Prove/disprove: $I \cup J$ is (always) an Ideal of $R$.

Question

Let $I$ and $J$ be the ideals of $R$. Prove/disprove: $I \cup J$ is (always) an Ideal of $R$.

Rough Sketch: Since, $I$ and $J$ are the ideals of $R$, we have $0_R \in I$ or $0_R \in J$. Hence, $0_R \in I \cup J$ Let, $x, y \in I$ or $x,y \in J$. Then, $x-y \in I$ or $x-y \in J$. Hence, $x-y \in I \cup J$. Now, let, $r \in R$. Then, $xr,yr \in I$ or $xr, yr \in J$. Then, $xr, yr \in I \cup J$. Also, $rx, ry \in I$ or $rx, ry \in J$. Hence, $rx, ry \in I \cup J$. Therefore, $I \cup J$ is an ideal of $R$. Is this correct? I need to make sure I am on the right track.

As the answer below shows, you can't necessarily conclude that $x-y\in I\cup J,$ given arbitrary $x\in I$ and $y\in J.$ You certainly can if $x,y\in I$ or if $x,y\in J,$ though. — Cameron Buie, Mar 08 '15 at 16:29

score 5 · Accepted Answer · answered Mar 08 '15 at 16:17

5

Consider $I=2\mathbb Z,J=3\mathbb Z$

then $I\cup J$ is not an ideal

If so then

$2,3\in I\cup J \implies 2.2-3.1\in I\cup J$ (since $I\cup J$ is an ideal) but $1\notin I\cup J$

NOTE:$I\cup J$ is an ideal is an ideal $\iff$ either $I\subseteq J$ or $J\subseteq I$

answered Mar 08 '15 at 16:17

Learnmore

31,062

How did you come up with that so quick? – Jellyfish Mar 08 '15 at 16:19
1

@abcd1234 That's a very standard example. In general, $I \cup J$ is almost never an ideal. – Crostul Mar 08 '15 at 16:20
@Crostul maybe, but since I am new to abstract algebra, it takes some time to come up with a counterexample so fast. – Jellyfish Mar 08 '15 at 16:21
Dont worry @abcd1234 you will find it easy once you start spending time with it – Learnmore Mar 08 '15 at 16:24
@learnmore your note makes sense. – Jellyfish Mar 08 '15 at 16:24
@learnmore thanks. – Jellyfish Mar 08 '15 at 16:25
Happy to help @abcd1234 Moreover please accept it if you find it useful – Learnmore Mar 08 '15 at 17:35

score 0 · Answer 2 · answered Mar 08 '15 at 16:21

0

Your proof is wrong. $x,y\in I\cup J$ does not imply $x-y\in I$ or $x-y\in J$ as $x$ and $y$ need not necessarily belong to the same ideal.

answered Mar 08 '15 at 16:21

user149418

2,386

score 0 · Answer 3 · answered Mar 08 '15 at 16:26

0

It is going to absorb multiplication, but it might not be closed under addition.

If you take the intersection this is no longer a problem, since the intersection of subgroups is always a subgroup (So it will be a subgroup in the additive sense. which implies closure under addition and substraction), and it will also absorb multiplication.

answered Mar 08 '15 at 16:26

Asinomás

105,651

@TheEmperorofIceCrear Just to make myself clear, when proving something is an ideal, we always have to make sure it is a subgroup under addition but not the multiplication? – Jellyfish Mar 08 '15 at 16:41
no, it doesn't have to be a subgroup under multiplication, you have to make sure it absorbs multiplication, that is that if $i$ is in the ideal and $r$ and $s$ are in the ring then $ris$ is in the ideal. If an ideal contains the $1$ then the ideal is the hole ring. Also note that elements in the ring don't have a multiplicative inverse in general. – Asinomás Mar 08 '15 at 16:45
Thanks. I get it now. – Jellyfish Mar 08 '15 at 16:46

Prove/disprove: $I \cup J$ is (always) an Ideal of $R$.

3 Answers3