I'm working on a problem from Pinter's Abstract Algebra and am wondering if someone can tell me if I'm on the right track.

Let $A$ be a finite integral domain. Prove that if there are distinct nonzero elements $a$ and $b$ in $A$ such that $125 \cdot a = 125 \cdot b$, then $A$ has characteristic 5.


As this is a finite integral domain, it has nonzero characteristic. By Theorem 20.2, that characteristic must be a prime number. By Theorem 20.1, all nonzero elements in an integral domain have the same additive order $n$. So, unity (1) has additive order $n$, as do $a,b \in A$.

Because $a,b$ have order $n$, $n \cdot a = 0$ and $n \cdot b = 0$. This implies that $n \cdot a = n \cdot b$. And since $125 \cdot a = 125 \cdot b$, and according to Theorem 10.5, 125 must be a multiple of $n$. The only prime factor of 125 is 5, so $char(A) = 5$.

Theorem 10.5: Suppose an element $a$ in a group has order $n$. Then $a^t = e$ iff $t$ is a multiple of $n$.
Theorem 20.1: All nonzero elements in an integral domain have the same additive order.
Theorem 20.2: In an integral domain with nonzero characteristic, the characteristic is a prime number.


The proof presented by our OP Alex Johnson seems fine to me, although I think there is a direct route to the desired result from first principles which avoids the need to develop the machinery of the stated theorems 10.5, 20.1, and 20.2, to wit:

It is a well-known and elementary result that a finite integral domain $A$ is in fact a field. Indeed, this may easily be seen as follows: for any $0 \ne a \in A$, we consider the function

$\theta_a:A \to A,\; \theta_a(r) = ar, \; \forall r \in A; \tag 1$

$\theta_a$ is easily seen to be injective, since for $r_1, r_2 \in A$ we have

$\theta_a(r_1) = \theta_a(r_2) \Longrightarrow ar_1 = ar_2 \Longrightarrow a(r_1 - r_2) = 0 \Longrightarrow r_1 - r_2 = 0 \Longrightarrow r_1 = r_2, \tag 2$

where we have used $a \ne 0$ in establishing this sequence of implications; now since $\theta_a$ is injective and $A$ is finite, we have that $\theta_a$ is also surjective; therefore

$\exists b \in A, \; \theta_a(b) = 1_A \Longrightarrow ab = ba = 1_A, \tag 3$

that is, $b$ is a multiplicative inverse of $a$; we have thus shown that every $a \in A$ has such an inverse, and thus that $A$ is indeed a field.

Now with

$a \ne b, \tag 4$

we further have

$a -b \ne 0, \tag 5$

so then

$125a = 125b \Longrightarrow 125(a - b) = 125a - 125b = 0$ $\Longrightarrow 5^3 = 125 = 0 \; \text{in} \; A \Longrightarrow 5 = 0 \; \text{in} \; A, \tag 6$

and now since $5$ is prime we have

$\text{char}A = 5, \tag 7$

as was to be shown. $OE\Delta$.

  • 1
    $\begingroup$ Thank you, Robert. Knowing that this was a finite integral domain should've sent me toward fields from the beginning, but as the theorems listed at the bottom of my post were easily pulled from the text, I went that direction instead. Again, thanks. $\endgroup$ – Alex Johnson Apr 18 '19 at 3:52
  • $\begingroup$ @AlexJohnson: thank you! I usually like working from first principles as much as possible, but--whatever works, right? And thanks for the "acceptance". Cheers! $\endgroup$ – Robert Lewis Apr 18 '19 at 4:19

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