is characteristics of integral domain ring also always a element of ring?

Question

In the proof of Theorem "The characteristic of an integral domain is either prime or zero" everyone assumed at first characteristics of ring $n = ab$. then try to prove n is not composite. My concern is if we write $n =ab$ it also means n is also an element of the ring by closure property.

But We have seen that ring like $\mathbb{Z}6$ or $\mathbb{Z}7$ their characteristics are not included as the element of ring. Even though $\mathbb{Z}6$ or $\mathbb{Z}7$ etc they are not integral domain but for the general case of shouldn't we assume $n$ could be out of the set of the ring? Or we should prove at first every integral domain characteristics ring is an element of the ring?

Answer 1

Your question doesn't really make sense. For every commutative ring with $1$, call it $A$, we have a unique ring homomorphism $\mathbb{Z} \to A$. Then when for $n \in \mathbb{Z}$ we refer to $n$ as an element of $A$, we implicitly assume the image of $n$ under this homomorphism. In other words, when we refer to $5$ as an element of $A$, we implicitly mean $1 + 1 + 1 + 1 + 1$, where $1$ is the unit of $A$. So there is no question of whether $5$ is in $A$ or not, just the definition as I explained. So $n$ means $1$ plus itself $n$ times, whatever this gives in $A$.

Hope this is clear.

Answer 2

As an alternative view to Sasha's answer, step back from rings to groups (where you are probably much less inclined to consider group elements as numbers), where we have the concept of order of an element $x$, i.e., the smallest positive integer $n$ such that $x^n:=\underbrace{x\circ x\circ\cdots\circ x}_n$ is the neutral element. We also have the concept of exponent of a group, i.e., the smallest positive integer $n$ such that $x^n$ is the neutral element for all elements $x$ of the group.¹ Note that in the definition of $x^n$, only $x$ is a group element whereas $n$ is instead a count of factors in the defining expression. For example, in the group of symmetries of a regular pentagon, we have reflections of order $2$, rotations of order $5$, and the neutral element of order $1$, and the exponent of the group is $10$.

As a matter of notation, we prefer to use $+$ for the group operation when dealing with abelian groups. In that case, we also write $n\cdot x$ for $x^n$ (i.e., for $\underbrace{x+x+\cdots+x}_n$. Note that "$\cdot$" is not from the group (which only has a "$+$"), but is as external to the abelian group as the exponential notation above is to the general group. The good point is that $\cdot$ defines an action of the group $\Bbb Z$ on our group, which just means that the rules that "should" hold for the combination of multiplication and addition, do hold, namely $$\tag1\begin{align}1\cdot x&=x,\\ n\cdot (x+y)&=n\cdot x+n\cdot y,\\ (n+m)\cdot x&=n\cdot x+n\cdot y,\\ (n\cdot m)\cdot x&=n\cdot(m\cdot x)\end{align} $$ (even though in the third equality, the two "$+$" symbols stand for different operations in different groups, and only one of the four "$\cdot$"'s in the fourth equality is a ring multiplication in $\Bbb Z$). The not so good point is that this very fact may lead to confusion when the two kinds of $+$ and the two kinds of $\cdot$ seem to be the same. In the case of a ring instead of an abelian group, we even get a third multiplication (so we have multiplication in the ring, multiplication in $\Bbb Z$, and the multiplicatively written action of $\Bbb Z$ on the ring. While thanks to rules like $(1)$ we have no problems working with all these different additions and multiplications at once, it is in times like in your question when we have to stand back and be aware of the differences between integers and ring elements, say.

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¹ In this language, the (non-zero) characteristic of a ring is nothing but the exponent of the additive group of the ring.