Prove that the zero ring is not a subring of $\mathbb{Z}$. Prove that the zero ring is not a subring of $\mathbb{Z}$.
My attempt:
Let $S$ denote the zero ring.
$1=0\in S$
$0-0=0\in S$
$0\cdot 0=0\in S$
So, $S$ is a subring of $\mathbb{Z}$. What am I missing?
 A: Conventions vary from text to text (and course to course), but many definitions require that a ring (and therefore a subring) have an identity element, and moreover that the identity element of a subring be the same element as the identity element of the larger ring.
For example, consider the ring $R=\mathbb{R} \times \mathbb{R}$, consisting of ordered pairs $(a,b)$, with addition and multiplication both defined componentwise, i.e. $(a,b)+(c,d)=(a+c,b+d)$ and $(a,b)\cdot(c,d)=(ac,bd)$.  Now $R$ is a ring with identity element $1_R=(1,1)$.  Consider now the subset $S \subset R$ consisting of ordered pairs of the form $(a,0)$.  $S$ is also a ring, with identity element $1_S=(1,0)$.  But even though $S$ is a ring that is a subset of $R$, it is not  a subring of $R$ because it does not contain the identity element of $R$.
Something similar is going on in your example.  The zero ring is a ring (vacuously), and is a subset of $\mathbb{Z}$, but it is not a subring of $\mathbb{Z}$ because it does not contain the identity element of $\mathbb{Z}$.
A: HINT
Note that $1_S = 0 \ne 1 = 1_\mathbb{Z}$ so your first line needs some adjustment
