Prove that R is a field.

Question

I have a question: Let R be an integral domain, if the only ideals are the trivial ideals, prove that R is a field.

So my proof for this was;

Let $a\in R$ be a non-zero element. Since we have that the only ideals are $(0)$ and $(1)$ we have that $a$ cannot be in $(0)$ since it's non-zero. Since $a$ isn't in $(0)$, it has to be in $(1)$ which implies that $a$ is a multiple of $1$ for some $b$ i.e. $ab = 1$

But this implies that $a$ is a unit so therefore $R$ is a field.

My professor told me there's a better way of doing this. Using the following theorem: Let $R$ be a ring that is commutative with identity and let $M$ be an ideal in $R$. Then $M$ is maximal iff $R/M$ is a field.

Personally, I have no idea how to use this theorem for this proof. It's obvious that $(0)$ is maximal so that means $R/(0)$ is a field. How do I go about showing R is a field using this?

I'm not sure the professor's way is "better". What you're giving is essentially the proof of the theorem he wants you to appeal to. (Though you should write "$1$ is a multiple of $a$" rather than "$a$ is a multiple of $1$"). — hmakholm left over Monica, Apr 28 '14 at 16:24
It's a matter of taste. I don't think that using $M$ maximal $\Rightarrow R/M$ is a field is a better way to do it. — Andrea Mori, Apr 28 '14 at 16:25
Ahh, I feel very silly for asking something that was very obvious. I did over think this, thanks everyone! — user146409, Apr 28 '14 at 16:31
related but I guess not a duplicate since this is essentially a proof verification/strategy question. — rschwieb, Apr 29 '14 at 09:58

score 3 · Answer 1 · answered Apr 28 '14 at 16:26

This isn't valid as written: Since $a \in (1)$, it means that $a$ is a multiple of $1$, so that there exists $b \in R$ such that

$$a = 1 \cdot b$$

But we already knew that, since we can always choose $b = a$, so this doesn't tell us anything new. In general, $(1) = R$ always.

For a different way to proceed, consider the ideal $(a)$. You've already noted that $(a) \ne (0)$, so $(a) = (1)$. Now use this statement to find an inverse of $a$.

Alternatively, to use the theorem that your professor mentioned, note that $(0)$ is maximal, so that $R/(0)$ is a field; but $R/(0)$ is isomorphic to $R$.

Ah, I rushed my proof quickly so I did make a mistake while writing this out. But my proof is pretty much identical to yours. Thank you! — user146409, Apr 28 '14 at 16:32

Prove that R is a field.

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