Left ideal in a ring $R$ is a subgroup

Question

How do I prove that if $I$ is an left ideal in a ring $R$ which only satisfies $0\in I$ and $$a\in I,r\in R\text{ then } ra\in I,$$ then already $$a-b\in I$$ for all $a,b\in I$? Please note that I'm a complete beginner to module theory. If this does not hold, is a left ideal assumed to be a subgroup of the ring $R$ in general? The reason why I think that $I$ should be a subgroup is the following parapraph in Rotman, Advanced modern algebra:

EDIT

Answer 1

You can define a left ideal of a ring $R$ as a subset $I\subseteq R$ that satisfies:

• $I$ is an abelian group under $+$.
• $ra\in I$ for every $r\in R$ and $a\in I$.

Alternatively you can define it to be a subset $I$ that satisfies:

• $I$ is not empty.
• $I$ is closed under addition.
• $ra\in I$ for every $r\in R$ and $a\in I$.

Note that $a\in I\implies-a=(-1)a\in I$ according to the last rule so that: $$a,b\in I\implies a-b=a+(-b)\in I$$ showing that also according to this definition $I$ is an abelian group.

Also note that $a\in I\implies 0=0a\in I$.

The conditions that you mention are not enough.

Answer 2

The conditions you give do not define ideals. That is, if $R$ is a ring, and $S\subseteq R$ is a subset such that:

• $0\in S$; and
• For all $x,r$, if $x\in S$ and $r\in R$, then $rx\in S$

then it does not follow that $S$ is closed under sums or differences.

For an explicit example, take $R=\mathbb{Z}$, and take $S=\{n\in\mathbb{Z}\mid 2|n\text{ or }3|n\}$; that is, $S$ is the set of all integers that are either multiples of $2$ or multiples of $3$.

This certainly satisfies your conditions: $0\in S$, and if $n$ is either a multiple of $2$ or a multiple of $3$, then so is $kn$ for any integer $k$.

However, $S$ is not closed under differences or sums: $2\in S$, and $3\in S$, but none of $2+3$, $2-3$, or $3-2$ are in $S$.

In short, you need to also explicitly require the subset to be closed either under differences, or under sums of $R$ has a unity. (If $R$ has a unity, then sums suffice because if $a,b\in S$, then $(-1)b\in S$, and hence $a-b=a+(-1)b\in S$).