Alternate expression for definition of an Ideal?

[PsychonautQQ]

If A is an additive subgroup of a ring R, A is said to be an ideal if Ra is contained in A for all a in A; that is, if every multiple of an element of A is again in A.

Is it true that A is an ideal of R if Ar is contained in A for all r in R? To me it seems like they are equivalent statements but I'm not sure.

Update: Second thought, I believe these are NOT equivalent statements. I'm looking at the proof that explains why if 1 is in A then A = R, and you could not employ this argument under the faulty definition I tried to push. Anyone want to shed further light on the difference between these statements?

 

[Stephen Tashi]

Is this just the difference between a "right ideal" and "left ideal"? http://mathworld.wolfram.com/RightIdeal.html

 

[PsychonautQQ]

Ah right, didn't mean to switch those around. Would it be fair to say that the following statements define the same ideal A?

Ar is contained in A for all r in R
aR is contained in A for all a in A

 

[lavinia]

Ah right, didn't mean to switch those around. Would it be fair to say that the following statements define the same ideal A?

Ar is contained in A for all r in R
aR is contained in A for all a in A

What makes you think that they are the same?

 

[jbunniii]

Ah right, didn't mean to switch those around. Would it be fair to say that the following statements define the same ideal A?

Ar is contained in A for all r in R
aR is contained in A for all a in A

Assuming that $A$ is an additive subgroup of $R$, I think either of these conditions implies that $A$ is a right ideal.

In either case, we need to verify that if $a \in A$ and $r \in R$, then $ar \in A$.

Suppose the first condition holds. If $a \in A$ and $r \in R$, then $ar \in Ar \subseteq A$, so $A$ is a right ideal.

Now suppose the second condition holds. If $a \in A$ and $r \in R$, then $ar \in aR \subseteq A$, so $A$ is a right ideal.