Explanation for the definition of an Ideal

Question

Let $R$ be a ring, ideal $\mathcal{I}$ is defined as a subset of $R$ satisfying: \begin{align*} &\forall x,y\in\mathcal{I}:x+y\in\mathcal{I},-x\in\mathcal{I}\tag{1}\\ &\forall x\in\mathcal{I},y\in R:xy,yx\in\mathcal{I}\tag{2} \end{align*} Now, some sources give equivallent definition of (1) given by $\forall x,y\in \mathcal{I}:x-y\in\mathcal{I}$. It seems obvious that this is equivalent, but could anyone elaborate this a bit more, how does $x-y\in\mathcal{I}$ imply that $x+y\in\mathcal{I}\wedge-x\in\mathcal{I}$ and converse?

Answer 1

If $x-y$ is always in $\mathcal I$, then in particular $x-x=0$ is in $\mathcal I$ (assuming that an ideal must be nonempty, which I think is generally required in addition to the axioms you're quoting).

Once $0\in\mathcal I$ we also have $0-x=-x\in\mathcal I$. Finally $y-(-x)=y+x$ is in $\mathcal I$.

The other direction is similar, but simpler, since $x-y=x+(-y)$.

Answer 2

For another perspective, an ideal $I$ of $R$ is simply a subgroup of the underlying abelian group $(R,+)$ for which the quotient group $R/I$ has a ring structure compatible with the quotient map $\pi :r\mapsto r+I$ (i.e. which makes it a ring homomorphism). Your question then amounts to the conditions on subgroups, and property (2) ensures that the multiplication on $R/I$ defined via the requirement that $\pi$ be a ring homomorphism is well defined.