Gabrijel Boduljak

Every finite integral domain is a field.

📅December 09, 2020 | ☕️5 minutes read

Claim: Every finite integral domain is a field.

Proof: Firstly, observe that a trivial ring cannot be an integral domain, since it does not have a nonzero element. Let FF be our finite integral domain. By the observation above, select any nonzero element. Say λF\lambda \in F. Let χ:FF\chi : F \to F be a map defined by χ(x)=λx\chi (x) = \lambda \cdot x, where \cdot is a multiplication in FF.

Now, it suffices to show that χ\chi is a surjective linear map. Why? If χ\chi is surjective, then there exists xFx \in F such that χ(x)=λx=1\chi(x) = \lambda \cdot x = 1, so λ\lambda has a right inverse, namely xx. But since FF is an integral domain, \cdot is commutative, so xx must also be a left inverse. Hence, λ\lambda is an unit. But since λ\lambda was arbitrary, every element of FF must be an unit. Hence FF must be a field.

We will prove that χ\chi is injective. Select x,yFx, y \in F. Suppose χ(x)=χ(y)\chi(x) = \chi(y). Now we have:

χ(x)=χ(y)λx=λyλ(xy)=0 \begin{aligned} \chi(x) &= \chi(y) & \\ \lambda \cdot x &= \lambda \cdot y \\ \lambda \cdot (x - y) &= 0 \\ \end{aligned}

Now, since λ0\lambda \neq 0 (by construction) and FF is an integral domain, we deduce (xy)=0(x - y) = 0. But then, x=yx = y, so χ\chi is indeed injective.

Now, since FF is finite and χ\chi is injective, χ\chi must be surjective. But then, by discussion in the second paragraph, FF must be a field. \blacksquare

I have decided that this proof deserves a post because it illustrates several methods in the abstract algebra. It is a good example of the power of generalisation. In this case, as it is in many others in this field, it was easier to show a more general statement. Instead of focusing how to find an inverse of an arbitrary element, we have shown χ\chi was surjective, which is a far more general statement which helped us to push the proof forward.

It goes without saying that this, albeit a basic result, deserves a respect because it such a general statement with a remarkably elegant proof.

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