Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been doing some reading about ideals and here is another question (to which I couldn't yet find or construct a counterexample).

Let $I, J$ be ideals in a ring $R$. Then $I\cup J$ is contained in $I+J$ but it may not be an ideal since it may not be closed under addition.

Can you give me a counterexample of ideals $I$ and $J$ in $R$ so that $I\cup J$ is not an ideal?

Note: $I\cup J\subseteq I+J$ since we can write $i \in I$ as $i+0\in I+J$ and similarly, we can write $j\in J$ as $0+j\subseteq I+ J$.

$$ $$ After reading so many great responses from this post, would this be a counter-example? Take $I=\left< 2\right>$ and $J=\left< x\right>$ in $R=\mathbb{Z}[x]$. Then $2\in I, x\in J$, but $2+x$ is not in either $I$ or $J$? After thinking about this counter-example, I don't think this is a good one: it only shows that $I\cup J$ is properly contained in $I+J$.

Thanks again for your time.

share|cite|improve this question
Is it ok to tag this commutative-algebra? – Rudy the Reindeer Jul 4 '12 at 19:48
Yes, please feel free to edit as needed. =) – math-visitor Jul 4 '12 at 19:49
up vote 7 down vote accepted

Let $R = \mathbb Z$ and $I = 2 \mathbb Z$ and $J = 5 \mathbb Z$. Then $2,5 \in I \cup J$ but $2 + 5 = 7 \notin I \cup J$.

share|cite|improve this answer
And to answer the second part of your question: Yes, $2+x \notin I \cup J$. – Rudy the Reindeer Jul 4 '12 at 19:44
Thank you Matt! Everyone on math.SE are so fast! – math-visitor Jul 4 '12 at 19:45
To add some details: $J$ looks like all polynomials with constant part zero. But $p(x) = x + 2$ has constant part $2$. Hence it's not in $J$. On the other hand, $I$ looks like all polynomials with even coefficients. But $p(x)$ has coefficient $1$ for $x$. – Rudy the Reindeer Jul 4 '12 at 19:47
@math-visitor You're welcome : ) We're all glad when we can help. – Rudy the Reindeer Jul 4 '12 at 19:47

In fact, if $I$ and $J$ are two ideals, $I\cup J$ is an ideal if and only if $I\subset J$ or $J\subset I$. Indeed, if neither of these two assertions is true, take $x\in J\setminus I$ and $y\in I\setminus J$. Then $x+y$ cannot be either in $I$ or in $J$.

So, you can pick a counter-example taking two ideals where one is not contained in the other.

share|cite|improve this answer
Thanks Davide. This is a great strategy! – math-visitor Jul 4 '12 at 19:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.