proof for members of ideals 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.
 A: 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.
A: 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.
A: 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.

