The characteristic of a subdomain is the same as the integral domain in which it is contained?

Question

Perhaps I have a misunderstanding of what a subdomain and an integral domain are, but I'm having a hard time figuring this out.

I'm asked to show that the characteristic of a subdomain is the same as the characteristic of the integral domain in which it is contained.

What was tying me up is: $\mathbb Z_7$ is an integral domain. $\mathbb Z_3$ is also an integral domain, and every element in $\mathbb Z_3$ is contained in $\mathbb Z_7$, so isn't $\mathbb Z_3$ a subdomain of $\mathbb Z_7$?

I assume it's probably fairly simple (a misunderstanding of a definition or something), but what am I missing here?

(Edit: I, apparently, had a lapse in brain functioning which resulted in a pretty bad misunderstanding of subrings. Once this was fixed, the proof came naturally.)

Answer 1

A subdomain $R \subseteq S$ is first of all a subring, i.e. the inclusion $i : R \hookrightarrow S$ is a ring homomorphism. Since ring homomorphisms send $1$ to $1$ by definition, $R$ and $S$ must have the same unity element $1$, so $p \cdot 1 = 0$ in $R$ iff $p \cdot 1 = 0$ in $S$, which shows that $R$ and $S$ have the same characteristic.

As for your example, $\mathbb{Z}_3$ and $\mathbb{Z}_7$ do not have the same unity element, and in fact not only is $\mathbb{Z}_3$ is not a subset of $\mathbb{Z}_7$, there is no injective ring homomorphism from $\mathbb{Z}_3$ to $\mathbb{Z}_7$.

Answer 2

Notice $1+1+1=0$ in $Z_3$ but $1+1+1\color{Red}{\ne}0$ in $Z_7$. Just because the symbols used to represent the elements of $Z_3$ are a subset of the symbols used to represent the elements of $Z_7$ doesn't mean that one structure sits inside the other structure. Not even remotely close. For a subset to be a subring its operations must be the same as those in the ring it sits inside.

If $A\to B$ is any homomorphism of unital rings, we have ${\rm char}(B)\mid{\rm char}(A)$. Can you prove this? Consider the fact that the element $\underbrace{1_B+\cdots+1_B}_{{\rm char}(A)}\in B$ is the image of $\underbrace{1_A+\cdots+1_A}_{{\rm char}(A)}=0_A$.

In particular if $B$ is a domain then its characteristic is either zero or a prime number (can you prove this fact?). If ${\rm char}(B)=p$ then $A$ is a domain and its characteristic is a multiple of $p$, so the characteristic of $A$ must also be $p$. I'll let you figure out the characteristic zero case.

Answer 3

More is true: the characteristic is preserved under all injective morphisms of rings*. This is simply because the characteristic of $R$ is the non-negative generator of the kernel of the unique ring morphism $\def\Z{\Bbb Z}\iota_R:\Z\to R$. If $f:R\to S$ is an injective ring morphism, then the kernel of $\iota_S=f\circ \iota_R$ is the same as that of $\iota_R$, and so (the larger ring) $S$ has the same characteristic as$~R$.

As for your confusion about $\Z_3$ and $\Z_7$, it is not true that every element of $\Z_3$ is also element $\Z_7$, since the elements of $\Z_n$ are congruence classes modulo$~n$, and the congruence class of (say) $2$ modulo$~3$ is not the same thing as the congruence class of $2$ modulo$~7$. What is true is that the usual representatives (in $\Z$) of elements in $\Z_3$ are also representatives of (other) elements of$~\Z_7$. But even if one would define these rings so that these are really elements rather than representatives, then $\Z_3$ is still not a subring of $\Z_7$ because the (addition, multiplication) operations on elements of $\Z_3$ differ from those on the same elements in$~\Z_7$. Indeed the characteristic shows that there cannot be any injective ring morphism (in particular no inclusion relation) $\Z_3\to\Z_7$. Actually this is even true without "injective".

*$~$I mean unitary rings of course; no characteristic is defined (like this) for non-unitary rings.