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?
Best Answer
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 $S\subseteq P$.