5
$\begingroup$

As I started typing this up, I realized I made a mistake somewhere but I cannot identify it.

Prove that any two fields having exactly four elements are isomorphic. [Hint: First prove that $1+1=0$, and then show that the nonzero elements form a cyclic group of order $3$ under multiplication]

First I will prove a few things about fields with four elements before demonstrating the isomorphism. Let $F= \{0,1,a,b\}$ be a field with four elements. First of all, since $F$ is a field, every nonzero element is a unit, meaning that the group of units $U(F)$ is a multiplicative group of $3$ which makes it isomorphic to $\Bbb{Z}_3$, and hence $U(F)$ is cyclic. Now I will prove that $1+1=0$. Note that $(1+1)^2 = (1+1)(1+1) = 1 + 1 + 1 + 1 = 0$, where the last equality follows because $F$ is an additive group of order $4$. Since $F$ is a field, there can be no nonzero divisors, implying that $1 + 1 = 0$.

Now I will prove that $a+b =0$, which can also be used to prove $1+1$ since $(-a)^3 = 1$ implies $-1 = 1$ (is this right?). If $a+b$ weren't zero, then it would be a unit and hence have a multiplicative order of $3$. That is, $(a+b)^3 = 1$ or $1 + 3a^2b + 3a b^2 = 0$. Multiplying by $a$ and then $b$ yields the two equations $a + 3b + 3a^2 b^2 = 0$ and $b + 3a^2 b^2 + 3a = 0$, and adding the two gives $4a + 4b + 6a^2 b^2 = 0$ or $6a^2 b^2 = 0$, which is a contradiction since neither $a=0$ nor $b=0$. Hence $a + b = 0$ or $b = -a$. Therefore $F = \{0,1,a,-a\}$ (Perhaps the mistake was made here?)

Now we compute some products and sums of elements in $F$. It is clear that $1+a = -a$, for $1+a=0$ implies $a=1$, a contradiction, and $1+a=a$ implies $1=0$, another contradiction...Hold on! Either $a^2 = 1$, $a^2 = a$, or $a^2 = -a$, but all lead to a contradiction...I must have made a mistake somewhere.

EDIT:

Okay. We still have $(1+1)^2 = 0$ implies $1+1 = 0$, and because $U(F)$ is a cyclic group, it's easy to see that $b = a^2$. Now let $F_1 = \{0,1,a,a^2\}$ and $F_2 = \{0',1',b,b^2\}$ be two fields with four elements. Define the map $f : F_1 \to F_2$ by $f(0)=0'$, $f(1)=1'$, $f(a) = b$, and $f(a^2)=b^2$. Clearly this is bijective, and so all that remains is to show that it is a ring homomorphism.

Let us first prove additivity. First note that $a+a = a(1+1) = a \cdot 0 = 0$; and $a+1 = a^2$ otherwise we obtain a contradiction; and finally $a + a^2 = a(a+1) = a \cdot a^2 = 1$. Hence $f(a + a ) = f(0) = 0' = b + b = f(a) + f(a)$; and $f(a+1) = f(a^2) = b^2 = b+1 = f(a) + f(1)$; and finally $f(a+a^2) = f(1) = 1 = b + b^2 = f(a) + f(a^2)$.

Now we prove that $f$ is multiplicative. Note that $f(1 \cdot x) = f(x) = f(1) f(x)$; and that $f(a \cdot a) = f(a^2) = b^2 = b \cdot b = f(a) f(a)$; and finally $f(a \cdot a^2) = f(a^3 ) = f(1) = 1 = b^3 = b \cdot b^2 = f(a) f(a^2)$.

How does this sound?

$\endgroup$
  • $\begingroup$ Why must $a+b$ have a multiplicative order of $3$? $\endgroup$ – Kenny Lau Sep 18 '17 at 12:35
  • $\begingroup$ @KennyLau Because if we assume $a+b \neq 0$, then it must be a unit and the units have an multiplicative order of $3$, unless I am mistaken. $\endgroup$ – user193319 Sep 18 '17 at 12:39
  • 5
    $\begingroup$ "$6a^2b^2=0$ is a contradiction since neither $a=0$ nor $b=0$" actually, it isn't a contradiction, because you missed out on $6$. Here, $6=0$. $\endgroup$ – Kenny Lau Sep 18 '17 at 12:41
  • 1
    $\begingroup$ @KennyLau Oh! Yes! You are right. I think I correctly prove that $b=a^2$ from the fact that $U(F)$ is cyclic. Give me some time to finish the proof and I will update my original post. $\endgroup$ – user193319 Sep 18 '17 at 12:46
  • 1
    $\begingroup$ @KennyLau Okay. I updated my post. $\endgroup$ – user193319 Sep 18 '17 at 13:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.