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.

Answer 1

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 $J\subseteq I$

Answer 2

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.

Answer 3

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.