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The question is as follows. Let $F$ be a finite field with unit $1$ not equal to zero. Let the function $f: F \to F$ be given by $f(x) = x^3$, where the $\operatorname{char}(F) = 3$. Prove it is a ring isomorphism.

I can prove it is a ring homomorphism by showing $f(xy) = (xy)^3 = x^3y^3 = f(x)f(y)$ by commutativity. Similarly, since the characteristic is three,

$$(x+y)^3 = x^3 + y^3$$

$$f(a+b) = f(a) + f(b)$$

$$f(1) = 1,\ \ \textrm{and}\ \ f(0) = 0$$

Now, how do I show $f$ is a bijection?

ThinkConnect
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3 Answers3

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That homomorphism is necessarily injective, but not necessarily surjective. For example, the field $F=\mathbb{F}_3(t)$ has characteristic $3$, and the homomorphism $\varphi:F\to F$ defined by $\varphi(a)=a^3$ for all $a\in F$ is not surjective because there is no $a\in F$ with $\varphi(a)=t$.

For a more in-depth analysis, take a look at my answer on the thread How to prove that the Frobenius homomorphism is surjective?

Community
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Zev Chonoles
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It's actually not an isomorphism in general, as Zev explained. But in the case that $F$ is finite it is:

It is a homomorphism, as you have shown, and any homomorphism of fields is injective. But also, any injective function from a finite set to itself is a bijection, hence $f$ is an isomorphism.

Noah Olander
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  • To show f is injective, I can say (given nonzero a,b) f(a) = f(b) <==> f(a)(f(b))^-1 = 1 <==> f(a)f(b^-1) = 1 <==> f(ab^-1) = 1 <==> ab^-1 = 1 <==> a = b. The reason f is a field is important is because b must have inverse. – ThinkConnect Jun 10 '15 at 01:49
  • Now, how do I show f is a surjection? That is, given a in F, how do i show there is b in F, such that a = b^3? – ThinkConnect Jun 10 '15 at 01:49
  • You haven't actually shown $f$ is injective in the comment. To do that it suffices to show ker $f = {0}$. So, suppose $f(a)=0$. This means that $a^3=0$, so that $a=0$ (fields are domains), and thus ker $f = {0}$. To show that $f$ is surjective, take another look at my answer - $f$ is an injective function from a finite set to itself. This means that it is also surjective. – Noah Olander Jun 10 '15 at 03:07
  • Thanks Noah, your response was very helpful! – ThinkConnect Jun 10 '15 at 04:53
  • If you don't mind me asking, why did my argument in the comment fail to show injectivity? Where was the error? – ThinkConnect Jun 10 '15 at 04:54
  • Sorry I should have mentioned, you didn't justify the step $f(ab^{-1})=1 \iff ab^{-1}=1$. – Noah Olander Jun 10 '15 at 05:02
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To get started, $\ker(f)$ is an ideal in the ring $F$.

ncmathsadist
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  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – hjhjhj57 Jun 10 '15 at 00:14
  • What are the ideals present in a field? Yes, it settles the question of injectivity. I am offering a substantial hint, and I trying to avoid giving away the answer and spoiling things for the OP. – ncmathsadist Jun 10 '15 at 01:05
  • I believe this is better suited as a comment. – hjhjhj57 Jun 10 '15 at 01:18
  • Since ker(f) is an ideal, f/ker(f) is isomorphic to f(F). The kernel is trivial by injectivity, so f(F) is isomorphic to f. Thus f(F) = F, proving surjectivity? – ThinkConnect Jun 10 '15 at 02:05