Elementary proof that there is no field with 6 elements

Question

I am a T.A at an introductory linear algebra course this semester, and before getting into vector spaces, we give them some basic examples and properties of fields. Since this is a first semester course, everything is very elementary and we do not have too many tools.

I'd be happy to give them some elementary proof that there is no field of size $6$ (or some other small, concrete number, like 10... I just do not really care about the general case). They already know that $\mathbb Z _6$ is not a field, but I want to show that there is no other possible field of this size.

By elementary, I mean something which can be taught to math students on the second week to their first semester (so no arguments using groups and such). I can start talking about characteristics of fields and then show that the size has to be some power of the characteristic, but I'd prefer not to get into all of this. I am simply wondering if anyone here knows of an elementary proof for some concrete size (like $6$).

Thank in advance!

Answer 1

Let $F$ be a field with six elements. Since $|F|=6$, the sequence $0$, $1$, $1+1$, $1+1+1$, ..., $6\times 1$ must have a repetition. If $n\times 1=m\times 1$ for $n>m$, then by subtraction we obtain $(n-m)\times 1=0$. Thus the first repetition must actually be $n\times 1=0$ in the list above for some $2\leq n\leq 6$.

If $n\times 1=0$, then $n\times a=0$ for all $a\in F$, and similarly if $m\times a=0$ for some $a\neq 0$ then $m\times 1=0$ by dividing by $a$. This shows that for any $a$, $a$, $a+a$, ..., $(n-1)\times a$ are all distinct.

(This idea is due to Eric Wofsey in the comments below). If $n>2$ then $a+a\neq 0$ for any $a\neq 0$, as we have seen above. This shows that $a\neq -a$ for all non-zero $a$. Thus, pairing $a$ with $-a$ for all $a\in F\setminus\{0\}$, we see that $|F\setminus\{0\}|$ is even. This contradiction means that $n=2$.

Thus $n=2$. Let $a\in F\setminus\{0,1\}$, and notice that $0,1,a,a+1$ are all distinct. (If $a+1=0$ then $a=1$, if $a+1=a$ then $1=0$, if $a+1=1$ then $a=0$.) Notice that adding $1$ moves the elements $\{0,1,a,a+1\}$ around. If $b\in F\setminus\{0,1,a,a+1\}$ then $F$ must be $$F=\{0,1,a,a+1,b,b+1\}.$$

Where is $a+b$? It cannot equal $0$ ($a=b$), it cannot equal $1$ ($a=b+1$), or $a$ ($b=0$) or $b$ ($a=0$), or $a+1$ ($b=1$) or $b+1$ ($a=1$). (All of these are by cancellation.) So $a+b\not\in F$, a contradiction.

Answer 2

Let F = {0, 1, -1, a, b, c}. All a, b and c must have unique multiplicative inverses, that is, they must be paired with eachother. There's only 3 elements therefore at least one does not have an inverse. You can fairly easily generalise this to even numbers perhaps? Sorry for lack of formalism, this is intended as a comment but my score isn't high enough

Edit: Thank you in comments for pointing out my grave error and thank you for the insights I think I actually have a proof now though I could be wrong again. As said below in comments, the above proves this for the case $1 \neq -1$ so lets consider the case $1=-1$ and take $F = \{0, 1, a_1, a_2, a_3, a_4\}$ to be a field for such a case.

Because of the existence of the multiplicative inverse, we know that the inverse of $a_1$ must be one of $a_2, a_3, a_4$. Let's pick one arbitrarily, say $a_3$, and consequently $a_2$ must be the inverse of $a_4$. Thus we have:

$a_1^{-1} = a_3$,

$a_2^{-1} = a_4$.

Now let's look at $a_1 a_4$. It is certain that $a_1 a_4 \neq a_1$ or $a_1 a_4 \neq a_4$ since that implies $a_4 = 1$ or $a_1 = 1$ which violates uniqueness of identity. Similarly, $a_1 a_4 \neq 0$ or $a_1 a_4 \neq 1$ since that would contradict uniqueness of zero element and inverse respectively. This means that $a_1 a_4 = a_2 \space or \space a_3$. It doesn't matter which one we pick, the proof is pretty much the same for both from here on out, so lets pick $a_1 a_4 = a_3$. From this equality and using a tad of manipulation:

$a_2 = a_1^2$

Because of the existence of the additive inverse, the inverse of $a_1$ must be one of $-a_2, -a_3, -a_4$. These 3 cases are considered briefly as follows (using $1 = -1$):

1.) $a_1 = -a_2 = a_2 = a_1^2$ which implies $a_1 = 1$

2.) $a_1 = -a_3 = a_3 = a_1^{-1}$ which implies $a_2 = a_1^2 = 1$

3.) $a_1 = -a_4 = a_2 = a_1^2$ which implies $a_1 = 1$

This contradicts uniqueness of multiplicative identity. Furthermore, since choices above were arbitrary and/or near identical, we are forced to conclude that F is not in fact a field.

I don't know if this is the right kinda proof for you as its a tad long but should be understandable enough right?

Answer 3

A modification of David Craven's fine answer.

Assume that such a field $F$ exists. Its multiplicative group has order five. Therefore (Lagrange) $F$ cannot have an element of multiplicative order two. In other words, we must have $-1=1$ implying that every element of $F$ is its own additive inverse.

Let $a\in F$ be distinct from $0$ and $1$. We know that $a+a=0$ and that $a+1$ is distinct from all of $0,1,a$. It follows that $\{0,1,a,a+1\}$ is an additive group. The remaining detail is that $a+(a+1)=(a+a)+1=0+1=1$. But, again by Lagrange, the group $(F,+)$ cannot have a subgroup of order four.