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!!

  • $\begingroup$ First question: do you know what the units in $F[[t]]$ are? $\endgroup$ Jan 27 '15 at 4:28
  • $\begingroup$ Yes, the units are the elements of $F[[t]]$ that have multiplicative inverses. $\endgroup$
    – Anonymous
    Jan 27 '15 at 4:31
  • $\begingroup$ Yes, but there’s a very simple way to recognize them from their form; do you know what that is? $\endgroup$ Jan 27 '15 at 4:32
  • 1
    $\begingroup$ Aren't they just the elements of $F[[t]]$ with nonzero constant term? $\endgroup$
    – Anonymous
    Jan 27 '15 at 4:36
  • $\begingroup$ Yes, they are. And if $a$ is the lowest exponent in $p$ with a non-zero coefficient, then you can factor out $t^a$ to get a power series $q$ with a non-zero constant term. Thus, $q$ will automatically be a unit. \\ First nonzero coefficient at degree $\ge n$ means that if $p(t)=\sum_{k\ge 0}c_kt^k$, then $c_k=0$ for all $k<n$. $\endgroup$ Jan 27 '15 at 4:39

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 $a$.


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 $(t^d)$, $d\ge 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.