2
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Check the dates, this post was upload in 2013, the other one 2018. $\endgroup$ – Rachel Sep 4 at 21:29
  • 1
    $\begingroup$ 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. $\endgroup$ – Mike Pierce Sep 4 at 21:39
2
$\begingroup$

(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 $b$).

(a) follows from (b) since a local ring has only trivial idempotents.

$\endgroup$
  • $\begingroup$ can you be more explicit in the part a) , and a little help for prove that $char(A/aA)=a$ .Please. $\endgroup$ – Rachel Oct 15 '13 at 22:08
  • $\begingroup$ i´m trying the same that in a) in my last proof , but i do not how to apply that the characteristic of A is n. $\endgroup$ – Rachel Oct 15 '13 at 22:25
  • $\begingroup$ What is your question? $\endgroup$ – Martin Brandenburg Oct 15 '13 at 22:47
  • $\begingroup$ 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. $\endgroup$ – Rachel Oct 15 '13 at 23:40
  • $\begingroup$ You don't see why $a \cdot 1 = 0$ in $A/aA$? Repeat the definition of a quotient ring. $\endgroup$ – Martin Brandenburg Oct 16 '13 at 1:44

Not the answer you're looking for? Browse other questions tagged or ask your own question.