Csdrhrt

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?

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 $J subset I$.

Suppose otherwise, i.e. $Icup J$ is an ideal and $I notsubset J$ and $J notsubset I$. Let $a in I setminus J$ and $bin Jsetminus I$. 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.

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.