Give examples of the following rings (concerning identities) Disclaimer: I do not assume rings to be unital (otherwise the question wouldn't make sense).


  Give examples of a ring $R$ with subring $S$ such that
  
  (1) $R$ has an identity but $S$ has not.
  
  (2) $S$ has an identity but $R$ has not.


Attempt:

(1) Take $R = \mathbb{Z}$ and $S = 2 \mathbb{Z} = (2)$

Suppose that $S$ has an identity $e$, then we must have (in $S$)

$$2e = 2$$

This is also true in $R$, and thus we can conclude $e = 1$. But $1 \notin S$. Contradiction. Thus $S$ has no idenity.

(2) Take $R$ the compactly supported functions $\mathbb{R} \to \mathbb{R}$. This is easily seen to be a subring of the set of all functions $\mathbb{R} \to \mathbb{R}$ (and in fact thus also gives an example for (1), as this ring has  identity $f = 1$). Suppose $R$ has an identity $g$. Take a function $l$ that is non-zero everywhere. Then we must have

$$lg = l \implies l(g-1) = 0 \implies \forall x \in \mathbb{R}: g(x) = 1 \implies g = 1$$

But $g \notin R$. Thus $R$ has no identity.

Define a function $k$ as the indicatorfunction $k= I_{[0,1]}$ which is $1$ on $[0,1]$ and $0$ elsewhere. This function is compactly supported and an idempotent. Hence, the ideal $(k) = kR$ is a subring with identity $k$ (since $(kr)k = k^2r = kr$).

Observation: If $R$ has an identity that also lives in $S$, this is also the identity for $S$. Thus the only way $S$ can be non-unital when $R$ is unital is that the idenity of $R$ does not live in $S$.

Is this correct?
Yes, both examples and the observation are correct.

Note for (2), that any idempotent element $e$ (even in a nonnecessarily commutative ring) gives rise to a subring with identity, namely $eRe$.
I think these examples are sufficient.


R has an identity but S has not.
R={0,1,2,3} under addition and multiplication modulo 4.
S={0,2} under the same operations
S  has an identity but R has not.
R={0,2,4,6,8,10} under addition and multiplication modulo 12.
S={0,4,8} under the same operations.(S is even a field)