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If $a^2+b^2$ is a nilpotent element of $\Bbb Z_n$ then $(a+bi)(a-bi)=a^2+b^2$ is also a nilpotent in $\Bbb Z_n[i]$. Does this imply that $a+bi$ or $a-bi$ is a nilpotent element in $\Bbb Z_n[i]$? Any idea would be helpful.

Adam Hughes
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User32912
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  • So I think $n=10$, $a=1$, $b=3$ is a counterexample but proving it is a lot less straightforward than I expected. – Cameron Williams Feb 22 '17 at 14:46
  • the main thing that i really want to prove is that an element $a+bi$ of $Z_n[i]$ is nilpontent if $a+b = 0 (mod m)$ and $a^2+b^2=0(mod m)$ where $m$ is the product of prime bases in the prime factorization of $n$. So basically $1+3i$ is not a nilpotent element of $Z_{10}[i]$ – User32912 Feb 22 '17 at 14:53
  • I feel like this ends up behind a question about rational multiples of pi as values of inverse trigonometric functions and there are results about that. See here: http://math.stackexchange.com/questions/79861/arctan2-a-rational-multiple-of-pi – Cameron Williams Feb 22 '17 at 14:56
  • how is this related to that? can you please explain it to me if thats okay with you? – User32912 Feb 22 '17 at 15:02
  • If it was nilpotent it would hit the line $y=x$ (with $x$ being an integer). If you write $a+bi$ in exponential form, that means that $m\theta = 2k\pi+ \frac{\pi}{4}$. – Cameron Williams Feb 22 '17 at 15:25
  • $a+ib$ is nilpotent iff $a+ib = 0$ in $\mathbb{Z}_p[i]$ for some prime $p| n$ ? – reuns Feb 22 '17 at 16:09
  • @CameronWilliams It's not straightforward, but easy enough: if $1+3i$ is nilpotent in $\mathbb Z_{10}[i]$ then it is so in $\mathbb Z_{5}[i]$. But the last ring is isomorphic to $\mathbb Z_5\times\mathbb Z_5$ hence it doesn't have non-zero nilpotents. – user26857 Feb 23 '17 at 13:40
  • @user26857 ohh that's nice! – Cameron Williams Feb 23 '17 at 13:54
  • @user1952009 $\hat a+\hat bi$ is nilpotent in $\mathbb Z_n[i]$ iff $\bar a+\bar bi=0$ in $\mathbb Z_p[i]$ for any prime $p\mid n$ iff $p\mid a$ and $p\mid b$ for any prime $p\mid n$. This shows that if $\hat a+\hat bi$ is nilpotent in $\mathbb Z_n[i]$ then $\hat a^2+\hat b^2$ is nilpotent in $\mathbb Z_n$. – user26857 Feb 23 '17 at 20:32
  • @user1952009 Actually I had to write "This shows that if $\hat a+\hat bi$ is nilpotent in $\mathbb Z_n[i]$ then $\hat a,\hat b$ are nilpotent in $\mathbb Z_n$". On the other side, I don't think that your claim $\hat a^2+\hat b^2$ is nilpotent iff $\gcd(a^2+b^2,n)>1$ is correct. (At least not $\Leftarrow$.) Instead it's true that $\hat a^2+\hat b^2$ is nilpotent iff $p\mid a^2+b^2$ for any prime $p\mid n$. This shows clearly that the OP's claim is wrong. – user26857 Feb 24 '17 at 06:52
  • @Isiah You are right: $n$ even is a case where such a phenomenon can occur since $\mathbb Z_2[X]/(X^2+1)=\mathbb Z_2[i]$ contains non-zero nilpotents (for instance, $1+i$), so my claim that $\hat a,\hat b$ are nilpotent in $\mathbb Z_n$ holds for $n$ odd. However if $n\ne 2^k$ we get $p\mid a$ and $p\mid b$ for any odd prime $p\mid n$. – user26857 Feb 24 '17 at 13:17
  • Now let's see what's going on if $n=2^k$: if $\hat a^2+\hat b^2$ is nilpotent in $\mathbb Z_n$ then $2\mid a^2+b^2$, so $a,b$ are both even or both odd. If they are even then $\hat a,\hat b$ are nilpotent in $\mathbb Z_n$, so $\hat a+\hat bi$ is nilpotent in $\mathbb Z_n[i]$. If both are odd, then $\hat a+\hat bi$ is also nilpotent in $\mathbb Z_n[i]$ since $(\hat a+\hat bi)^2=\hat a^2-\hat b^2+2\hat a\hat b i$ and $a^2-b^2$ and $2ab$ are both even, hence $\hat a^2-\hat b^2$ and $2\hat a\hat b$ are nilpotent in $\mathbb Z_n$. – user26857 Feb 24 '17 at 13:29
  • How about in $Z_6[i]$, $Z_{10}[i]$, $Z_{12}[i]$? Is there any way to determin nilpotent elements when 2 is one of the prime factors of n? – User32912 Feb 24 '17 at 13:48
  • @Isiah I don't understand: now do you want to determine all nilpotent elements in $\mathbb Z_n[i]$? – user26857 Feb 24 '17 at 14:46
  • @user1952009 You should refresh a little the basic algebra. This way can find out that the nilpotents of $\mathbb Z_n$ are those classes whose representatives are divisible by all primes $p\mid n$. – user26857 Feb 24 '17 at 15:16
  • @user26857 yes of course, I was thinking to the zero-divisors... tks – reuns Feb 24 '17 at 15:18

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