Find a subring of $\Bbb Z \oplus \Bbb Z$ that is not an ideal of $\Bbb Z \oplus \Bbb Z$.

Question

Find a subring of $\Bbb Z \oplus \Bbb Z$ that is not an ideal of $\Bbb Z \oplus \Bbb Z$.

I can't see any way a subring of $\Bbb Z \oplus \Bbb Z$ can NOT be an ideal. Subrings of $\Bbb Z \oplus \Bbb Z$ are of the form $n\Bbb Z \oplus k\Bbb Z$ where $n$ and $k$ are integers. So for something to not be an ideal in $\Bbb Z \oplus \Bbb Z$ there must be an $x = (j_1, j_2)$ in $\Bbb Z \oplus \Bbb Z$ and an $y = (b_1n, b_2k)$ in $n\Bbb Z \oplus k\Bbb Z$ such that $xy$ or $yx \notin$ $n\Bbb Z \oplus k\Bbb Z$. But this doesn't make sense because it seems any integer pair can be reached.

Anyone have any ideas?

Answer 1

The subring $R = \{(m,m)\mid m \in \Bbb Z\}$ is not an ideal. This is easily checked since $(1,1)\in R$ but $(0,1)\cdot (1,1) = (0,1) \notin R$.

Answer 2

I cannot leave comments unfortunately, so I am responding in this answer to @Mathematicing's comment that "$R=\{(m,m) : m \in\Bbb Z^+\}$ would have worked equally well.". In fact, this is not a subring because it is not closed under subtraction: consider $(1, 1) - (2, 2) = (-1, -1)$.

Answer 3

Any subring would I presume contain $(1,1)$, the ring's unity (this is not the only possible interpretation).

Then any proper subring will do.

The proper subrings are the $R_n=\{(a,b)\in\Bbb Z^2\mid a\equiv b\pmod n\},$ for $n\gt1.$ See @Qiaochu's proof here.

$R_\infty$ is @Arthur's example.