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UPSC 2020 Maths Optional Paper 2 Q3a — Step-by-Step Solution

15 marks · Section A

Fields and finite fields · Algebra · asked 6× in 13 yrs · Read the full method →

Question

Let $R$ be a finite field of characteristic $p(>0)$. Show that the mapping $f:R\to R$ defined by $f(a)=a^p$, $\forall a\in R$ is an isomorphism.

Technique

Frobenius endomorphism; Freshman’s Dream $(a+b)^p=a^p+b^p$ from $p\mid\binom{p}{k}$; injectivity via trivial kernel of a field homomorphism; surjectivity via finiteness (pigeonhole).

Solution

Let $R$ be a finite field with $\operatorname{char} R=p$ (a prime, since the characteristic of a field is $0$ or prime, and here $p>0$). We show $f(a)=a^p$ is a field isomorphism $R\to R$ (the Frobenius automorphism).

Step 1 — $f$ preserves multiplication and identity

For all $a,b\in R$,

(commutativity of the field), and $f(1)=1^p=1$.

Step 2 — $f$ preserves addition (the “Freshman’s Dream”)

By the binomial theorem in a commutative ring,

For $1\le k\le p-1$, the binomial coefficient $\binom{p}{k}=\dfrac{p!}{k!\,(p-k)!}$ is divisible by $p$: indeed $p\mid p!$ and $p\nmid k!$, $p\nmid(p-k)!$ since $p$ is prime and $k,p-k<p$. So $\binom{p}{k}\equiv 0$ in a ring of characteristic $p$, killing all middle terms. Hence

Together with Step 1, $f$ is a ring homomorphism of $R$ into itself fixing $1$.

Step 3 — $f$ is injective

$f$ is a nonzero homomorphism of a field $R$. Its kernel $\ker f$ is an ideal of the field $R$; the only ideals of a field are $\{0\}$ and $R$. Since $f(1)=1\ne 0$, $\ker f\ne R$, so $\ker f=\{0\}$. Thus $f$ is injective.

(Equivalently: $a^p=0\Rightarrow a=0$ because a field has no nonzero nilpotents/zero-divisors.)

Step 4 — $f$ is surjective (uses finiteness)

$R$ is finite, and $f:R\to R$ is an injective map from a finite set to itself. An injective self-map of a finite set is automatically surjective (pigeonhole). Hence $f$ is bijective.

Step 5 — Conclusion

$f$ is a bijective ring homomorphism $R\to R$ preserving $+,\,\cdot,\,1$, i.e. a field isomorphism (in fact an automorphism of $R$).

Answer