# Characteristic $n$ and local rings [duplicate]

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Prove that:

a) If A is a local ring then A has characteristic zero or a power prime.

Proof.

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.

## marked as duplicate by Mike Pierce, The Count, Lee David Chung Lin, nmasanta, YuiTo ChengSep 5 at 4:42

• Check the dates, this post was upload in 2013, the other one 2018. – Rachel Sep 4 at 21:29
• Right, but an older post can be marked as a duplicate of a newer post (heck, it's happened to me). It's more about which one should be the "canonical version," of the question on MathSE. The other one has more thorough answers, and a more terse question statement, so I choose that one. – Mike Pierce Sep 4 at 21:39

(b) Since $a\mathbb{Z}$ and $b\mathbb{Z}$ are coprime, the same is true for $aA$ and $bA$. The Chinese Remainder Theorem implies that $A \cong A/aA \times A/bA$. Now prove that $\mathrm{char}(A/aA)=a$ (resp. for $b$).
• can you be more explicit in the part a) , and a little help for prove that $char(A/aA)=a$ .Please. – Rachel Oct 15 '13 at 22:08
• Well since $char(A)=n$ then $n1=0$ so $A+n1=A$ , and $aA + an1 = aA$ he i stuck how prove that $a(a+aA)=0$ but i can see this. – Rachel Oct 15 '13 at 23:40
• You don't see why $a \cdot 1 = 0$ in $A/aA$? Repeat the definition of a quotient ring. – Martin Brandenburg Oct 16 '13 at 1:44