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Let $A$ be a ring with unity such that $\operatorname{char} A=8$. Let $I$ be an ideal of $A$. Show that $\operatorname{char}(A/I) \neq 0$ and $\operatorname{char}(A/I)\leq 8$.

What I think I know:

Characteristic of a ring is smallest number of times one must use the ring's multiplicative identity element (1) in a sum to get the additive identity element (0)

So in $A$ we have $1+1+1+1+1+1+1+1=0$. From this, can I deduce $A$ is finite? Since characteristic of a ring and order of a group are related?

An ideal a subset which has the properties "closed with respect to subtraction" and "$I$ absorbs products from $A$".

So let $h,i\in I$ and $a\in A$; then $h-i \in I$ and $ai\in I$.

Where I get a bit lost is with the quotient ring. I tried using the Fundamental Homomorphism Theorem to show if there is an isomorphism $\phi$ from $A/I$ to the image of $\phi$ then $\operatorname{char}(A/I)$ must be $\leq 8$. But I am not sure if that's even the way to go, and if it is, I could use some help turning that idea into a formal proof.

Also, I am not sure if this can be applied to the characteristic of rings. But I do know that, from Group Theory, the order of $A/I$ divides the order of $A$, if $I$ is the kernel of $\phi$. However, I do not know if $I$ is the kernel. Is there another relationship I am unaware of by chance?

Any help, even just getting this started would be greatly appreciated!

Thank you!

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Consider the element $1+I$ in $R/I$. Is it clear that this is the multiplicative identity in $R/I\,$? Then to say something about the characteristic, we need to look at $$\sum_1^n (1+I)$$

$$= \left(\sum_1^n 1\right)+I\quad\text{(why?)}$$

But we know that this is the zero element of $R/I$ if $n = 8$ since the characteristic of $R$ is 8. This tells us two things: First, the quotient ring has positive characteristic and second, it is at most 8.

Also, the characteristic of the ring and the order are not related as you suggest since you can have polynomial rings like $\mathbb{Z}_8[x]$; this ring has characteristic 8 and is infinite.

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  • $\begingroup$ I have doubt , Why at most $8$ ? why not $8$ ,as according to your answer $\sum_{1}^{n} 1=e$ then its chararcteristic should be $8$ $\endgroup$ – TheStudent Nov 28 '19 at 4:41
  • $\begingroup$ @TheStudent It could be that there is a number $n$ smaller than 8 that gives you the identity. For instance, if you had a ring of characteristic 4, then it's still the case that $\sum_1^8 1 = e$, but 8 is not the characteristic. $\endgroup$ – Chris Brooks Nov 28 '19 at 5:18
  • $\begingroup$ but $1$ is in $R$ so n should be $8$ $\endgroup$ – TheStudent Nov 28 '19 at 5:34
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For any ring $A$, there is a unique ring homomorphism $\phi_A \colon \mathbb Z \to A$, since the generator $1 \in \mathbb Z$ must be sent to $1 \in A$. Another way of describing the characteristic of $A$ is as the (non-negative) generator of the kernel of $\phi_A$. Since the map $\phi_{A/I}\colon \mathbb Z \to A/I$ factors as $$ \mathbb Z \stackrel{\phi_A}{\longrightarrow} A \longrightarrow A/I , $$ we have $\ker \phi_A \subset \ker \phi_{A/I}$, and so the characteristic of $A/I$ divides the characteristic of $A$. In particular, for $\operatorname{char} A > 0$ we have $\operatorname{char} A/I \leq \operatorname{char} A$ and $\operatorname{char} A/I \neq 0$.

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