This question already has an answer here:
a) If A is a local ring then A has characteristic zero or a power prime.
Suppose M is the unique maximal ideal of A then $A/M$ is Field in particular integral domain then $Char( A/M ) = 0$ or $p$ with some $p$ prime.
If $Char( A/M ) = 0$ then $\forall n \in \mathbb N, $ , $\forall a+M \in A/M $ we have that $ n(a + M )= na + M \ne M$ then $ na \notin M$ $\forall n \in \mathbb N $, so $ n 1 \notin M $, this implies $n 1 \ne m $ $ \forall m \in M $ $ \forall n \in \mathbb N $ then $ n 1 \ne 0 $ $ \forall n \in \mathbb N $ finally $Char ( A )= 0$
Suppose that $Char( A/M ) = p $ and $Char( A ) = n$ then$ n 1 \ne m$ $ \forall m \in M $ , so $ M + n 1 = M $ and since $Char( A/M ) = p $ this implies that $ p | n$ then $n = pq$ , so we have $ n= p^l m$ where $ (p,m)=1$ but with this there are $ x,y\in \mathbb Z$ such that $ 1= px 1 + my 1$, so $ px 1 $ is unity or $my 1$ is unity.
Since $p 1 \in M $ then $px 1 \in M $ this implies $px 1 \notin A^* $ then $ my 1 \in A^* $ ,so for $ a \in A$ such that $my 1 a = 1$ then $m 1$ is unity with this we have that $o(m1) = o(1)$ and since $p^l(m1)=0$ this implies that $$p^lm | p^l$$ then $$ m=1 $$.
Finally we get that $Char( A ) = p^l $ for some integer l and p prime.
Is this correct?
b) Let A a ring commutative with identity and with characteristic n. If $n= ab$ with $ (a,b)=1$ then A is isomorphic to the direct product of two rings,one of them is characteristic a and the other one is characteristic b.
I cannot prove this can someone help me please.