If $F$ is a finite domain, it's a field.

Question

Show that a finite domain $F$ is a field.

Let $I$ a proper ideal of $F$ and let $a\in I$. In particular, $a$ is not invertible, otherwise $I$ wouldn't be proper.

I would like to show that $I=(a)=(0)$, but without success.

You're starting the wrong way: you just need to prove that every nonzero element is invertible. — egreg, Dec 02 '15 at 22:01

score 5 · Answer 1 · answered Dec 02 '15 at 20:44

5

Hint: Use the fact that $f_a(x)=ax$ is bijective

answered Dec 02 '15 at 20:44

Tsemo Aristide

87,475

The fact that is bijective is that $\ker f_a={0}$ and that $F$ finite ? With this hint, I say that $1\in I$ and thus $I=A$. Is it finish ? – MSE Dec 02 '15 at 20:49
but $f$ bijective implies $f$ surjective, thus $1$ is in the image of $f$,... go from there – Tsemo Aristide Dec 02 '15 at 20:51
I know, but I want to show that $f_a$ is really bijective. Then $1\in I$ and thus $I=1$. Is it finish ? – MSE Dec 02 '15 at 20:52
What does it mean for $F$ to be a domain? @MSE – Thomas Andrews Dec 02 '15 at 20:53
that $ab=0\implies a=0$ or $b=0$. @ThomasAndrews – MSE Dec 02 '15 at 20:55
So if $f_a(b)=f_a(c)$ then ... – Thomas Andrews Dec 02 '15 at 20:56
then $b=c$, then $f$ is injective. But I can also says that $ax=0\implies x=0$ is $a\neq 0$. And thus $\ker f={0}$ what also prove injectivity. But where do you want me to go ? – MSE Dec 02 '15 at 20:57
One doesn't even need that $R$ has a unity at the outset. – egreg Dec 02 '15 at 21:15

score 1 · Accepted Answer · answered Dec 02 '15 at 20:57

1

Let continue your idea. Suppose $I$ is an ideal and it contains $a \neq 0$. Then $aF \subset I$. However as $F$ is a domain $af \neq af'$ for $f \neq f'$. Thus $|aF| = |F|$ and $|F| \le |I| $. Since $|I| \le |F|$ is trivial, we have $|I|=|F|$. Now since $F$ is finite this means $I=F$.

answered Dec 02 '15 at 20:57

quid

42,135

and thus the only ideal is $F$ and thus finish ? – MSE Dec 02 '15 at 21:00
Yes, almost. The only non-zero ideal is $F$. We assume $I$ nonzero and shiwed it is $F$. So the only two ideals are ${0}$ and $F$ and thus this is a field. – quid Dec 02 '15 at 21:01
The idea is exactly the same as in Tsemo Aristide's answer. – egreg Dec 02 '15 at 21:14
@egreg I do not agree that this is exactly the same idea. But even if it were, what's the point of your comment? – quid Dec 02 '15 at 21:23
@quid You use “injective implies surjective” and the map is just the same. It's simply the same, but with some frills added. – egreg Dec 02 '15 at 21:27
@egreg the argument framed in this way uses (and needs) the notion of ideal and the fact the a domain with only two ideal is a field. Using this prior knowledge it is not necessary to observe "Since $1$ is in the image there is $b$ such that $ab=1$ and so $a$ is invertible."¨and OP knowns and wanted to use that. So I showed them that this is possible. By contrast the other answer does not involve these notions, which in isolation could even count as advantage, but needs that additional step. . – quid Dec 02 '15 at 21:47
Those are the frills. I think the OP started the wrong way, that's all. – egreg Dec 02 '15 at 21:59
@egreg I do not agree that this is "the wrong way." If one has that prior knowledge, the argument seems more direct. I actually think it is an interesting idea to set it up this way, which is why I wanted to show it worked. As opposed to the OP being forced into the approach "everybody" takes. – quid Dec 02 '15 at 22:06

score 0 · Answer 3 · answered Dec 02 '15 at 22:17

Since $F$ is finite, it has a minimal non zero ideal $I$. If $a\in I$, $a\ne0$, then $\{0\}\ne aF\subseteq I$; by minimality, $I=aF$. Now consider $a^2$; since $F$ is a domain, then $a^2\ne0$, so $\{0\}\ne a^2F\subseteq aF$ and, by minimality, $a^2F=aF$. In particular $a=a^2x$, for some $x\in F$. Therefore $1=ax$ and $a$ is invertible. So $aF=F$ and therefore $F$ has no nonzero proper ideal.

Note that this actually proves that an artinian domain is a field, because the only requirement is that the ring has a minimal non zero ideal.

Of course a more direct proof is available, using finiteness: for $a\ne0$, the map $x\mapsto ax$ is injective, hence surjective. Thus $1=ax$ for some $x\in F$.

If $F$ is a finite domain, it's a field.

3 Answers3