If I and J are two ideals, I∪J is an ideal if and only if I⊂J or J⊂I. How can we prove the subset part?

Question

If $I$ and $J$ are two ideals, $I \cup J$ is an ideal if and only if $I \subset J$ or $J \subset I$. How can we prove the subset part?

What would the proof look like to prove that either $I$ is a subset of $J$ or vice versa?

Answer 1

By the contrapositive, suppose that $I$ and $J$ are both ideals such that neither $J \subset I$ nor $I \subset J$. Thus, there exist elements $i \in I \setminus J$ and $j \in J \setminus I$. Now, this implies that $i+j$ is not in any of the ideals: if it were $i+j \in I$, then we would have that $j = (-1)i + (i +j) \in I$, and likewise for $J$. Hence $i,j \in I \cup J$ but $i+j \not \in I \cup J$, which proves that $I \cup J$ is not an ideal.

A direct approach: suppose that the union is an ideal and that $I \not \subset J$, so that we can choose $i \in I \setminus J$. Now, for each $j \in J$, we know that $i+j \in I \cup J$ so either $i+j \in I$ or $i+j \in J$. It can't be that $i+j \in J$ because it would imply $i \in J$. Hence $i+j \in I$ and thus $j = (-1)i + (i+j) \in I$, which proves that $J \subset I$.

Answer 2

Suppose otherwise, i.e. $I\cup J$ is an ideal and $I \not\subset J$ and $J \not\subset I$. Let $a \in I \setminus J$ and $b\in J\setminus I$. Consider $c = a + b$. Then $c \in I \cup J$. If $c \in I$, then $b = c - a \in I$ a contradiction. On the other hand, if $c \in J$, then $a = c - b \in J$ a contradiction.

Answer 3

If

$I \subset J \tag 1$

then

$J \subset I \cup J \subset J, \tag 2$

whence

$I \cup J = J, \tag 3$

so $I \cup J$ is the ideal $J$; likewise, if

$J \subset I, \tag 4$

then

$I \subset I \cup J \subset I, \tag 5$

and $I \cup J = I$, and $I \cup J$ is the ideal $I$; so $I \cup J$ is an ideal provided one of (1), (4) holds.

Now suppose $I \cup J$ is an ideal but that

$I \not \subset J, \; J \not \subset I; \tag 6$

then

$\exists i \in I, \; i \notin J; \; \exists j \in J, \; j \notin I; \tag 7$

we note that

$i, j \in I \cup J, \tag 8$

whence, since $I \cup J$ is an ideal,

$i + j \subset I \cup J, \tag 9$

which yields

$i + j \in I, \tag{10}$

or

$i + j \in J; \tag{11}$

in the former case,

$j = (i + j) - i \in I, \tag{12}$

whilst in the latter,

$i = (i + j) - j \in J; \tag{13}$

but either of these conclusions contradicts (7), whence we find that $I \cup J$ cannot be an ideal in $R$ unless at least one of

$I \subset J, \; J \subset I \tag{14}$

holds.

Note Added in Edit, Tuesday 24 March 2020 12:33 PM PST: This result apparently holds if $I$ and $J$ are left ideals, right ideals, or two sided ideals, or even if they are ideals of different types. Indeed, it appears this is really a statement about subgroups of $\langle R, +, 0 \rangle$, the abelian group of elements of $R$. End of Note.