# Proving ring of FPS is not a field

Following on from my question of the proof of the ring of a FPS is an integral domain:

Proving the ring of formal power series over a finite field is integral domain.

I would like to extend to that to finding a proof of showing that the ring of formal power series over a finite field of order prime is not a field.

Just confused really, with proving it was an integral domain, it seemed to be fine.

Thanks for any help

• How do you invert an element of zero constant term that is nonzero? (for a specific example, think of the formal power series of sine).
– Mark
Nov 23, 2016 at 20:09
• The ring of formal power series in never a field. Can you find a non-invertible element? Nov 23, 2016 at 20:14
• I would think that it would be when the denominator of some constant is non-zero Nov 23, 2016 at 20:41
• Or perhaps if we were to have some constant, not equal to 0 over a field then that would also be invertible. Nov 23, 2016 at 20:46
• Sorry English isn't my 1st language. I read it is invertible. If it were non-invertible then if we had (a+b) where a and b are both elements which belong to the same ring. Nov 23, 2016 at 20:57

You could find a noninvertible element, as the comments suggest, or you could (equivalently) find a nonzero ideal that isn't the whole ring itself. Consider the homomorphism \begin{align*} \phi : k[\![T]\!]&\to k\\ \sum_{i = 0}^\infty a_i T^i&\mapsto a_0. \end{align*} The homomorphism above is clearly surjective, and it's not hard to verify that the kernel is $(T)$. In a field, the only ideals are $(0)$ and $(1)$. But $(0)\subsetneq (T)\subsetneq (1)$ ($T\not\in (0)$, and $1\not\in (T)$, because $\phi(1) = 1$), so $k[\![T]\!]$ cannot be a field.