# Given distinct nonzero elements $a$ and $b$ in $A$…prove $A$ has characteristic $x$.

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.

Proof:

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.

## 1 Answer

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. $$OE\Delta$$.

• 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. – Alex Johnson Apr 18 '19 at 3:52
• @AlexJohnson: thank you! I usually like working from first principles as much as possible, but--whatever works, right? And thanks for the "acceptance". Cheers! – Robert Lewis Apr 18 '19 at 4:19