11
$\begingroup$

A question in Rotman's Advanded Modern Algebra asks to prove the question in the title. I'm convinced of my proof, but a subquestion asked to prove that $1+1$ is zero, and for this I proceeded on a case by case basis; that is, assuming for the sake of contradiction that $1+1 \ne 0$, given that we now know that the field $F = \{0,1,1+1,a\}$, I proved that this structure cannot be a field.

However, I think this solution is quite ugly and I was wondering if is a nicer but still elementary solution (avoiding the fact that char($F$) = $2$).

PS I'm not sure if this is a suitable question. If it's not, feel free to remove it.

$\endgroup$
  • $\begingroup$ Provided $1+1=0$, what are your arguments? $\endgroup$ – Berci Sep 3 '18 at 17:40
8
$\begingroup$

Since there is an odd number of non-zero elements in the field, at least one of them must be its own additive inverse (because those who aren't come in pairs). So let's call that element $a$.

Since $a\ne 0$, it has a multiplicative inverse $a^{-1}$. But then we have $$1 + 1 = a a^{-1} + a a^{-1} = (a+a) a^{-1} = 0 a^{-1} = 0$$

Note that this in turn implies that all elements of the field are their own additive inverse, since $$x+x = 1x + 1x = (1+1)x = 0x = 0$$ Or in short, any finite field with an even number of elements must be of characteristic $2$.

$\endgroup$
  • $\begingroup$ Sorry to everyone who answered that I respond only now. All answers were wonderful but maybe this one captured the intent of my question the most. $\endgroup$ – Jan De Meyer Nov 29 '18 at 13:48
6
$\begingroup$

The underlying Abelian group $(F, +)$ has 4 elements, so the additive order of $1$ divides 4. If it were 4, the element $1+1$ would not be invertible.

$\endgroup$
  • $\begingroup$ I thought $1+1=0$ is not invertible. $\endgroup$ – Chris Custer Sep 3 '18 at 19:34
  • $\begingroup$ Yes of course. But besides $0$, in a field, there is no noninvertible element. $\endgroup$ – Berci Sep 3 '18 at 19:42
  • $\begingroup$ Of course. You are arguing by contradiction. My mistake. An inverse for $1+1$ would require a new element... $\endgroup$ – Chris Custer Sep 3 '18 at 20:01
  • 6
    $\begingroup$ We could adjust this argument like this: $(1+1)^2=(1+1)(1+1) = 1+1+1+1 = 0$ where the last equality must hold since order of the underlying additive group is $4$. $\endgroup$ – Ennar Sep 3 '18 at 20:12
4
$\begingroup$

We need $0$ and $1$. Call the other two elements $a$ and $b$. Then $ab=1, a^2=b$ and $b^2=a$ (otherwise $a^2=1 \implies a=b$ etc...) So the multiplication table is fixed.

But so is the addition table: $1+a=b$, necessarily. (Also, $b+1=a$.) For if $1+a=a,0$ or $1$ we get a contradiction. For instance, $1+a=0\implies a=-1\implies a=1$, since $1+1\not =1,a$ or $b$. Again, we would have a contradiction: say $1+1=a$, then $1+b=b,0$ or $1$, in which case $1=0$ or $b=a$ or $0$.

So our hand winds up being forced.

$\endgroup$
  • $\begingroup$ I'm not following $a^2 = b^2 = 1 \Rightarrow a = b = 1$. (After showing $ab=1$, I do see $a^2 = 1 \Rightarrow a = b$, which is enough of a contradiction.) The second paragraph gets rather confusing with nested suppositions and contradictions. $\endgroup$ – aschepler Sep 4 '18 at 0:38
  • $\begingroup$ Well we do have $ab=1$, because the other choices are not feasible. I regretted how complicated it is; but the OP wanted an elementary proof, and it's the most elementary way I could see to approach it... $\endgroup$ – Chris Custer Sep 4 '18 at 0:50
  • $\begingroup$ @aschepler I adjusted it. Thanks. $\endgroup$ – Chris Custer Sep 4 '18 at 1:07
3
$\begingroup$

Another path is to consider the multiplicative group $F^\times$. Since it has order $3$, it has to be cyclic. Thus, the multiplication table is fixed. This will also fix the addition table.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.