Subset of $\mathbb{Z} \times \mathbb{Z}$

Question

I have a past exam question that is as follows:

Let $k$ be a fixed integer and $S = \{(a,ka)|a \in \mathbb{Z}\}$ be a subset of $\mathbb{Z} \times \mathbb{Z}$. Prove that $S$ is a subgroup of $\mathbb{Z} \times \mathbb{Z}$ under addition. For which value(s) of $k$ is $S$ a subring of $\mathbb{Z} \times \mathbb{Z}$? For which value(s) of $k$ is $S$ an ideal of $\mathbb{Z} \times \mathbb{Z}$? Justify your answers.

$S$ is a subgroup of $\mathbb{Z} \times \mathbb{Z}$ under addition:

Proof

Associativity: $(a,ka)+((b,kb)+(c,kc)) = (a,ka) + (b+c,kb+kc) = (a+b+c,k(a+b+c))$

$((a,ka)+(b,kb))+(c,kc)=(a+b,ka+kb)+(c,kc)=(a+b+c,k(a+b+c))$

Inverse: $(a,ka) + (-a,-ka) = (0,0)$

Identity: $(a,ka) + (0,0) = (a,ka)$

Hence $S$ is a subgroup

For $k = 0,1$ $S$ is a subring:

Proof

$Let \; k = 0,(a,0)(b,0)=(ab,0), \; let \; ab=c, (ab,0) = (c,kc)=(c,0)$ $(a,0)+(b,0)= (a+b,0), \; let \; a+b=c, (c,0) = (c,kc)$

$Let \; k = 1,(a,a)(b,b)=(ab,ab)=(c,kc)=(c,c) $ $(a,a)+(b,b) = (a+b,a+b), \; let \; a+b=c, (c,kc)=(c,c)$ Hence $S$ holds additive and multiplicity closure, and is therefore a subring.

$S$ is only an ideal of $\mathbb{Z} \times \mathbb{Z}$ at $k=0$ $(s,ks)(b,c)=(b,c)(s,ks)=(sb,ksc)=(sb,0sc)=(sb,0)=(a,o)$, which can't have form $(a,ka)$ with the second term equalling anything other than zero. Not sure if this is a proper solution.

Answer 1

For some authors it is required that a ring contains a multiplicative identity element. I'm not sure if that's the case for you, but my guess would be no, since you seem to imply that ideals are subrings. [Exercise: Any ideal containing the identity element is the entire ring.]

If not, then yes: you need only show multiplicative closure. The distributive law follows from the fact that you are already inside an ambient ring.

Formally,

A subset $I$ is an ideal of a ring $R$ if is a (additive) subgroup of $R$ with the following property: For every $i\in I$, and every $r\in R$, then both $ri\in I$ and $ir\in I$.

Note that this is a strictly stronger condition than multiplicative closure. This is because $r$ is allowed to be any element in $R$, even elements which are not in the ideal $I$. Talking about an "ideal ring", therefore, is discouraged, since every ring is an ideal in itself. But it becomes a meaningful (and rather strong) condition when considering subrings of an ambient ring.