3

I have the following problem that I need to solve:

For which of $m \in \{5, 6\}$ is $\mathbb Z/m\mathbb Z$ a field?

I know how to do this for $m = 5$. I know that $[1][1] = [1]$, $[2][3] = [1]$, $[4][4] = [1]$.

So every non-zero element of $\mathbb Z/5\mathbb Z$ has an inverse, and so $\mathbb Z/5\mathbb Z$ is a field.

However, I am not sure how to do this for $m = 6$.

bzc
  • 8,622
Amber
  • 175
  • I have replaced some of your stuff with symbolic notation. Just make sure that what I edited in actually means the same as what you meant, since I don't know this particular math itself very well. – AJMansfield Jul 29 '13 at 18:51
  • 1
    Hint $2 \cdot 3=6$. – N. S. Jul 29 '13 at 18:52

3 Answers3

4

Hint(s): What is $[2][3]$ in $ \mathbb{Z}/6\mathbb{Z}$?

For a more general result, see if you can prove that $\mathbb{Z}/m\mathbb{Z}$ is a field if and only if $m$ is prime. (a hint for proving this: since $m$ is prime, every nonzero element of $\mathbb{Z}/m\mathbb{Z}$ is relatively prime to $m$!)

Alex Wertheim
  • 20,278
  • 1
    i might be wrong but its it 0? – Amber Jul 29 '13 at 18:45
  • [2][2] = [4], [3][5] = [15] = [3], [5][5] = [25] = [1], – Amber Jul 29 '13 at 18:54
  • 1
    Precisely. So what are the inverses of $[2]$ and $[3]$ in $\mathbb{Z}/6\mathbb{Z}$? (You could cut this shorter by noting that every field is an integral domain, and since $\mathbb{Z}/6\mathbb{Z}$ has zero divisors, it is not an integral domain and hence cannot be a field). – Alex Wertheim Jul 29 '13 at 19:08
2

Note that if $m$ is composite, say $m=p\cdot q$; then $p,q\not \equiv 0 \pmod m$ in $\Bbb Z/m\Bbb Z$ yet $p\cdot q\equiv 0\pmod m$ so $\Bbb Z/m\Bbb Z$ is not even a domain, that is, there are zero divisors!

Pedro
  • 122,002
1

Since $\mathbb{Z}$ is commutative ring with unity and A is ideal of R=$\mathbb{Z}$ then use this result R/A is field if and only if A is maximal ideal.

Here $A=\left\langle 6 \right\rangle$(generated by 6) is not maximal ideal.

I feel still you are not understand

here $\left\langle 6 \right\rangle \subset \left\langle 2 \right\rangle \subset \mathbb{Z}$ so it is not maximal it implies that$ \mathbb{Z}/ 6\mathbb{Z}$ is not field.

Additional Ans. of below comments:

For example, Let us consider $\mathbb{Z}/ 4\mathbb{Z}=\{0+4\mathbb{Z},1+4\mathbb{Z},2+4\mathbb{Z},3+4\mathbb{Z}\}$

If possible suppose there is $a+4\mathbb{Z} \in \mathbb{Z}/4\mathbb{Z}$

such that $(2+4\mathbb{Z})(a+4\mathbb{Z})=(1+4\mathbb{Z})$

$(2+4\mathbb{Z})(a+4\mathbb{Z})=2.a+2.4\mathbb{Z}+4\mathbb{Z}.4\mathbb{Z}+a.4\mathbb{Z}$

but all the other elements except '$2.a$' will absorbed in $4\mathbb{Z}$

So we get $2.a+4\mathbb{Z}=(1+4\mathbb{Z})$

it implies that $2.a-1 \in 4\mathbb{Z}$

that is not possible.

therefore $2+4\mathbb{Z}$ has no inverse in $\mathbb{Z}/4\mathbb{Z}$

Siddhant Trivedi
  • 2,298