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Is it possible for a field to exist containing both $\mathbb{Z}/{5}$ and $\mathbb{Z}/{7}$ as subfields?

I know that there is a result that says that every field must have a subfield isomorphic to either $\mathbb{Z}/{p}$ for some prime $p$ or $\mathbb{Q}$, but it doesn't tell us for which primes $p$, or whether it can have two subfields of prime order.

$\mathbb{Z}/{35}$ contains both $\mathbb{Z}/{5}$ and $\mathbb{Z}/{7}$, but it's not a field.

Could anyone point me in the right direction here?

Thank you for your time and patience.

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    $\begingroup$ $1+1+1+1+1=0=1+1+1+1+1+1+1$. Therefore $1+1=0=1+1+1+1+1$. Hence $1+1+1=0=1+1$. It follows that $1=0$, and then everything is $=0$. $\endgroup$ – user553213 Apr 28 '18 at 21:22
  • $\begingroup$ @deyore are you assuming the existence of such a field at the beginning? $\endgroup$ – ALannister Apr 28 '18 at 21:25
  • $\begingroup$ Sylow theorem's $\endgroup$ – Tiger Blood Apr 28 '18 at 21:28
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    $\begingroup$ Dear OP, I'm sorry, but I couldn't help but replace $\mathbb Z_p$ by $\mathbb Z/p$, since in very many contexts $\mathbb Z_p$ (for prime $p$) is the ring of $p$-adic integers... If you object, please do change back. $\endgroup$ – paul garrett Apr 28 '18 at 23:15
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    $\begingroup$ $1=15-14=0-0=0$ $\endgroup$ – Pierre-Yves Gaillard Apr 28 '18 at 23:25
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The characteristic of a field is the smallest positive integer $n$ such that $n\times 1=0$, or $\infty$ if there is no such integer. If it is not infinite, then it has to be a prime number.

A couple of relevant results:

  1. $\Bbb F_p$ has characteristic $p$ for any prime $p$.
  2. If $K$ is a subfield of $L$, then they both have the same characteristic.

As a consequence, there can be no field containing $\Bbb F_5$ and $\Bbb F_7$ as subfields.

Edit: $\Bbb F_p$ here is just the field notation for $\Bbb Z_p$, when you want to emphasize that you are talking about the field rather than the group.


Edit to include the answers to your questions in comments:

Proof of 2. above. Call $\phi$ the injective field homomorphism $K\to L$. Then $\phi(n\times 1_K)=n\times 1_L$ for any $n\in \Bbb N$. This implies, on the one hand, that if $n\times 1_K=0$ then $n\times 1_L=0$. On the other hand, since $\phi$ is injective, it also implies that if $n\times 1_L=0$ then $n\times 1_K=0$.

An infinite field may have finite characteristic. The field of fractions of $\Bbb Z_2[X]$ has characteristic $2$, but it is infinite.

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  • $\begingroup$ That is not a shorter way. That is making a claim without proving it. $\endgroup$ – user553213 Apr 28 '18 at 21:28
  • $\begingroup$ @ArnaudMortier is $L$ also a field? (Possibly a stupid question, but humour me please) $\endgroup$ – ALannister Apr 28 '18 at 21:29
  • $\begingroup$ @ALannister yes. $\endgroup$ – Arnaud Mortier Apr 28 '18 at 21:29
  • $\begingroup$ Even more, it took you over 6 lines to make the unproven claim. I took 2 to actually prove it. $\endgroup$ – user553213 Apr 28 '18 at 21:31
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    $\begingroup$ @ALannister Call $\phi$ the injective field homomorphism $K\to L$. Then $\phi(n\times 1_K)=n\times 1_L$ for any $n\in \Bbb N$. This implies, on the one hand, that if $n\times 1_K=0$ then $n\times 1_L=0$. On the other hand, since $\phi$ is injective, it also implies that if $n\times 1_L=0$ then $n\times 1_K=0$. $\endgroup$ – Arnaud Mortier Apr 29 '18 at 1:11
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The ring $\def\Z{\mathbb{Z}}\Z/35\Z$ contains $\Z/5\Z$ and $\Z/7\Z$ as subrings only if you don't require subrings share the identity with the ring.

However, a subfield of a field must share the identity, because fields have a single nonzero idempotent, namely $1$: indeed, from $x^2=x$ we can deduce, in a field, $x=0$ or $x=1$.

On the other hand, there is at most one unital ring homomorphism $\chi\colon\Z\to R$, for every ring $R$ (with $1$, of course), because the map $\chi$ is determined by $\chi(1)=1$, being a group homomorphism.

Actually, $\chi$ exists, but this is not needed for the argument here.

If both $\Z/5\Z$ and $\Z/7\Z$ are (unital) subrings of $R$, there would be two distinct homomorphisms $\Z\to R$. Contradiction.

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Hint: If a field has an element $x\neq 0$ with $px=0$ then for all elements $y$ of the set $py=0$.

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  • $\begingroup$ I'm not sure how that helps. $\endgroup$ – ALannister Apr 28 '18 at 21:24
  • $\begingroup$ You forgot to impose that $x\neq0$. $\endgroup$ – José Carlos Santos Apr 28 '18 at 21:24
  • $\begingroup$ @NickA. are you saying the same thing that deyore is saying in their comment above? $\endgroup$ – ALannister Apr 28 '18 at 21:26
  • $\begingroup$ @ALannister Yes, exactly. I didn't see the commen before I posted. Basicaly, he proved my "claim". $\endgroup$ – Nick A. Apr 28 '18 at 21:36

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