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I had a query regarding the characteristic of the direct sum of two rings. That is, given two rings R and S, with characteristics m and n respectively, show that: $$char(R \oplus S)=lcm(m,n)$$ That is, it is the least common multiple of the two characteristics of the rings given. The definition of characteristic of a ring that I have been using is as follows: The characteristic of a ring $R$ is the least positive integer $n>0$ such that $nx=0$ for all $x$ in $R$. If no such integer exists, the characteristic of $R$ is said to be $0$.

As a side note: I do know how to do this for the specific case when $R=\mathbb{Z}_m$ and $S=\mathbb{Z}_n$ where it is clear that $char(R)=m$ and $char(S)=n$. Here is how I would approach this specific case:

Claim: $$char({Z}_m \oplus {Z}_n)=lcm(m,n)$$ Let $k=lcm(m,n)$ and $Q={Z}_m \oplus {Z}_n$. Suppose $q$ is the characteristic of $Q$. As $(1,1)$ generates the ring $Q$, it is clear that $k(1,1)=(k,k)=(0,0)$ and hence $kx=0$ for all $x\in Q$. That is, $q$ must divide $k$ (by definition of characteristic) and hence $q\le k$. Then, using the definition of characteristic, the following holds. $$q\cdot 1\equiv 0\pmod m$$ $$q\cdot 1\equiv 0\pmod n$$ This implies (by the definition of modular arithmetic), that $m$ divides $q$ and $n$ divides $q$. Then, by the definition of characteristic and least common multiple, the smallest integer for the above to hold is consequently $lcm(m,n)$. Thus, $q=k=lcm(m,n)$.

I know I cannot directly apply this specific case to the more general one because it is not certain that the rings $R$ and $S$ are indeed generated by 1 element - So the same argument cannot be used. If someone could give me a push in the right direction, I would be most grateful. I have been looking at this concept for a few weeks and am interested in proving it.

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    $\begingroup$ The characteristic of a ring is the order of $1_R$ (the identity of $R$) as an element of the additive group $(R,+)$. Then use math.stackexchange.com/questions/724348/… $\endgroup$
    – user26857
    Mar 24, 2016 at 23:59
  • $\begingroup$ Great. This is a great alternative method to the other ones suggested. Thank you. $\endgroup$
    – Mohabat
    Mar 25, 2016 at 1:54

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There's a perhaps better way to define the characteristic of a ring (I assume rings have an identity $1$). For each ring $R$, the map $$ \chi_R\colon\mathbb{Z}\to R,\qquad n\mapsto n1 $$ is the only ring homomorphism $\mathbb{Z}\to R$. Then there is a unique $k\ge 0$ such that $\ker\chi_R=k\mathbb{Z}$ and $k$ is the characteristic of $R$ (the proof is straighforward).

Now, if $R$ and $S$ are rings, we have $$ \chi_{R\times S}(n)=n(1,1)=(n1,n1) $$ so $\ker\chi_{R\times S}=\ker\chi_R\cap\ker\chi_S$.

If $a\mathbb{Z}$ and $b\mathbb{Z}$ are ideals of $\mathbb{Z}$, then $$ a\mathbb{Z}\cap b\mathbb{Z}=\operatorname{lcm}(a,b)\mathbb{Z} $$ (where I use $\operatorname{lcm}(a,0)=0$, together with $\gcd(a,0)=a$, so the relation $\operatorname{lcm}(a,b)\gcd(a,b)=ab$ holds without restrictions, provided $a,b\ge0$).

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  • $\begingroup$ Thank you greatly for your detailed response. It lays out a very clear explanation. I voted you up! Thanks again. $\endgroup$
    – Mohabat
    Mar 25, 2016 at 1:52
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I know I cannot directly apply this specific case to the more general one because it is not certain that the rings $R$ and $S$ are indeed generated by 1 element - so the same argument cannot be used.

The underlying additive group being cyclic isn't really important. If your definition of characteristic is the order of the multiplicative identity in the underlying additive group, then your argument goes through without a hitch. If you want to speak of not necessarily unital rings, then the argument needs to be augmented slightly to work. Firstly, the definition of characteristic is then the unique nonnegative principal generator of the annihilator ideal in $\mathbb{Z}$, treating the underlying additive group of $R$ as a $\mathbb{Z}$-module. Secondly, write

$$\begin{array}{ll} & k(x,y)=(0,0) \quad \forall x\in R,y\in S \\ \iff & (kx=0 ~~\forall x\in R)~\mathrm{and}~(ky=0~~\forall y\in S) \\ \iff & n\mid k~\mathrm{and}~m\mid k \\ \iff & \mathrm{lcm}(n,m)\mid k. \end{array}$$

Thus, $\mathrm{Ann}_\mathbb{Z}(R\oplus S)=(\mathrm{lcm}(n,m))$.

More generally if $A$ is a ring then $\mathrm{Ann}_A(M\oplus N)=\mathrm{Ann}_A(M)\cap \mathrm{Ann}_A(N)$ for $A$-modules $M,N$.

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    $\begingroup$ Thank you for taking the time to answer. I understand this uses a higher-level understanding of "modules" which I have not encountered in my experience before. $\endgroup$
    – Mohabat
    Mar 25, 2016 at 1:53
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    $\begingroup$ Thank you for also mentioning that the generator-argument is unnecessary as long as both rings have unity - You provided insight into my problem. $\endgroup$
    – Mohabat
    Mar 25, 2016 at 12:51

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