TITLE: Show $100+t$ is reducible in $\mathbb{Z}[[t]]$
QUESTION [8 upvotes]: $\mathbb{Z}[[t]]$ is the ring of formal power series with integer coefficients, and the problem asks to show that $100+t$ is reducible. This means I need to find two power series, $\sum_{i=0}^\infty a_i t^i$ and $\sum_{j=0}^\infty b_j t^j$ with $a_i,b_j\in\mathbb{Z}$, that when multiplied give $100+t$. Since the integers are an integral domain, I know that neither of these power series may be finite, but I'm having a really hard time figuring out how to come up with two series that will telescope properly. Any suggestions?

REPLY [5 votes]: First, solve it modulo $t$. Then modulo $t^2$. Then....
Splitting $100 = 25 \cdot 4$ is the only choice, because that is the only way to factor $100$ into relatively prime parts. this is important for reasons that will be clear if you try to apply the method below)
So, I've chosen

$f \equiv 25 \pmod t$
$g \equiv 4  \pmod t$

To set some terminology, let

$f_n$ be the truncation of $f$ to degree $n$
$g_n$ be the truncation of $g$ to degree $n$
$a_n$ be the coefficient of $t^n$ in $f$
$b_n$ be the coefficient of $t^n$ in $g$

Now, to get the coefficient on $t$:

$(f_0 + t a_1) (g_0 + t b_1) = 100 + t (4 a_1 + 25 b_1) \pmod {t^2}$

So I need do an extended GCD and decide upon $a_1 = -6$ and $b_1 = 1$.
Now, induct. By hypothesis, we may assume
$$ f_{n-1} g_{n-1} \equiv 100 + t \pmod {t^n} $$
and going up a degree gives
$$f_{n-1} g_{n-1} \equiv 100 + t + t^n c_n \pmod{ t^{n+1}} $$
for some value of $c_n$.
$$ \begin{align}
f_n g_n &= (f_{n-1} + a_n t^n) (g_{n-1} + b_n t^n) & \pmod {t^{n+1}}
\\ &= 100 + t + t^n (c_n + a_n g_{n-1}(0) + b_n f_{n-1}(0)) & \pmod{ t^{n+1}}
\\ &= 100 + t + t^n (c_n + 4 a_n  + 25 b_n ) & \pmod{ t^{n+1}}
\end{align} $$
From which it's clear that we can solve for $a_n$ and $b_n$. e.g. $a_n = 6 c_n$ and $b_n = -c_n$
Since we obtain $f_n$ and $g_n$ for all $n$, we get a solution for $f$ and $g$.
EDIT: Notice at this point, we can observe that one of the solutions has $a_n = -6 b_n$ for $n > 0$, and so we could write $100 = (25 + 6 t h(x)) (4 - t h(x))$, and then segue into Bill Dubuque's method.