3
$\begingroup$

Does it contradict the axioms of a field? I think not. If not so there need to be $a \in Z_7$ and so $3+a=0$ and $3\cdot a=1$ but I can not find this $a \in Z_7.$

$\endgroup$
13
$\begingroup$

If $a+b=0$, then $b=-a$. If simultaneously $1=ab$, then $1=-a^2$ which is equivalent to $a^2=-1$. This is possible in $\Bbb{Z}_p$, if and only if $p=2$ or $p\equiv1\pmod4$.

So you're right. This cannot happen in $\Bbb{Z}_7$.

$\endgroup$
  • $\begingroup$ What do you say about: math.stackexchange.com/a/863700/103441 Thanks $\endgroup$ – gbox Jul 10 '14 at 21:55
  • 3
    $\begingroup$ @gbox: You were discussing field axioms. Therefore I assumed that you are only interested in the case, where $\Bbb{Z}_p$ is a field. In other words I assume that $p$ is a prime. Kaj's extra examples (+1) concern the cases where that is not the case. If $m$ is not a prime, there is an element $a$ such that $a^2=-1$ if and only if $m$ is not divisible by four, and all its odd prime factors are $\equiv1\pmod4$. $\endgroup$ – Jyrki Lahtonen Jul 10 '14 at 22:11
  • $\begingroup$ Thanks a lot! I understood your proof for $1=-a^2$ but where did $a^2=-1$ came from? $\endgroup$ – gbox Jul 10 '14 at 22:19
  • 1
    $\begingroup$ Multiply the previous equation by $-1$. Ahh. It is phrased a bit oddly. Let me clarify. $\endgroup$ – Jyrki Lahtonen Jul 10 '14 at 22:19
1
$\begingroup$

Absolutely! I wrote a Python program a while ago to generate some examples for general quotient rings $\mathbb{Z}_n$. Here are some:

$2$ and $3$ $\pmod{5}$

$3$ and $7$ $\pmod{10}$

$5$ and $8$ $\pmod{13}$

$4$ and $13$ $\pmod{17}$

Anyone interested can find my Python code here. In general, let's suppose a pair of such elements exists in the ring $\mathbb{Z}_m$. They will satisfy the system $x+y \equiv 0 \pmod{m}$ and $xy \equiv 1 \pmod{m}$, and from this we discover that such a pair exists if and only if there exists an $x \in \mathbb{Z}_m$ such that $x^2 \equiv -1 \pmod{m}$.


To expand on your specific question regarding $\mathbb{Z}_p$ when $p$ is prime, note that if there is an element $x$ such that $x^2 \equiv -1 \pmod{p}$, then in particular $x^4 \equiv 1 \pmod{p}$. That is, there must exist an element of order $4$ in the multiplicative group $\mathbb{Z}_p^\times$. Now, it is a theorem that $\mathbb{Z}_p^\times \cong \mathbb{Z}_{p-1}$ when $p$ is prime. Finally, since $\mathbb{Z}_{p-1}$ is cyclic, it will contain an element of order $4$ $\iff$ $4|(p-1)$. So this phenomenon occurs in $\mathbb{Z}_p \iff$ $p \equiv 1 \pmod{4}$.

$\endgroup$
  • $\begingroup$ Thanks! I do not know Python. but I did not understand your question $\endgroup$ – gbox Jul 10 '14 at 21:51
  • 1
    $\begingroup$ @gbox, the claim is that there will exist two elements that are each others additive and multiplicative inverse in $\mathbb{Z}_m$ whenever $m = a^2 + 1$. This is a sufficient condition, but not a necessary condition (useful for generating examples). I edited my response with a necessary and sufficient condition. $\endgroup$ – Kaj Hansen Jul 10 '14 at 22:01
  • 1
    $\begingroup$ @gbox, I added some detail to my response so that, hopefully, it ties in with Jyrki's response and makes a bit more sense. $\endgroup$ – Kaj Hansen Jul 11 '14 at 4:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.