, then $A \cap B = \emptyset$. However, there is always ONE element of $A$ guaranteed to be in $I$ AND $J$. Can you think of what it is? Again, just because two elements of two sets are in both of them, does not guarantee that the sum of these two elements is in EITHER set. Suppose we take: $A = \{1,2,4\}$ and $B = \{1,2,5\}$. Then $1,2 \in A \cap B$, but $1 + 2 = 3$ is not in EITHER set. You are failing to invoke the proper properties of ideals in your proof. You're just stating that what you HOPE to be true IS true. That just won't do. THIS is what you have to prove for (a): 1. $I \cap J$ is an additive subgroup of $A$. 2. If $a \in A$, and $x \in I \cap J$, that $ax,xa \in I \cap J$. I'll show you how a proof of (2) starts. Let $a \in A$, and $x \in I \cap J$. Since $I \cap J \subseteq I$, we have $x \in I$. Since $I$ is an ideal, $ax \in I$. What do you suppose comes next? 
July 11th, 2014, 06:28 AM  #3  
Member Joined: Mar 2013 Posts: 71 Thanks: 4 
Good morning, Deveno! Thanks for writing. Concerning item a: Quote:
Quote:
(2) $a,b \in I \cap J \rightarrow a,b \in I$ and $a,b \in J$. But $I,J$ are ideals, therefore $a+b \in I$ and $a+b \in J$. Hence $a+b \in I \cap J$. This is not true for any set, but it is true for an ideal. After reading your post I redo the other step the following way: (3) $a \in I \cap J, n \in A \rightarrow a \in I, a \in J, n \in A \rightarrow na \in I, na \in J$ (by hypothesis that both I and J are ideals)$ \rightarrow na \in I \cap J$. (1),(2),(3) together imply $I \cap J$ is an ideal, similarly for item b. Do you think now it is convincent?  
July 11th, 2014, 07:19 AM  #4 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 
Yes, that is what was missing: you need to invoke the CLOSURE properties of ideals. This may have been what you were thinking, but until you write it down, no one else knows that.


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demonstration, ideals, involving, ring 
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