# A commutative ring is a field iff the only ideals are $(0)$ and $(1)$

Let $R$ be a commutative ring with identity. Show that $R$ is a field if and only if the only ideals of $R$ are $R$ itself and the zero ideal $(0)$.

I can't figure out where to start other that I need to prove some biconditional statement. Any help?

• Think about what happens to ideals when they contain a unit.
– sxd
Commented Jan 21, 2012 at 23:21

From some of the comments above you seem a little confused. Since you said you are not familiar with proofs I will try to write this out in a way that you can understand.

You are trying to prove the equivalence of the following statements:

$P:$ A commutative ring $R$ with $1$ is a field.

$Q:$ The only ideals of $R$ are $(0)$ and $(1)$.

Let us look at statement $Q$ closely. Well it is saying that the only ideals of $R$ are the zero ideal (which has only one element, zero) and $(1)$. What is $(1)$? Well by definition of an ideal if you multiply anything in the ideal $(1)$ by anything in $R$, you should get something back in $(1)$ again. But then $1$ is the multiplicative identity of $R$ so multiplying everything in $R$ by $1$ just gives everything in $R$ back again. This means that $(1)$ must be the whole ring $R$.

Now suppose we want to prove $P \implies Q$. Let $I$ be an ideal of a ring $R$. Here "ring" means "commutative ring with a unit". Now here are some things you should know:

(1) $I$ is non-empty

(2) $I$ must at least contain the element $0$ (Why?)

(3) If $I$ has more than one element this means that at least one non-zero element $a$ of the ring must be in $I$. (Why?)

Therefore if $I$ contains only $0$, $I = (0)$. If $I$ does not only contain zero, then by (3) above it contains at least one non-zero element $a$ of $R$. Now recall that we are trying to prove $P \implies Q$. We already know $P$. Therefore this means by definition of a field that $a^{-1}$ exists in $R$.

But then by definition of an ideal $I$, $a^{-1} a = 1$ must be in $I$. Therefore $1 \in I$ so that $I$ must be the whole ring $R$. Hence $I = (1)$. This establishes $P \implies Q$.

Now for the converse:

To show $Q \implies P$ it suffices to show that non-zero every element $a \in R$ contains a multiplicative inverse. So let $a$ be a non-zero element of $R$. The trick now is to consider the principal ideal generated by $a$ (which we denote by $(a)$ ).

Now by assumption of $Q$, since the only ideals of $R$ are $(0)$ and $(1)$, this means that $(a)$ being an ideal of $R$ must be either $(0)$ or $(1)$. Now $(a)$ cannot be $(0)$ for $a \neq 0$. So $(a) = (1)$. But then this means that $1$ is a multiple of $a$, viz. there exists a non-zero $c$ such that

$$ac = 1.$$

However this is precisely saying that $a$ has a multiplicative inverse. Since $a$ was an arbitrary non-zero element of $R$, we are done. Q.E.D.

Does this help you? I can discuss more if you need help.

• Don't we have to add for the statement $Q$ that $1\neq 0$? Commented Sep 21, 2019 at 13:20

Let $a\in R$, $a\neq 0$. Then the smallest ideal of $R$ that contains $a$ is $Ra=\{ra\mid r\in R\}$.

(Verify that the set $\{ra\mid r\in R\}$ is an ideal of $R$, and that it contains $a$; it's not hard. Then think about why any ideal that contains $I$ must contain this set.).

If $R$ is a field, then what is $Ra$ for any $a\neq 0$? If $I$ is any ideal of $R$, and $a\in I$ with $a\neq 0$, what does that tell you?

Conversely, if the only ideals of $R$ are $(0)$ and $R$, what is $Ra$? What does that tell you about $a$?

• Since the asker had no idea where to start, I suspect they may not know that $Ra$ is an ideal. Drew sam, can you prove that?
– user23211
Commented Jan 21, 2012 at 23:05
• Ummm, for the first part, I am not sure what you mean. You just multiply the field by that value? Is it just the identity? Commented Jan 21, 2012 at 23:07
• @drew sam: You're just not putting things together, and are hoping I will put them together for you. It's not my homework, it's yours. Here is what you know (for one direction): If $R$ is a field, then every nonzero element has a multiplicative inverse. If $I$ is an ideal, and it's not the zero ideal, then you need to show that $I=R$. If $I$ not the zero ideal, then it contains a nonzero element. It also contains the product of any element in the ring by an element of the ideal. And if it contains the identity, then it contains everything. If it contains everything, then it is all of $R$. Commented Jan 21, 2012 at 23:34
• I am sorry, I do not expect you to do it, that's not what I wanted. I have a learning disability to it takes me really long to understand things. I am sorry I will try and work harder Commented Jan 21, 2012 at 23:36
• @drew sam: If that's the case, then let me suggest that this website and quick interaction through comments is not an appropriate forum for you; you should try to find one-on-one, personal help at your institution instead, or try to think through things before posting quick follow-up questions as you have done here. Commented Jan 21, 2012 at 23:37

It is given that the $0$-ring $\{0\}$ and the whole ring $R$ are ideals of any ring $R$. (Here I assume $R = \mathbb{Q}$, the rationals)

Suppose by contradiction that there is another ideal $N$ of $Q$ that is not $\{0\}$ or all of $Q$.

$N \neq \{0\}$ implies there is a nonzero rational number $a$ contained in $N$. Since $\frac{1}{a}$ is contained in $\mathbb{Q}$, the ideal $N$ must contain $(\frac{1}{a})\cdot a = 1$.

$N \neq Q$ implies that there is a rational number $b$ which is not contained in $N$. But $N$ contains $1$, so $N$ must also contain $b\cdot 1 = b$ by the definition of an ideal, contradicting the assumption that $N$ is not all of $\mathbb{Q}$.

Therefore there are no proper, non-trivial ideals of $\mathbb{Q}$.

• For some basic information about writing math at this site see e.g. here, here, here and here. Commented Nov 2, 2012 at 19:57
• Also this answers only one direction of the question (only for $\mathbb{Q}$, but that generalises easily). Commented Nov 2, 2012 at 20:00

By definition $$\rm R$$ is a field $$\!\iff\!$$ every $$\rm\color{#0a0}{nonzero}\ r\in R$$ is a $$\rm\color{#c00}{unit}$$ (invertible). Below we sketch how to mechanically translate this into the sought corresponding (extremal) properties of ideals.

\begin{align}\rm Hint\!:\ \ field\ \rm R\! &\iff \rm \, \ \color{#0a0}{0 \,=\, r}\, \ \ {\rm or}\ \ \, \color{#c00}{r\,\ unit}\!\\[.2em] &\iff \rm\ \, \color{#0a0}{0\ \ \ |\,\ \ r}\ \ \, or\ \ \ \color{#c00}{r\ \ \ |\ \ \, 1}\\[.2em] &\iff\rm \color{#0a0}{(0)\!\supset\! (r)} \ or \ \color{#c00}{(r)\!\supset\! (1)}\\[.2em] &\iff\rm \color{#0a0}{(0)\supset I}\,\ \ or\,\ \ \color{#c00}{I\supset (1)}\end{align}\qquad\qquad\qquad\qquad

Or it is special case $$\rm\: I = 0\:$$ of: $$\,$$ field $$\rm\: R/I \iff I\:$$ maximal  (proof).

• Do you mean that a is in r? Commented Jan 21, 2012 at 23:21
• @drew I see no "a" above. r denotes any element of R. Commented Jan 21, 2012 at 23:24
• How $0|r$ is possible in field? Commented Jan 5, 2015 at 14:16
• @Groups By definition of divisibility: $\,\ 0\mid r \iff \exists s\!:\ 0\cdot s = r\iff r = 0.\$ Therefore we have that $\ 0\mid r\,$ or $\,r\mid 1\iff r = 0\,$ or $\,r\,$ is invertible,  i.e. $\ r\ne 0\,\Rightarrow\,r\,$ invertible. $\$ Commented Jan 5, 2015 at 14:55
• Always interesting answers! Commented Aug 20, 2022 at 1:01

Let $$R$$ be a commutative ring with $$1$$. Then $$R$$ is a field if and only if $$R$$, $$(0)$$ are the only ideals of $$R$$.

First suppose $$R$$ is a field and let $$a \in R \setminus \{0\}$$ and suppose

$$(0) \subsetneq (a) \subsetneq R.$$

As $$a \in R \setminus \{0\}$$ and $$R$$ is a field, there exists $$a^{-1} \in R$$ such that, by the ideal property,

$$1=a^{-1}a \in (a)$$

Thus $$1 \in (a)$$ forcing $$(a)=R$$.

Conversely, suppose the only ideals are $$(0)$$ and $$R$$ itself and let $$a \in R \setminus \{0\}$$. As $$a$$ is nonzero, $$(a)=R$$ thus there exists an $$r \in R$$ such that

$$ar=1$$

thus $$R$$ is a field as $$a$$ has a multiplicative inverse.