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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 non-null sets is not necessarily non-null. 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? Thanks from walter r July 11th, 2014, 06:28 AM   #3
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Good morning, Deveno! Thanks for writing. Concerning item a:
Quote:
 The intersection of two non-null sets is not necessarily non-null. For example, if A={1} and B={2}, then A∩B=∅. However, there is always ONE element of A guaranteed to be in I AND J. Can you think of what it is?
(1) Do you mean $0 \in I, 0\in J \rightarrow 0 \in I \cap J$?

Quote:
 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.
At this point I disagree.
(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. Thanks from walter r Tags demonstration, ideals, involving, ring Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Lolyta Abstract Algebra 1 September 30th, 2013 04:35 AM cartesy Algebra 3 November 6th, 2012 05:24 AM Fingolfin Math Events 10 August 26th, 2012 08:21 AM forcesofodin Abstract Algebra 1 April 8th, 2010 08:49 AM cartesy Number Theory 1 December 31st, 1969 04:00 PM

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