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

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.)

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Welcome to math.SE: Good question, generally I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Sameh Shenawy Feb 7 '14 at 8:26
Thanks for the welcome and tips. :) Henceforth I'll be sure to include more information regarding any questions I have and have edited my post appropriately. – AmagicalFishy Feb 7 '14 at 8:39

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 $\mathbb{Z}_7$.

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Oh. I'm all messed up, aren't I? A subring of $\mathbb Z_7$ wouldn't cease to be modular 7, for example; it would just be a set of elements in $\mathbb Z_7$ that could still be considered a ring on their own mod. 7. – AmagicalFishy Feb 7 '14 at 8:03
Right, every subring of $\mathbb{Z}_7$ still has the operations of addition and multiplication modulo $7$ (although it turns out in this case that $\mathbb{Z}_7$ has no proper subrings) – zcn Feb 7 '14 at 8:05

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.

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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.

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Non-unitary rings can still have a characteristic (smallest $n$ with $na = 0$ for all $a$). – Tobias Kildetoft Feb 7 '14 at 10:52
@TobiasKildetoft: You are right; actually I never work with non-unitary rings. In any case the definition I used here will not work for non-unitary rings; nor indeed is the result I state true, probably. – Marc van Leeuwen Feb 7 '14 at 10:54
Indeed (otherwise take $2\mathbb{Z}/4\mathbb{Z}\subseteq \mathbb{Z}/4\mathbb{Z}$). I am not sure if it might still be salvaged in case of integral domains without unit. – Tobias Kildetoft Feb 7 '14 at 10:57