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The question pertains to the question in the title. That is,

Show that if $n$ is not prime then $\Bbb Z /n\Bbb Z$ is not a field.

I realise that this has many duplicates, but my question is rather this;

Why does it not suffice to say that if $n$ is not prime then $\lvert\Bbb Z /n\Bbb Z \rvert = n \neq p^m$ for some prime $p$ and integer $m$ and so is not a field by definition?

I realise this somewhat misses the reason why it's not a field, but would it suffice as an answer to this question?

Chill2Macht
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Edward Evans
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  • @DietrichBurde Did you read the question? – Edward Evans Oct 23 '16 at 19:37
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    You are right, a finite field must have $n=p^m$ elements. But also $\mathbb{Z}/p^m$ is not a field for $m>1$, but has $p^m$ elements. And you have to prove that a finite field must have $p^m$ elements, too - which is as difficult as proving the duplicate. And yes, I read the question, and still think, the main task is to show both directions, like in the duplicate. – Dietrich Burde Oct 23 '16 at 19:37
  • Basically to prove that a finite field has $p^{m}$ elements requires knowing about prime subfields which in turn requires knowing when is $\mathbb{Z}_{n}$ is a field (so you are assuming your result, essentially, to prove it, this way). – Kevin Bowman Oct 23 '16 at 20:50

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Because this is not enough to say that $\mathbb Z/9\mathbb Z $ is not a field, as $9$ is in fact of the form $p^m$ for a prime $p$ and positive integer $m$.

Asinomás
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A major reason why you cannot use this argument is that you seem to saying that because you know every finite field must have $p^{m}$ elements you get a contradiction.

However how do you know that you don't need this result in order to prove that finite field must have $p^{m}$ elements.

So does the proof of finite field must have $p^{m}$ elements, require the knowledge of the prime subfield and does this in turn require knowing that $\mathbb{Z}_{n}$ is a field iff $n$ is prime...

Kevin Bowman
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