# The smallest commutative subring of $\mathbb{Z}\times \mathbb{Z}$

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?

• Does your definition of subring require that the ring's multiplicative identity be a member of the subring? Oct 16, 2019 at 17:20
• Yes, the subring must contain a unity.
– user535444
Oct 16, 2019 at 17:22
• 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. Oct 16, 2019 at 17:22
• Sorry, now I see it. The subring must contain the unity of the entire ring.
– user535444
Oct 16, 2019 at 17:26

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 $$S\subseteq P$$.

• Which you can express as $2 \mid (b-a)$. 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.