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.

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

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 $2$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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.

  • $\begingroup$ 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

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.


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