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?

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.
