Ideals of formal power series ring I need help understanding the following solution for the given problem. 
The problem is as follows: Given a field $F$, the set of all formal power series $p(t)=a_0+a_1 t+a_2 t^2 + \ldots$ with $a_i \in F$ forms a ring $F[[t]]$. Determine the ideals of the ring.
The solution: Let $I$ be an ideal and $p \in I$ such the number $a := \min\{i|a_i \neq 0\}$ is minimal. We claim $I=(t^a).$ First, $p=t^aq$ for some unit $q$, hence $(t^a) \subset I$. Conversely, any $r \in I$ has first nonzero coefficient at degree $\geq a$, hence $t^a s$ for some $s \in F[[t]]$, and so $r \in (t^a)$.
My questions: Why the claim $I=(t^a)$? Why does $q$ have to be a unit? What does "first nonzero coefficient at degree $\geq a$ mean? And I don't understand the last part of the proof!!
 A: Let 
$$p(t) = a_0 + a_1 t + a_2 t^2 + \cdots \\
q(t) = b_0 + b_1 t  + b_2 t^2 + \cdots $$
Then we have
$$p(t) \cdot q(t) = c_0 + c_1 t + c_2 t^2 + \cdots $$
where
\begin{eqnarray}
c_0 &=& a_0  b_0 \\
c_1 &=&a_0 b_1 + a_1 b_0 \\
c_2 &=& a_0 b_2 + a_1 b_1 + a_2 b_0\\
c_3 &=& a_0 b_3 + a_1 b_2 + a_2 b_1 + a_1 b_0\\
&\ldots \ldots
\end{eqnarray}
Therefore
$p(t) \cdot q(t) =1$ if and only if we have the infinite sequence of equalities:
\begin{eqnarray}
1 &=& a_0  b_0 \\
0 &=&a_0 b_1 + a_1 b_0 \\
0 &=& a_0 b_2 + a_1 b_1 + a_2 b_0\\
0 &=& a_0 b_3 + a_1 b_2 + a_2 b_1 + a_1 b_0\\
&\ldots \ldots
\end{eqnarray}
Therefore, if $p(t)$ has an inverse $q(t)$ then $a_0 \cdot b_0=1$ and so $a_0$ is invertible. Conversely, if $a_0$ is invertible then in the above system we can solve inductively for $b_0$, $b_1$, $b_2$, $\ldots $ and therefore $f(t)$ is invertible.
For example:
 \begin{eqnarray}  
 b_0 &=& \frac{1}{a_0} \\  
 b_1 &=& - \frac{a_1}{a_0^2}\\  
 b_2 &=& - \frac{a_2}{a_0^2} + \frac{a_1^2}{a_0^3}\\  
 b_3 &=& -\frac{a_3}{a_0^2}+ 2 \frac{a_1 a_2}{a_0^3}- \frac{a_1^3}{a_0^4}\\ 
\ldots \ldots 
 \end{eqnarray} 
Let us define the order $o(p(t))$ of a power series $p(t)= \sum_n a_n t^n $ as follows:
$o(p(t)) = \min \{ n \ | \ a_n \ne 0\}$ if $p(t) \ne 0$ and $o(0) = \infty$. 
From the above we have $p(t)$ invertible if and only if $o(p(t))=0$.  Moreover, for any $p(t) \ne 0$ we have$$p(t) = t^{o(p(t))} \cdot \bar p(t)$$ with $\bar p(t)$ invertible. $\tiny{ \text{(a prime factor decomposition )}}$ 
It is easy to check that $o(p(t)\cdot q(t) ) = o(p(t))+ o(g(t))$ and $p(t) \mid q(t)$ $\tiny{\text{(divides )}}$ if and only if $o(p(t)) \le o (q(t))$ $\tiny{ \text{(like for numbers) }}$
Let $I$ a nonzero ideal. Let $d$ the smallest order of nonzero elements in $I$.$\tiny{\text{(any set of natural numbers has a smallest element)}}$ Let $p(t) = t^d \cdot \bar p(t) \in I$. For any other $q(t)$ in $I$ we have $q(t) = t^e \cdot \bar q(t)$ with $e \ge d$ and so $t^d \mid q(t)$. We conclude that $I \subset (t^d)$. Moreover, $t^d = \frac{1}{\bar p(t)} \cdot p(t) \in I$. Therefore $I = (t^d)$. 
Hence all the ideals of $k[[t]]$ are $0$ and $(t^d)$, $d\ge 0$. 
A: We’re supposing that $I$ is an ideal in $F[[t]]$. For each $p(t)=\sum_{k\ge 0}a_kt^k\in I$, let $$a_p=\min\{k:a_k\ne 0\}\;,$$ so that
$$p(t)=a_{a_p}t^{a_p}+a_{a_p+1}t^{a_p+1}+\ldots=t^{a_p}\left(\underbrace{a_{a_p}+a_{a_p+1}t+a_{a_p+2}t^2+\ldots}_{q(t)\in F[[t]]}\right)\;.\tag{1}$$
Among all elements of $I$, choose $p\in I$ so that $a_p$ is as small as possible, and let $a=a_p$. $(1)$ shows that there is a $q(t)\in F[[t]]$ with a non-zero constant term such that $p(t)=t^aq(t)$, and the claim is that $I=(t^a)$.
Since $q(t)$ has a non-zero constant term, $q(t)$ is a unit in $F[[t]]$, and therefore $t^a\in I$; clearly this implies that $(t^a)\subseteq I$.
Now suppose that $r(t)\in I$, say $r(t)=\sum_{k\ge 0}b_kt^k$. Recall that $p$ was chosen so that $a_p$, the exponent on the first non-zero term of $p(t)$ was as small as possible for any member of $I$. That means that $a_r\ge a_p=a$. But $a_r$ is the exponent on the first non-zero term of $r(t)$, so $b_k=0$ for all $k<a_r$. And since $a\le a_r$, clearly $b_k=0$ for all $k<a$. But that means that every non-zero term of $r(t)$ has an exponent of $a$ or more, which means that we can factor out $t^a$: there is some $s(t)\in F[[t]]$ such that $r(t)=t^as(t)$. Of course this means that $r(t)\in(t^a)$, so we’ve now shown that $I\subseteq(t^a)$.
Putting the two pieces together, we get $I=(t^a)$. Thus, every ideal of $F[[t]]$ is of this form for some $a$.
