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

Yes so my question shown in the first image;

I would like to prove that the ring of formal power series over the finite field of order prime, s is an integral domain.

So if it is not clear the first image is the question in its entirety and page 1 of my working out and the 2nd image is also my working out.

If anyone could verify my work, or tell me if I have got it all wrong lol it would be welcome.

• okay sorry, for further questions I will upload the problem written and attach a written solution if that is okay? Thanks for your comment. – nomad609 Nov 23 '16 at 23:59
• The images are no longer available. – Dietrich Burde Apr 16 '17 at 18:14

This question can be done much easier by proving the converse: $$fg\neq0 \iff f\neq0 \land g\neq0\tag1$$ where $f,g\in K[[x]]$, $K$ is finite field of order $p$.

Let $a_0$ be constant of $f$ and $b_0$ be constant of $g$. Clearly $a_0b_0$ is constant of $fg$ and $$p|a_0b_0\iff p|a_0 \lor p|b_0\quad\text{or}\quad p\nmid a_0b_0\iff p\nmid a_0 \land p\nmid b_0$$ This means $$(a_0b_0\not\equiv0\mod p) \iff (a_0\not\equiv0\mod p)\land (b_0\not\equiv0 \mod p)$$ So $(1)$ follows.

• Ah very nice, I like your method :) Thank you for sharing. May I ask you if my method is also acceptable - obviously yours is prefered, but just trying to consolidate in this area. – nomad609 Nov 23 '16 at 9:50
• You try to prove it directly through $fg=0\iff f=0\lor g=0$, which is ok. But in this way you have to prove all coefficients of $fg$ are $0$, which clearly takes more steps. – Math Wizard Nov 23 '16 at 17:27