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?

  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 $p$.

  • $\begingroup$ "...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
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    $\begingroup$ @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

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