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EDIT: Let $A$ be an ideal $S$ be any subset of commutative ring $R$. Define $AS$ to be of the form $\sum{as}$ for every $a\in A, s\in S$. Then $AS$ is an ideal. This is easy to see.

Let $A,B$ be ideals of commutative ring $R$. I'm trying to determine the ideal $(A+B)(A\cup B)$. I'm getting $$(A+B)(A\cup B)=A+B$$ which I feel is unlikely as $A\cup B$ need not be $R$. How I'm getting here is

  1. $(A+B)(A\cup B)=A(A\cup B)+B(A\cup B)$.

  2. $A(A\cup B)+B(A\cup B)=(A\cup AB)+(B\cup AB)=A+B$.

Which of these steps is wrong?

Thanks in advance!

EDIT: Now I feel, $A(A\cup B)\supseteq AB\cup AC$, which brings me to the even more unlikely conclusion that $(A+B)(A\cup B)\supseteq A+B$.

  • What do you mean by $IS$ when $I$ is an ideal and $S$ is a subset(!) of a ring? – Martin Brandenburg Oct 30 '13 at 10:14
  • When both $I$ and $S$ are ideals, $IS=\sum{is}$, for all $i\in I$ and $s\in S$, which makes $IS$ an ideal. –  Oct 30 '13 at 10:15
  • For $R = \Bbb Z, A = 2\Bbb Z, B = 3\Bbb Z$, we have $A + B = \Bbb Z$ while $(A + B)(A\cup B) = A \cup B$ doesn't contain $1$. Are you sure you mean union $\cup$? Because the union of ideals isn't necessarily an ideal (it's not in general closed under addition, so it's not even a subring). The intersection $\cap$ of two ideals, however, is an ideal. Is that what you meant? – Arthur Oct 30 '13 at 10:15
  • @Arthur- Yes I do mean $\cup $. –  Oct 30 '13 at 10:18
  • @MartinBrandenburg- Terribly sorry I just got what you meant. –  Oct 30 '13 at 10:41
  • Although I suppose that $IS$ could still be an ideal were it defined in the fashion: every element of $IS$ is of the form $\sum{is}$ for all $i\in I$ and $s\in S$. –  Oct 30 '13 at 10:45
  • I edited the description. Kindly take a look at the modified question. Thanks. –  Oct 30 '13 at 11:07

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The first non-sequitur that caught my eye is:

$A(A\cup B)+B(A\cup B)=(AA\cup AB)+(BB\cup AB) \neq (A\cup AB)+(B\cup AB)$.

In general, $AA \neq A$. For example, consider any nonzero proper ideal in $\mathbb Z$, e.g., $(2)$.