Is this set a subring of $\mathbb{Z}\times\mathbb{Z}$?

Question

Is the set $S = \{(x,-x) : x \text{ is an integer}\}$ a subring of $\mathbb{Z}\times\mathbb{Z}$?

I am not sure where to start here. Is $\mathbb{Z}\times\mathbb{Z}$ a matrix? It doesn't seem like $S$ is closed under multiplication.

Answer 1

$\mathbb{Z}\times\mathbb{Z}$ is the set of ordered pairs of integers.

$(2,-2)$, $(1,2)$, $(-3,3)$, $(0,0)$, $(1,1)$ are all elements of the set $\mathbb{Z}\times\mathbb{Z}$. (The "$\times$" symbol between the two $\mathbb{Z}$ does not mean multiplication!)

The ring operations on $\mathbb{Z}\times\mathbb{Z}$ are:

1. $(a,b) + (c,d) = (a+b,c+d)$
2. $(a,b)*(c,d) = (ac, bd)$

The set $S$ is a subset of $\mathbb{Z}\times \mathbb{Z}$ that must be of the form $(x,-x)$.

Of the elements I listed above, only $(2,-2)$, $(-3,3)$ and $(0,0)$ are elements of $S$.

In order for $S$ to be a subring, it needs to be closed under the ring operations of $\mathbb{Z}\times\mathbb{Z}$. As a number of other posters pointed out, $S$ is not closed under multiplication.

$(2,-2)*(2,-2) = (4,4) \not\in S$.

I didn't have to choose $(2,-2)$. The two elements don't even need to be the same.

$(3,-3) * (-4,4) = (-12,-12) \not\in S$.

Since $S$ fails to be closed, it is not a subring.

Answer 2

If $R$ is a ring and $\emptyset \ne S\subset R$ then $S$ of $S$ is a subring if and only if $a,b\in S$ implies that $a-b\in S$ and $ab\in S$. If you are working with rings having an identity $1$ for multiplication, then next to that a third condition must also be satisfied: $1\in S$. Several comments make clear that $S$ is not closed under multiplication in the case mentioned in your question.