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It is stated in Wikipedia that if $I$ is an ideal of $R$, that is $I\triangleleft R$, then $_R I$ is a submodule of $_R R$.

Here, I assume $R$ is commutative. Despite the notation, I mean ideal is the left and right ideal. But it can interpreted as left or right ideal only if necessary. The point of my question is not emphasising on left or right, I am just curious whether the statement works for the converse. Hopefully it is clear. But if there is something wrong in my interpretation, please kindly let me know.

I would like to know whether the converse is true, but it is not stated in Wikipedia. That is for any submodule $_R N$ of $_R R$, then $N$ is an ideal of $R$? In other words, is it true that if $_R N\subseteq _R R$ then $N\triangleleft R$?

If this is a well known useful property, then it should be stated, but I could not find in some books that I read and other sources.

Is it something to proof or is it defined? If it is something to proof, could anyone give some ideas about the proof?

Thanks!

user71346
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    "Ideal" by itself often means both-sided so this may not be true when $R$ isn't commutative – Cocopuffs Jun 10 '13 at 07:52
  • @Cocopuffs. Yes, you are right. Sorry not mentioning it clearly. I assume R is commutative. I will add that to the post. – user71346 Jun 10 '13 at 07:55

2 Answers2

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I think it's just by definition of submodule, because if $_RN$ is a submodule of $_RR$ you are asking that $_RN$ is closed for multiplication on the right by elements of $_RR$ and it's just the condition for $_RN$ to be an ideal of $_RR$. So you can conclude that submodules of a ring are all and only ideals of the ring.

Simone
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You need to show two things: that $(N,+)$ is a subgroup of $(R,+)$ and that for all $r \in R$ and $x \in N$: $rx = xr \in N$.

But both of these follow immediately from the module property, the first because addition on $N$ is well-defined, and the second because $N$ has a scalar multiplication by $R$.

Cocopuffs
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