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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.

-nomad609

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

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  • $\begingroup$ 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. $\endgroup$ – nomad609 Nov 23 '16 at 9:50
  • $\begingroup$ Thank you for your reply. May I ask you elucidate what you mean please. Shoudl I just follow what you did instead - or instead just merge the two ideas together. I am unsure because your method seems cleaner. Thank you for your help :) $\endgroup$ – nomad609 Nov 23 '16 at 16:23
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    $\begingroup$ 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. $\endgroup$ – Math Wizard Nov 23 '16 at 17:27
  • $\begingroup$ Ah I see what you mean. Thanks :L $\endgroup$ – nomad609 Nov 23 '16 at 17:41
  • $\begingroup$ Certainly. Kind Regards $\endgroup$ – nomad609 Nov 23 '16 at 18:15

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