# Prove that if the set of ideals is $\{\{0\}, R \}$, then $R$ is a field.

Let $R$ be an integral domain. Prove that if the set of ideals is $\{\{0\}, R \}$, then $R$ is a field.

$\{0 \}$ and $R$ are the trival ideals of $R$. Let $I$ ba an ideal of $R$ and $a\in I$ where $a$ is not a zero element. And let $b\in R$. Because $R$ is a integral domain which is a commutative ring with unity has no zero divisor, then there exist $a^{-1}\in R$, and $a^{-1}b\in R$. So $a(a^{-1}b)=b\in I$, hence $I=F$; therefore, $\{0\}$ and $R$ are ideals of $R$, thus $R$ is a field.

Does the argument above right? If not, can anyone give me a hit to write a better one ? Thanks

• About the word trivial: The word trivial is very common in mathematics, so its use can get a bit muddy. But, with ideals in rings, the meaning is pretty consistent: the trivial ideal is $\{0\}$. An ideal $I$ that isn't the entire ring $R$ is called proper, so $R$ itself is sometimes called the improper ideal (this usage is less common, though). Commented Jun 7, 2015 at 3:17
• I agree that "the trivial ideal" (singular) refers to the zero ideal, but it is also as common to refer to "the trivial ideals" (plural) to be the zero ideal and the whole ring. Commented Mar 1 at 21:11

No, that's not quite right.

The only difference between a field and a domain is that in a field, every $a \neq 0$ has an inverse $b$ such that $ab=1$. Hence, to show $R$ is a field, we have to take any nonzero element $a \in R$ and show it has an inverse.

Since the prompt says something about ideals, we have to think of an ideal having something to do with $a$ that will help us prove this, and there is a natural one to think of: $(a)$, the ideal generated by $a$. Since $a\neq 0$, $(a) \neq \{0\}$, hence by hypothesis, $(a)=R$. In particular, $1 \in (a)$, but what does this mean?

• I=F since 1 is unity and 1 is in I Commented Jun 7, 2015 at 2:58
• Right, $1$ is in the ideal $(a)$, but you want to show $1$ is a multiple of $a$, so now you'll need to remember what the definition of the ideal $(a)$ is... Commented Jun 7, 2015 at 3:01

Perhaps another useful (if less intuitive) perspective is to use the following characterization: A proper ideal $I$ in a commutative ring $R$ is maximal if and only if $R/I$ is a field. In our case, $\{ 0 \}$ is maximal. Indeed there are no other proper ideals!

• True, although I should probably point out: How does one prove that $\mathfrak{a}$ is maximal if and only if $R/\mathfrak{a}$ is a field? Commented Jun 7, 2015 at 3:41
• Yes I might be circular here. I have seen some authors use this as the definition of maximal, though. Commented Jun 7, 2015 at 3:43

Here is @Kooper 's approach, but with a better argument.
Let R be an integral domain(and of course, a commutative ring with unity), such that its only ideals are $$\{0, R\}$$. By Zorn's lemma, we know that every non-unit element is contained in one maximal ideal, but the only maximal ideal is $${0}$$, so we get that the only non-unit element in R is $$0$$, and by definition, a field is a commutative ring with unity, such that every non-zero element is a unit. Since the only element is $$0$$ we get what we wanted. $$\square$$