# An integral domain of characteristic $k>0$ is a vector space over $\mathbb{Z}_k$?

Problem statement:

Suppose that $R$ is an integral domain of characteristic $k > 0$. Show that $R$ can be considered as a vector space over $\mathbb{Z}_k$.

This seems like a trivial problem, and it probably is, I think I just need to be sure of what is meant by ''a vector space over $\mathbb{Z}_k$''. That just means when considering scalar multiplication compatibility with field multiplication (and the remaining 3 field axioms), the scalars are from $\mathbb{Z}_k$, correct?

As a vector space over $\mathbb{Z}_k$, $R$ has to be an abelian group under addition, which is trivially satisfied.

Besides that, we have the field axioms: $a(b\mathbf{v}) = (ab)\mathbf{v}$, $1\mathbf{v} = \mathbf{v}$, $a(\mathbf{u}+\mathbf{v})=a\mathbf{u} + a\mathbf{v}$, $(a+b)\mathbf{v}=a\mathbf{v}+b\mathbf{v}$. Are these trivially satisfied by the fact that $R$ is an integral domain?

Hope you can help me clarify.

• Vector space? Unless $k$ is prime, the best you can do is view it as a module. Apr 24 '14 at 18:07
• OK, a more careful reading reveals $R$ an integral domain, $\text{char} R = k$, so $k$ must be a prime, and$\Bbb Z_k$ a field, no? Apr 24 '14 at 18:10
• A ring $R$ is a module over $S$ iff there is a homomorphism $S \xrightarrow{\varphi} R$. Notice that the axioms are then automatically satisfied. Apr 24 '14 at 18:11
• @1950RobertLewis Yes, that is how I see it. Apr 24 '14 at 19:40
• We're getting to the bottom of this! Apr 24 '14 at 19:50

The first thing you need to realize is that $R$ being an integral domain, $k = \text{char}R$ is prime. For if not, then $k = mn$, with $2 \le m, n < k$. Then, letting $1_R \in R$ be the unit element, which exists since $R$ is an integral domain, we have $(m1_R)(n1_R) = (mn1_R) = k1_R = 0 \in R$; but $m1_R, n1_R \ne 0$, since $m, n < k$, contradicting the hypothesis that $R$ have no zero divisors. So $k$ is prime, and as such, $\Bbb Z_k$ is a field.
I think that, once it is understood that $k$ is prime and $\Bbb Z_k$ is a field, it is a relatively straightforward matter to see that $R$ is a vector space over $\Bbb Z_k$. Indeed, "the scalars are from $\Bbb Z_k$", to quote our OP Bo Schmidt. The easiest way for me to see how this all works is to first observe that $R$ in fact contains a subfield isomorhic to $\Bbb Z_k$, namely $P = \{0, 1_R, 2 \cdot 1_R, . . . , (k - 1) \cdot 1_R \}$, the "prime subfield" of $R$ as it were; such being the case, we see that $R$ is a vector space over $P$ by virtue of the natural operations of addition and multiplication inherent in its structure as a ring; the ring multiplication serves as scalar multiplication for elements in the prime subfield $P$. So $R$ can be considered a vector space over $\Bbb Z_k$ solely by virtue of its intrinsic properties; we simply identify $\Bbb Z_k$ with $P$ by means of the (unique) isomorphism $\phi: \Bbb Z_k \to P$. Then the vector space structure of $R$ over $\Bbb Z_k$ is given by $\alpha v \equiv \phi(\alpha)v$ for $\alpha \in \Bbb Z_k$ and $v \in R$. Of course, it may seem like we are picking logical nits to go so far as to set up a formal isomorphism such as $\phi$ when the structural equivalence of $\Bbb Z_k$ and $P$ is so nearly self-evident; on the other hand, conceptually $\Bbb Z_k = \Bbb Z/(k)$ and $P$ as a subfield of $R$ arise in different contexts, so formally $\phi$ is a useful clarification.
Are there any other actions of $\Bbb Z_k$ on $R$ besides that given intrinsically by $P$ which yield other vector-space structures on the pair $(\Bbb Z_k, R)$? The answser is "no" if we admit the axiom $1_{\Bbb Z_k}v = v$ for $v \in R$, for then $(a1_{\Bbb Z_k}) v = a(1_{\Bbb Z_k}v) = av = a(1_Rv) = (a1_R)v$ for $a \in \Bbb Z$, suitable reductions mod $k$ being understood, and of course we take $a1_{\Bbb Z_k}$ to mean $a$ copies of $1_{\Bbb Z_k}$ added together, and similarly for $a1_R$ etc. This shows that vector-space actions of $\Bbb Z_k$ are always determined by the ring multiplication by elements of $P$ under the correspondence $1_{\Bbb Z_k} \leftrightarrow 1_R$.