Let $\mathcal R$ and $\mathcal S$ be non zero commutative rings with unity. Then:
- 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 $char(\mathcal R)=0$?
- 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?
- 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?
Best Answer
To do even better, you could add an example of a ring with a characteristic that is a composite number.
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.
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.
"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 $p$.