## Suzie-Q

Problem Statement: Let $I$ be an integral domain. Show that $I$ is a field if and only if $I$ has no nontrivial ideals.

Proof: First assume that $I$ is a field and let $J$ be an ideal of $I$. If $1\in J$ then it follows that $J=I$ since $x1=x\in J$ for every $x\in I$. Suppose then that $1\not\in J$ and let $j\in J$, then there exists a $j^{-1}\in I$ since by assumption $I$ is a field. But since $J$ is an ideal it follows that $jj^{-1}\in J$, but since $1\not\in J$ this is only possible if $j=0$ and so it follows that $J=\{0\}$. Thus the only ideals of $I$ are the trivial ideals.

Now assume that the only ideals of $I$ are $J=\{0\}$ and $I$. Fix a non-zero $x\in I$ and define $\phi_{x}:I\rightarrow I$ by $\phi_{x}(y)=xy$. $\phi_{x}$ is a homomorphism with $\ker\phi_{x}=J=\{0\}$ since $I$ is an integral domain and so we have no zero divisors. So we have that $\phi_{x}$ is one to one and thus $\phi_{x}$ is onto since it maps $I$ to itself. Now, since $1\in I$ it follows that there exists $y\in I$ such that $\phi_{x}(y)=1\in I$. By definition of $\phi_{x}(y)$ this implies that $xy=1$ and so $y=x^{-1}$. Thus every non-zero element in $I$ is a unit and so $I$ is a field.

$\Box$

Reflection: This technique of using a homomorphism to find an inverse is a really helpful tool. This only works since we can force $\ker\phi_{x}=\{0\}$ which is due to the fact that $I$ is an integral domain.