3
$\begingroup$

Let $\mathcal R$ and $\mathcal S$ be non zero commutative rings with unity. Then:

  1. char($\mathcal R$) is always a prime number

$\mathbb{Q,R}$ contradicts this. ($char(\mathcal R)=0$)

2.If $\mathcal S$ is a quotient ring of $\mathcal R$ then either $char(\mathcal S)$ divides $char(\mathcal R)$ or $char(\mathcal S)=0$

True. This is the argument I used: If ring homomorphism exits between $\mathcal R$ and $\mathcal S$ then $char(\mathcal S)|char(\mathcal R)$, which is true for a ring and its quotient ring. Are there any other arguments that can be used to justify this answer?

Also, if $\mathcal S$ is a quotient ring of $\mathcal R$ am I right in assuming that $char(\mathcal S)=0$ happens only when $char(\mathcal R)=0$?

  1. If $\mathcal S$ is a subring of $\mathcal R$ containing $1_\mathcal R$ then $char(\mathcal R)=char(\mathcal S)$

True. $1_\mathcal R$ can't behave differently in a subring.

Edit: Any examples where subring has a different characteristic from that of the ring?

  1. If $char(\mathcal R)$ is a prime number, then $\mathcal R$ is a field.

Statement is wrong (But true the other way around, Field$\implies char(\mathcal R)=0$ or prime)

Are my justifications correct?

$\endgroup$
3
$\begingroup$
  1. To do even better, you could add an example of a ring with a characteristic that is a composite number.

  2. What you have written here is not very convincing, and even looks a bit like you are begging the question. How about you work in terms of the homomorphism $\phi$ and argue using $1$ and $\phi(1)$? If you are not granted the identity, then you can still use this strategy on elements.

  3. Your reason is good. In $\Bbb Z/(10)$ the ideal generated by $(6)$ is a ring with identity which has a different characteristic than the containing ring.

  4. "Statement is wrong" is not really a justification. It is easy to produce an example considering that a product of rings with characteristic $p$ also has characteristic $p$.

$\endgroup$
  • $\begingroup$ "...and argue using 1 and $\phi(1)$" -can you throw some more light into it? $\endgroup$ – Jesse P Francis Jun 12 '15 at 15:01
  • $\begingroup$ Also, I can't think of product of rings with characteristic p having characteristic p! I said its wrong in the light of the fact that integral domains also has characteristic p or 0. $\endgroup$ – Jesse P Francis Jun 12 '15 at 15:29
  • 1
    $\begingroup$ @JessePFrancis For the first comment, you're just supposed to work with the fact that $n\cdot\phi(1)=\phi(n\cdot 1)$ $\endgroup$ – rschwieb Jun 12 '15 at 15:38
  • $\begingroup$ @JessePFrancis For the second comment: you are telling me that you can't think of a single ring $R$ with characteristic $p$ and then forming $R\times R$? $\endgroup$ – rschwieb Jun 12 '15 at 15:40
  • $\begingroup$ thank you! Got it cleared! $\endgroup$ – Jesse P Francis Jun 12 '15 at 16:57

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.