TITLE: A commutative ring is a field iff the only ideals are $(0)$ and $(1)$
QUESTION [46 upvotes]: 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?

REPLY [3 votes]: 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 $\mathbb{Q}$.