Existence of multiplicative identity for a subring generated by a set 
Let $S$ be a subring of $R$, a ring with identity. For any arbitrary element $a\notin S$, the subring generated by $S\cup\{ a\}$ is represented by $⟨S,a⟩$. If $a\in\operatorname{cent}(R)=\{a\in R\mid ar=ra, \ \forall r\in R\}$, establish that $$⟨S,a⟩=\{r_0 +r_1 a+r_2 a^2 +,\ldots , r_n a^n\mid n\in\mathbb{Z}_+ ,r_i \in S\}:=Y$$

My attemp of solution:
$⟨S,a⟩=\cap\{X\mid X \ is \ a \ subring \ of \ R \ and \ X\supseteq S\cup\{a\}\}$. Then it's clear that $$\sum_{k=0}^{n} r_{k}a^{k}$$ is contained in $⟨S,a⟩$, for all $n\in\mathbb{Z}_{+}$ and every choice of $r_{k}\in S$. Now, it's also clear that $Y$ is a subring of $R$, then if we show that $⟨S,a⟩\subseteq Y$ we have done. Now, if $s\in S$, we can take $r_{0}=s$ and $r_{i}=0$ for all $i≥1$. My idea, for $a$, was to take $r_{0}=r_{2}=\ldots =r_{n}=0$ and $r_{1}=1$. But the fact that $R$ has an identity element doesn't mean that $S$ has one, and I have to take $r_{i}$ in $S$, hence I can't finish the proof. Maybe I lost something related to the fact that $a\in \operatorname{cent}(R)$ (I use it only to prove that if I take something like $a^{k}s\in ⟨S,a⟩$ is also in $Y$).
 A: Actually, showing that $\langle S, a\rangle\subseteq Y$ is the easier part.
If you prove that $Y$ is a ring, then the relation is trivial, because $\langle S, a\rangle$ is the intersection of all subrings of $R$ that include $S$ and $a$. Therefore, $\langle S, a\rangle = Y\cap (\text{a bunch of other subrings})$, so clearly, $\langle S,a\rangle \subseteq Y$ (because $A\subseteq A\cap B$ is always true for all sets).

Edit:
actually, I think you are correct that if $1\notin S$, we may have problems and the statement cannot be proven.
Take

*

*$R=\mathbb Z[X]$, i.e. the polynomials with integer numbers,

*$S=\langle X\rangle$, i.e. all polynomials with a zero free coefficient (it is easy to see that this is indeed a ring)

*$a=1$.

Then, you have

*

*$Y = S$ ($S\subseteq Y$ is trivial, and $Y\subseteq S$ is not much harder, since $r_0+r_1a + \cdots + r_na^n = r_0 + r_1 + \cdots r_n\in S$ so long as all $s_i$ are elements of $S$).

*$a\notin S$ (since $1$ is a polynomial with a nonzero free coefficient)

meaning that the statement is false.

Edit 2:
Based on the standard definitions of rings, a subring must also include the multiplicative identity, so the above point is moot.
See here.
