Union and sum of ideals is not ideal
Emily Wong 
For the union case, I'm trying $2\mathbb Z \cup3\mathbb Z$, the two are ideal of $\mathbb Z$ but in their union, when I multiply some element of $\mathbb Z$ by $2\mathbb Z \cup3\mathbb Z$ I still get it inside the union. However, I think that the union cannot be a subring, since $2+3 = 5$ which is not in the union, therefore one of the axioms broke. Is my reasoning right? I showed a counter example therefore it shouldn't be true.
For the sum case, It's harder, because I cannot do it in the way I did for the union. Any ideas?
$\endgroup$ 52 Answers
$\begingroup$The union of Ideals is not an ideal, but there is something weaker when the union for ideals works: If $(I)_{j \in J}$ is a nestled ordered family of ideals ($ I_1 \subseteq I_2 \subseteq ... \subseteq I_j \subseteq...)$ of a ring $\mathcal{A}$ then the union $\bigcup_{j \in J} I_j$ is an ideal. But the sum of ideals is always an Ideal, in fact let $I,J$ be ideals of the ring $\mathcal{A}$ you can define the sum of two ideals being the set $I + J := \{ i + j | i \in I, j \in J\}$ It is easy to show that is an ideal of $\mathcal{A}$ let $h \in I + J$ and let $a \in A$. If $h$ in an element of $I+J$ then is of the form $h=i+j$ $\Rightarrow$ $ah=a(i+j)=ai + aj$, but $ai \in I$ and $aj \in J$ because we suppose that $I,J$ are ideals, so $ah \in I + J$, for the arbitrary choice of the element $a \in \mathcal{A}$ you can conclude that $I + J$ is an ideal (left, right, two-sided) of $\mathcal{A}$.
$\endgroup$ 1 $\begingroup$For the sum you could take the product ring $(\mathbb R^2, +, \cdot)$ with the operations:
$$(x, y)+(z, w):=(x+z, y+z)\ \textrm{e}\ (x, y)\cdot (z, w)=(xz, yw).$$Then:
$$I:=\{(x, 0): x\in\mathbb R\}\ \textrm{e}\ J:=\{(0, y): y\in\mathbb R\}$$ are ideals but $I\cup J$ is not an ideal. For instance, $(1, 0)\in I$ and $(0, 1)\in J$ but $(1, 0)+(0, 1)=(1, 1)\not\in I\cup J$. Therefore, $I\cup J$ is not a soubgroup and therefore is not an ideal.
$\endgroup$
