We prove the equivalences $(1) \Leftrightarrow (2)$ and $(2) \Leftrightarrow (3)$.

$(1) \implies (2)$

Suppose that $R$ is a field. Let $I$ be an ideal of $R$.
If $I=(0)$, then there is nothing to prove.
So assume that $I\neq (0)$.

Then there is a nonzero element $x$ in $I$.
Since $R$ is a field, we have $x^{-1}\in R$.

Since $I$ is an ideal, we have
\[1=x^{-1}\cdot x\in I.\]
This yields that $I=R$.

$(2) \implies (1)$

Suppose now that the only ideals of $R$ are $(0)$ and $R$.
Let $x$ be a nonzero element of $R$. We show the existence of the inverse of $x$.
Consider the ideal $(x)=xR$ generated by $x$.

Since $x$ is nonzero, the ideal $(x)\neq 0$, and thus we have $(x)=R$ by assumption.
Thus, there exists $y\in R$ such that
\[xy=1.\]

So $y$ is the inverse element of $x$.
Hence $R$ is a field.

$(2)\implies (3)$

Suppose that the only ideals of $R$ are $(0)$ and $R$.
Let $S$ be any ring with $1$ and $f:R\to S$ be any ring homomorphism.
Consider the kernel $\ker(f)$. The kernel $\ker(f)$ is an ideal of $R$, and thus $\ker(f)$ is either $(0)$ or $R$ by assumption.

If $\ker(f)=R$, then the homomorphism $f$ sends $1\in R$ to $0\in S$, which is a contradiction since any ring homomorphism between rings with $1$ sends $1$ to $1$.
Thus, we must have $\ker(f)=0$, and this yields that the homomorphism $f$ is injective.

$(3) \implies (2)$

Suppose that statement 3 is true. That is, any ring homomorphism $f:R\to S$, where $S$ is any ring with $1$, is injective.
Let $I$ be a proper ideal of $R$: an ideal $I\neq R$.
Then the quotient $R/I$ is a ring with $1$ and the natural projection
\[f:R\to R/I\]
is a ring homomorphism.

By assumption, the ring homomorphism $f$ is injective, and hence we have
\[(0)=\ker(f)=I.\]
This proves that the only ideals of $R$ are $(0)$ and $R$.

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