Proving that a field isomorphism exists

Question

I have a question that I need some helping solving:

Show if $G$ is a field with exactly three elements, then there is a field isomorphism $f: \Bbb Z_3 \to G$. Note that $\Bbb Z_3$ is defined $\Bbb Z_3=(\{0,1,2\}, +, .)$ where $x+y=(x+y) \mod 3$ and $xy=xy \mod 3$.

Answer 1

If $G$ is a field with three elements, then the elements must be $\{0, 1, \alpha\}$, where $0$ is the additive identity, $1$ is the multiplicative identity, and $\alpha$ is a third element.

• Show that $\alpha = 1 + 1$.

• Define $f : G \to \mathbb{Z}_3$ by $f(0) = 0$, $f(1) = 1$, $f(\alpha) = 2$. Show $f$ is an isomorphism.

Answer 2

Note that $G$ consists of $-1,0,1$. One idea for verifying this is to assume for the sake of contradiction $1=-1$ and call the remaining element $x$. Then $x=1/x$ so $x+1=x+x^2=x(1+x)$ and either $1+x=0$ or $x=1$, we have a contradiction.

Define $\varphi:\mathbb Z_3\to G$ by $\varphi(i)=i\cdot 1_G$. In your own work, you will have to carefully define $i\cdot g$, where $g\in G$. Check that $\varphi$ is a homomorphism and that it is injective. Since any injective function between finite sets of the same size is a bijection we are done.

Answer 3

A slightly more abstract approach: G and Z₃ are both fields of order 3. A field is a group both under addition and multiplication. Now both G and Z₃ have additive subgroups of order 3, and since there is only one group of order 3 (to within isomorphism), it must be that the additive group in G is isomorphic to the additive group in Z₃. Same with the multiplicative group.So the 2 fields are isomorphic.

What we did here was show there must be an isomorphism without actually constructing it. One nice consequence of this is that the same proof works for fields of any finite prime order.

Answer 4

Another answer proceeds by showing that the additive groups are isomorphic and the multiplicative groups are isomorphic. For the convenience of the readers of this question, here is a counterexample to show that this is not in general sufficient for the fields to be isomorphic. (Thus in my opinion this other answer is totally bogus.)

Let $F_1 = \mathbb{Q}(\sqrt{-7})$ and $F_2 = \mathbb{Q}(\sqrt{-11})$. These are nonisomorphic fields. However, the additive groups of each one are two dimensional $\mathbb{Q}$-vector spaces, hence are isomorphic. Each multiplicative group is isomorphic to the direct product of a cyclic group of order $2$ with a free commutative group of countable rank. This is especially easy to see here since the ring of integers of each of $F_1$ and $F_2$ is a PID, but in fact for any number field $K$, the group $K^{\times}$ is isomorphic to the product of the finite cyclic group of roots of unity in $K$ with a free commutative group of countable rank.

I will also add that in my opinion it is always better to construct an explicit isomorphism than to argue that two structures are isomorphic, if possible. In the case of a finite field $F$ of prime order $p$, it is not only possible but easy: as for any ring, there is a unique ring homomorphism $\varphi: \mathbb{Z} \rightarrow F$. Since $F$ is finite of order $p$, $\varphi$ is surjective and its kernel is $p\mathbb{Z}$, so $\varphi$ induces an isomorphism $\varphi: \mathbb{Z}/p\mathbb{Z} \stackrel{\sim}{\rightarrow} F$. If now $F_1$ and $F_2$ are two finite fields of order $p$, then $\varphi_2 \circ \varphi_1^{-1}: F_1 \rightarrow F_2$ is an explicit isomorphism. (Of course this is really just a slightly fancier/cleaner version of the straightforward argument involving $f(1+\ldots+1) = f(1) + \ldots + f(1) = 1 + \ldots + 1$.) It is easy to see that this isomorphism is unique.