Subrings of integral domains have the same identity element. If $R$ is an integral domain and a subring $S$ has identity $1_S$, how would you show that $1_S=1_R$ (here $1_R$ is the identity of the ring $R$)? I am unsure about what an integral domain really is and how the subring comes into play here.
 A: If $1_S$ is a multiplicative identity of $S$, then in particular $1_S1_S=1_S$. But $1_S=1_S1_R$. It follows that
$$1_S(1_S-1_R)=0.$$
In an integral domain, if $ab=0$, then $a=0$ or $b=0$. It follows that $1_S=0$ or $I_S-1_R=0$, meaning $1_S=1_R$. 
So we need to eliminate the possibility $1_S=0$. That is eliminated by the assumption that $S$ is a non-trivial subring of $R$.
Remark: The condition that a product of two non-zero objects is non-zero is a key part of the definition of integral domain. It is a property common to many familiar rings, such as the ring of integers, the ring of complex numbers, the ring of polynomials with rational coefficients, and many others. 
But there are plenty of interesting commutative rings with unit which are not integral domains. For example, consider the ring whose elements are $0,1,2,\dots,11$, under addition and multiplication modulo $12$. Then $(3)(4)=0$, but neither $3$ nor $4$ is equal to $0$. 
A: Hint $\,$ If not, the quadratic $\rm\,x^{\color{#c00}{\large 2}}\! - x\,$ has $\,\color{#c00}3\,$ roots $\rm\,0,\:1_S,\:1_R\,$ in the domain $\rm\,R,\:$ contradiction.
Note $\,$ Generally a ring R is a domain iff every nonzero polynomial over R has no more roots than its degree. This special case has easy proof: $\rm\,x(x\!-\!1)=0\:\Rightarrow\:x=0\:$ or $\rm\:x=1,\:$ by $\rm\,R\,$ a domain.
It fails in nondomains, e.g. in $\rm\,R = \Bbb Q^2\,$ the subring $\rm\:S = \Bbb Q\times \{0\}\:$ has $\rm\:1_S\! = (1,0)\neq 1_R\! = (1,1).$
