In Richard Borcherds' excellent online commutative algebra lecture series he talks about the Weierstrass preparation theorem in lecture 10 and uses it to prove that the ring of formal power series is a unique factorization domain. He then gives a warning that the ring of convergent power series is not a UFD and gives an example of a power series with infinitely many zeros to show why this is the case.
Both of these arguments seem to make sense on their own, but isn't the ring of convergent power series a subset of the ring of formal power series? In which case why does this not contradict that the ring of formal power series is a UFD? I know I must be missing something simple here.
Best Answer
As already mentioned in the comments, a subring of a UFD does not have to be a UFD. I like the following illuminating example:
Let $k$ be a field and consider the ring of polynomials $k[t]$ and its subring $k[t^2,t^3]$, which consists of all polynomials with zero linear term. It is not a UFD: the element $t^6$ has two factorizations into a product of irreducible elements $$t^6=t^2\cdot t^2\cdot t^2=t^3\cdot t^3.$$
Of course, $t^6$ has an essentially unique factorization in $k[t].$ The issue is in the absence of $t$ - by killing it off, we made two irreducible elements $t^2$ and $t^3$ "incomparable".