Every Integral Domain Artinian Ring is a Field
Problem 437
Let $R$ be a ring with $1$. Suppose that $R$ is an integral domain and an Artinian ring.
Prove that $R$ is a field.
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Definition (Artinian ring).
A ring $R$ is called Artinian if it satisfies the defending chain condition on ideals.
That is, whenever we have ideals $I_n$ of $R$ satisfying
\[I_1\supset I_2 \supset \cdots \supset I_n \supset \cdots,\]
there is an integer $N$ such that
\[I_N=I_{N+1}=I_{N+2}=\cdots.\]
Proof.
Let $x\in R$ be a nonzero element. To prove $R$ is a field, we show that the inverse of $x$ exists in $R$.
Consider the ideal $(x)=xR$ generated by the element $x$. Then we have a descending chain of ideals of $R$:
\[(x) \supset (x^2) \supset \cdots \supset (x^i) \supset (x^{i+1})\supset \cdots.\]
In fact, if $r\in (x^{i+1})$, then we write it as $r=x^{i+1}s$ for some $s\in R$.
Then we have
\[r=x^i\cdot xs\in (x^i)\]
since $(x^i)$ is an ideal and $xs\in R$.
Hence $(x^{i+1})\subset (x^i)$ for any positive integer $i$.
Since $R$ is an Artinian ring by assumption, the descending chain of ideals terminates.
That is, there is an integer $N$ such that we have
\[(x^N)=(x^{N+1})=\cdots.\]
It follows from the equality $(x^N)=(x^{N+1})$ that there is $y\in R$ such that
\[x^N=x^{N+1}y.\]
It yields that
\[x^N(1-xy)=0.\]
Since $R$ is an integral domain, we have either $x^N=0$ or $1-xy=0$.
Since $x$ is a nonzero element and $R$ is an integral domain, we know that $x^N\neq 0$.
Thus, we must have $1-xy=0$, or equivalently $xy=1$.
This means that $y$ is the inverse of $x$, and hence $R$ is a field.
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