Show that ideal is a subring

Solution 1:

If your book defines (like most books do I think) rings as unitary rings, in other words requires them to contain a neutral element for multiplication, then ideals are almost never subrings. Indeed, for unitary rings one requires subrings of$~R$ to contain the element $1\in R$ (rather than some element that is neutral for mulitplication restricted to the subring), and an ideal that contains $1$ must contain its multiples by elements of $R$, which gives all of $R$; the only ideal of $R$ that is a subring is $R$ itself.

If on the other hand your book says nothing of neutral elements for multiplication, then it deals with a larger category of structures that are often called "rngs", to indicate the (possible) lack of $1$. Then indeed all ideals are subrngs.

Finally I note that sometimes a non-trivial ideal can be considered a ring in itself by choosing a different element to be $1$ (this was the reason for my parenthesised remark above). For instance in $\def\Z{\mathbf Z}\Z/10\Z$ you can make the ideal of "even" classes into a ring (using the same addition and multiplication) by electing the class of $6$ to be the neutral element for multiplication. This situation arrives typically in rings that can be decomposed as a product of rings; in the current example $\Z/10\Z\equiv (\Z/5\Z)\times(\Z/2\Z)$ by the Chinese remainder theorem. The factors in such a decomposition are naturally quotients of the original ring, but also correspond to an ideal of that ring (namely the kernel of projecting to the other factor(s)); such an ideal is not a subring, but can be made into a ring by choosing the projection of $1$ onto the ideal as new neutral element.