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?
 A: 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.
A: 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 $a$?
A: 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 $\mathbb{Q}$.
A: 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).
