I want to find the smallest commutative subring of $\mathbb{Z}^{2}$ (operations defined entrywise) with element $(2,0)$. Let's call it $S$.

$S$ must contain elements $(1,1)$, $(0,0)$, $(2,0)$ and be closed under multiplication and addition. Therefore I define it as a set containing $ \sum_{i=1}^{r} (k_{i},k_{i})^{ n_{i} } (2l_{j},0)^{ m_{j} } $ $ \forall r \in \mathbb{N}, k_{i}, l_{i} \in \mathbb{Z},n_{i}, m_{i} \in \mathbb{N} $, which is closed under addition, multiplication and contains $(1,1)$, $(0,0)$ and $(2,0)$.

Now in order for such $S$ to be the smallest subring of $\mathbb{Z}^{2}$ containing $(2,0)$, every subring of $\mathbb{Z}^{2}$ containing $(2,0)$ must contain $S$. I think it is true by the definition of $S$.

Is $S$ really the smallest subring of $\mathbb{Z}^{2}$ with $(2,0)$ and is there a better way to describe it?

  • 2
    $\begingroup$ Does your definition of subring require that the ring's multiplicative identity be a member of the subring? $\endgroup$ Oct 16, 2019 at 17:20
  • $\begingroup$ Yes, the subring must contain a unity. $\endgroup$
    – user535444
    Oct 16, 2019 at 17:22
  • $\begingroup$ That's not quite what I asked. As my answer below illustrates, it's possible for an element that isn't the unity of the entire ring to nonetheless be a unity for a subring. $\endgroup$ Oct 16, 2019 at 17:22
  • $\begingroup$ Sorry, now I see it. The subring must contain the unity of the entire ring. $\endgroup$
    – user535444
    Oct 16, 2019 at 17:26

2 Answers 2


Your description is correct, but it's rather overcomplicated. Note that $(k_i,k_i)^{n_i} = (k_i^{n_i},k_i^{n_i})$, which is also an element of the form $(k,k)$. And similarly, when $m_i>0$, $(2l_i,0)^{m_i} = (2^{m_i}l_i^{m_i},0)$, and $2^{m_i}l_i^{m_i}$ is also even, so this is again an element of the form $(2l,0)$. Finally, we have $(k_i,k_i)(2l_i,0) = (2l_ik_i,0)$, which is again of the form $(2l,0)$. Of course, when $m_j = 0$, $(2l,0)^0 = (1,1)$. So we've reduced your description to a sum of elements of the form $(k,k)$ and elements of the form $(2l,0)$. But a sum of elements of the form $(k,k)$ again has the form $(k,k)$, and a sum of elements of the form $(2l,0)$ again has the form $(2l,0)$. So in total:

You could significantly simply your presentation to $\{(k,k) + (2l,0)\mid k,l\in \mathbb{Z}\}$.

But there's an even cleaner description: Your ring is just $$\{(a,b)\mid a\text{ and }b\text{ have the same parity}\}$$ (i.e. $a$ and $b$ are both even or both odd).

To see this, write $S$ for the smallest subring containing $(2,0)$, and write $P$ for the ring defined above. Now if $a$ and $b$ have the same parity, then $(a-b) = 2k$ for some $k\in \mathbb{Z}$, so we can write $(a,b) = b(1,1) + k(2,0)$, and thus $(a,b)\in S$. So $P\subseteq S$. Conversely, it is easy to check that $P$ is closed under addition and multiplication and contains $(1,1)$ and $(2,0)$, so it is a subring of $\mathbb{Z}^2$ containing $(2,0)$, and thus $S\subseteq P$.

  • $\begingroup$ Which you can express as $2 \mid (b-a)$. $\endgroup$ Oct 16, 2019 at 17:26

If $(1, 0) \in S$, then necessarily $\forall n \in \Bbb Z (n, 0) \in S$. It's easy to verify that $S= \{ (n, 0)~\vert~n \in \Bbb Z \} \cong \Bbb Z$ is in fact a subring, so it must be the smallest subring that contains $(1, 0)$.

I'm assuming here that your definition of a subring requires that in include an element that operates as a multiplicative identity in the subring. If that's not the case, then $S= \{ (2n, 0)~ \vert ~ n \in \Bbb Z \} $ works.

Edited to add: On the other hand, if your definition of subring requires you to include the ring's multiplicative identity, then the answer provided above by Alex Kruckman is the correct answer.


You must log in to answer this question.