Translate of an ideal by $a$ coincides with the ideal iff $a \in I$

How can I show the following?

Show that, if $a$ and $b$ are elements of a ring $R$ and $I$ is an ideal of $R$, then $$a+I=0+I \iff a \in I$$

I am so interested to know the proof.Thanks and good night all!

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Note that this is really a statement about groups (the underlying additive abelian group of the ring, here): if $H < G$ then $aH = H$ if and only if $a \in H$. –  Dylan Moreland Feb 1 '12 at 23:04
@DylanMoreland Hence the title using the word "Translates" to convey the idea of cosets. =) –  user21436 Feb 1 '12 at 23:07

Consider the definition of $a+I$:

$$a+I=\{a+r\mid r\in I\}$$

Clearly, we have that $$0+I=\{0+r\mid r\in I\}=\{r\mid r\in I\}=I.$$

Because $I$ is an ideal, $0\in I$. Thus, for any $a\in R$, one of the elements of $a+I$ is $a+0=a$. If we knew that $a+I=0+I=I$, then $a$ is an element of $I$.

Now try showing the opposite direction on your own :)

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Beat me to it. But, I was also editing the question! +1 –  user21436 Feb 1 '12 at 23:01