An equivalence regarding polynomials and rings

Question

Let $\theta \in \mathbb{C} \setminus \mathbb{Z}$. For every $k\in \mathbb{N}^{*}$ consider $\mathbb{Z}[\theta]_{k}=\{f(\theta) | f\in \mathbb{Z}[X], \deg f \le k\}$.
Prove that the following statements are equivalent:
1)$\exists n \in \mathbb{N}$ and $a_0, a_1, ..., a_{n-1} \in \mathbb{Z}$ such that $\theta^n+a_{n-1}\theta^{n-1}+...+a_1\theta + a_0=0$
2)$\exists k_0 \in \mathbb{N}$ such that ($\mathbb{Z}[\theta]_k$, $+$, $\cdot$) is a ring for $k\ge k_0$ and it isn't a ring for any $1\le k_0 <k$ ($k_0 \ge 2$).
Note: $+$ and $\cdot$ are the regular addition and multiplication of complex numbers.
I don't really know how to actually approach this. I know that condition $1)$ basically states that there is a monic polynomial of degree $n$ which has $\theta$ as a root, but I don't know how this relates to $2)$. I think that a reductio ad absurdum proof may work, but I don't know how to get started on it.

Answer 1

Any of the sets $\mathbb{Z}[\theta]_k$ is an abelian group under addition, since for each element $$ \sum_{i = 0}^k a_i \theta^i $$ it includes its additive inverse $$ \sum_{i = 0}^k (-a_i) \theta^i, $$ and it is closed under addition: $$ \sum_{i = 0}^k a_i \theta^i + \sum_{i = 0}^k b_i \theta^i = \sum_{i = 0}^k (a_i + b_i) \theta^i. $$

Also, the multiplication from $\mathbb{C}$ is associative and distributes over addition. So we only have to check if $\mathbb{Z}[\theta]_k$ is closed under multiplication to determine if it is a ring.

Now, suppose condition (1) holds: we have $n \in \mathbb{N}$ and integer coefficients $a_0, \dotsc, a_{n-1}$ such that $\theta^n + \sum_{i = 0}^{n-1} a_i \theta^i = 0$; equivalently, $\sum_{i = 0}^{n-1} a_i \theta^i = -\theta^n$. Then there is a least $n$ with this property, so WLOG assume $n$ to be minimal.

Then any power $\theta^k$ with $k \geq n$ can be rewritten as a combination of lower powers $\{1, \theta, \dots, \theta^{n-1}\}$. So, for $k \geq n-1$, any sum $\sum_{i = 0}^{k} b_i \theta^i$ can be written as $\sum_{i = 0}^{n-1} b_i' \theta^i$, and any product $$ \Bigl(\sum_{i = 0}^{n-1} b_i \theta^i\Bigr) \Bigl(\sum_{i = 0}^{n-1} c_i \theta^i \Bigr) = \sum_{i=0}^{2n-2}\Bigl(\sum_{j+l=i}b_j c_l\Bigr) \theta^i $$ can similarly be rewritten, so $\mathbb{Z}[\theta]_k$ is closed under multiplication.

On the other hand, for $k < n-1$, consider the two elements $\theta^k$ and $\theta$; if their product $\theta^{k+1}$ is in $\mathbb{Z}[\theta]_k$, then it is equal to $\sum_{i=0}^k a_i \theta^i$ for some integer coefficients $a_i$, and we have an equation $\theta^{k+1} - \sum_{i=0}^k a_i \theta^i = 0$; but this contradicts the minimality of $n$.

So condition (2) holds, with $k_0 = n-1$: $\mathbb{Z}[\theta]_k$ is a ring if and only if $k \geq n - 1$.

Now, conversely, assume condition (2): there exists $k_0 \in \mathbb{N}$ such that $\mathbb{Z}[\theta]_k$ is a ring if and only if $k \geq k_0$. Then $\mathbb{Z}[\theta]_{k_0}$ is closed under multiplication, so $\theta^{k_0} \cdot \theta = \theta^{k_0+1}$ is an element of $\mathbb{Z}[\theta]_{k_0}$, hence $$ \theta^{k_0+1} = \sum_{i=0}^{k_0} b_i \theta^i $$ for some integer coefficients $b_i$. Then $n = k_0 + 1$ and $a_i = -b_i$ satisfy condition (1).

Answer 2

First of all note that $\mathbb{Z}[T] = \cup_{k > 0} \mathbb{Z}[T]_k$, and that this union is increasing, and that $\mathbb{Z}[\theta]$ (resp. $\mathbb{Z}[\theta]_k)$ are the images of $\mathbb{Z}[T]$ under $T \mapsto \theta$. Moreover we claim that the condition (2) is equivalent to $\mathbb{Z}[\theta] = \mathbb{Z}[\theta]_{k_0}$ for some $k_0$.

First we prove this claim,first suppose (2) is true. The inclusion $\mathbb{Z}[\theta]_{k_0} \subset \mathbb{Z}[\theta]$ is clear. Now as an abelian group $\mathbb{Z}[\theta]_k$ is generated by $\{\theta^i, i \leq k\}$, so if $\mathbb{Z}[\theta]_{k_0}$ is a ring then it contains $\{\theta^i, i \leq k\}$ for all $k \geq k_0$ and thus $\mathbb{Z}[\theta]_k \subset \mathbb{Z}[\theta]_{k_0}$, taking union over all $k$ yields the claim. Conversely if $\mathbb{Z}[\theta] = \mathbb{Z}[\theta]_{k_0}$ for some $k_0$, taking $k_0$ minimal gives shows that for all $k < k_0, \mathbb{Z}[\theta]_{k}$ is not a ring and for all $k \geq k_0, \mathbb{Z}[\theta] = \mathbb{Z}[\theta]_k = \mathbb{Z}[\theta]_{k_0}$ is a ring.

Now for (1) $\implies$ (2), suppose $f \in \mathbb{Z}[T]$ is monic of degree $n > 1$ such that $f(\theta) = 0$, then the natural map is surjective $$\pi :\mathbb{Z}[T]/(f) \rightarrow \mathbb{Z}[\theta]$$ and as $\mathbb{Z}[T]/(f) = \pi(\mathbb{Z}[T]_n)$ (note that for this to be true we use that $f$ is monic) we have $\mathbb{Z}[\theta] = \mathbb{Z}[\theta]_n$ and by the previous point this is equivalent to (2).

For (2) $\implies$ (1), fix $k_0$ as in (2), then as $\mathbb{Z}[\theta]_{k_0 + 1} = \mathbb{Z}[\theta]_{k_0}$, we have $\theta^{k_0 + 1} \in \mathbb{Z}[\theta]_{k_0}$ and thus there is a polynomial $f \in \mathbb{Z}[T]$ of degree $\leq k_0$ such that $\theta^{k_0 +1} = f(\theta)$ then $g = T^{k_0 + 1} - f$ is a monic polynomial such that $g(\theta) = 0$.