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I am wondering if anyone could point me to a good method to go about determining if the following sets of integers constitute a field or not: $\mathbb Z_3$, $\mathbb Z_4$, $\mathbb Z_5$, and $\mathbb Z_6$. I have come to the conclusion that the ternary integers do not because there is not multiplicative inverse for the even integers. But, as for the others, I feel that there must exist a method for deciding without going through and checking commutativity, Identity, additive and multiplicative inverses, associativity. Thanks in advance for any assistance.

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  • $\begingroup$ *NO multiplicative inverse I meant to type... $\endgroup$ – court Aug 20 '13 at 4:15
  • $\begingroup$ There is quick method for $Z_q^*$ simply if q is a prime number. $\endgroup$ – SomeOne Aug 20 '13 at 4:24
  • $\begingroup$ Ah man! ;). That's okay... I've still got to prove it! Thank you guys. $\endgroup$ – court Aug 20 '13 at 4:27
  • $\begingroup$ What do ya mean simplify if q is prime? Like, if it's prime then it's not a field? $\endgroup$ – court Aug 20 '13 at 4:28
  • $\begingroup$ If the ternary is a field, what is the multiplicative inverse of 1/2? $\endgroup$ – court Aug 20 '13 at 4:32
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If $s=mn$ is composite, $m,n$ are not zero in $\Bbb Z_s$, but $mn$ is. This means $\Bbb Z_s$ is not even an integral domain if $s$ is composite, let alone a field. If $p$ is prime, $p\mid ab\implies p\mid a$ or $p\mid b$. This means $\Bbb Z_p$ is an integral domain. Moreover, since if $1\leq q<p$ then $p\not\mid q\implies (p,q)=1$, this means all nonzero elements have inverses (use Bezout's theorem), so it is a field.

As it has been brought up, every finite integral domain is a field. The proof is not hard: if $a$ is nonzero consider the map from $D$ to itself given by $x\mapsto ax$. If $ax=ay$ then $a(x-y)=0$. Since $a\neq 0$, $x=y$, so this map is injective, whence it is bijective, since $D$ is finite. Thus there must exist $a'$ that gets mapped to $1$. This means $aa'=1$. Since $a$ was arbitrary, we're done.

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  • $\begingroup$ Thanks, I'll see where this takes me... Thanks again! $\endgroup$ – court Aug 20 '13 at 4:18
  • $\begingroup$ Thank you again Peter. I think I see. $\endgroup$ – court Aug 20 '13 at 4:37
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Think about this: An element $x$ in ${\bf Z}_n$ is invertible if and only if $x$ is relatively prime to $n$.

He explained to you that if $n$ is not prime, then ${\bf Z}_n$ is not a integral domain. It turns out that every finite integral domain is a field.

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    $\begingroup$ "It turns out that every finite integral domain is a field." +1 A proof is not hard to write down: pick $a\in D$ nonzero and look at the map $x\mapsto ax$ $\endgroup$ – Pedro Tamaroff Aug 20 '13 at 4:35
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    $\begingroup$ Precisely. You can show injectivity(cancellation property) and this shows surjectivity (finiteness). Thus there is an $a$ such that $ax=1$ $\endgroup$ – LASV Aug 20 '13 at 4:40

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