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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?

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    In some definitions, a ring (and any subring) must contain a multiplicative identity. – 2017-01-10
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    Perhaps by *ring* here it is meant *ring with unity*? – 2017-01-10
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    Put another way, I think it would be good to include, for us, your definition of *subring* (and possibly *ring* as well) since conventions with the multiplicative identity vary. – 2017-01-10

2 Answers 2

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HINT Note that $1_S = 0 \ne 1 = 1_\mathbb{Z}$ so your first line needs some adjustment

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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 $\mathbb{Z}$.