Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

If $I$ is an ideal, could you show that if $ x\in I$ and $y\notin$ I, then $x+y \notin I$? It seems like an intuitively obvious statement and yet my rigor is failing me. So if you could show me all the steps of the proof that would be much appreciated.

share|cite|improve this question
If $x + y \in I$ and $x\in I$ then $x + y - x = y \in I$ contradiction. – Tobias Kildetoft Jul 6 '12 at 6:38
This is really a fact about groups. – Dylan Moreland Jul 6 '12 at 6:38
@Tobias Perhaps you could write that as an answer so the question can be marked as answered? – Alex Becker Jul 6 '12 at 6:41
up vote 4 down vote accepted

Assume for the sake of contradiction that $x\in I$, $y\not\in I$ and $x+y\in I$. Since $I$ is a subgroup of the additive group of the ring, we have that $y = (x+y) - x \in I$ which is a contradiction. As mentioned by Dylan Moreland, this is a general fact about subgroups of any group.

share|cite|improve this answer

Another way which I find useful in dealing with things like this: Suppose that $x \in I$, $y \notin I$ but $x + y \in I$. Then saying that $x + y \in I$ implies that $x + y = m$ for some $ m \in I$. Rearrange to write $y = x - m$. Now $x \in I$ and $m \in I$. Since $I$ is an ideal and $m \in I$ we must have

$$(-1)\cdot m = -m$$

being in $I$. Furthermore $I$ being closed under addition means that $x + (-m) = x - m \in I$. However recall $y = x-m$ so this means $y \in I$ which contradicts the assumption that $y \notin I$.

$\hspace{6in}$ Q.E.D.

share|cite|improve this answer

This is a special case of the following complementary view of a subgroup.

Theorem $\ $ Let $\rm\,G\,$ be a nonempty subset of an abelian group $\rm\,H,\,$ with complement set $\rm\,\bar G = H\backslash G.\,$ Then $\rm\,G\,$ is a subgroup of $\rm\,H\iff G + \bar G\, =\, \bar G. $

Proof $\ $ $\rm\,G\,$ is a subgroup of $\rm\,H\iff G\,$ is closed under subtraction, so, complementing

$\begin{eqnarray} & &\ \ \rm G\text{ is a subgroup of }\, H\ fails\\ &\iff&\ \rm\ G\ -\ G\ \subseteq\, G\,\ \ fails\\ &\iff&\ \rm\ g_1\, -\ g_2 =\,\ \bar g\ \ \ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\ &\iff&\ \rm\ g_2\, +\ \bar g\ \ =\,\ g_1\ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\ &\iff&\ \rm\ G\ +\ \bar G\ \subseteq\ \bar G\ \ fails\qquad\ {\bf QED} \end{eqnarray}$

Instances of this are ubiquitous in concrete number systems, e.g. below. For many further examples see some of my prior posts here.

enter image description here

share|cite|improve this answer

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.