Basic Ring Theory - 12

[Tsakanikas Nickos]

Let \( A \) be an Artinian ring. Show that every prime ideal of \( A \) is maximal. Additionally, conclude that the Nilradical is equal to the Jacobson radical.

[Papapetros Vaggelis]

Lemma

If a commutative ring \(\displaystyle{A}\) is an \(\displaystyle{\rm{Artinian}}\) ring and \(\displaystyle{x\,y\neq 0\,,\forall\,x\,,y\in A-\left\{0\right\}}\),

then, \(\displaystyle{A}\) is a field.

Proof of Lemma

It is sufficient to prove that \(\displaystyle{U(A)=A-\left\{0\right\}}\). Let \(\displaystyle{a\in A-\left\{0\right\}}\).

Then, \(\displaystyle{a^{n}\neq 0\,,\forall\,n\in\mathbb{N}}\) and \(\displaystyle{a^{k}\mid a^{k+1}\,,\forall\,k\in\mathbb{N}}\).

Consider the decreasing sequence of ideals of \(\displaystyle{A}\),

\(\displaystyle{...\langle{a^m\rangle}\subseteq \langle{a^{m-1}\rangle}\subseteq ...\subseteq \langle{a^2\rangle}\subseteq \langle{a\rangle}}\).

Since \(\displaystyle{A}\) is Artinian ring, there exists \(\displaystyle{n\in\mathbb{N}}\) such that

\(\displaystyle{\langle{a^n\rangle}=\langle{a^{n+1}\rangle}=...}\). Then,

\(\displaystyle{a^n\in\langle{a^{n+1}\rangle}\implies a^n=s\,a^{n+1}\implies 1=s\,a}\).

The same progress for \(\displaystyle{s\neq 0}\) gives \(\displaystyle{1=r\,s}\) for some \(\displaystyle{r\in A}\).

So, \(\displaystyle{r=r\cdot 1=r\cdot (s\,a)=(r\,s)\,a=1\cdot a=a\implies 1=s\,a=a\,s\implies a\in U(A)}\).

Application 1

If \(\displaystyle{P}\) is a prime ideal of the Artinian ring \(\displaystyle{A}\) (which is commutative)

then, the ring \(\displaystyle{A/P}\) is an Artinian ring and since \(\displaystyle{P}\) is prime,

\(\displaystyle{(a+P)\cdot (b+P)=a\,b+P\neq P\,,\forall\,a\,,b\in P}\).

According to the Lemma, \(\displaystyle{A/P}\) is a field and \(\displaystyle{P}\) is maximal.

Application 2

Since the ring \(\displaystyle{A}\) is Artinian, the Jacobson radical is nilpotent, that is \(\displaystyle{J(A)^n=\left\{0\right\}}\),

so, \(\displaystyle{a\in J(A)\implies a^n=0\implies a\in \sqrt{0}}\), so, \(\displaystyle{J(A)\subseteq \sqrt{0}}\).

Conversely, suppose that there exists maxiaml ideal \(\displaystyle{M}\) of \(\displaystyle{A}\)

such that \(\displaystyle{\sqrt{0}\subseteq M}\) is not true. Then,

\(\displaystyle{M\subset M+\sqrt{0}\subset A\implies M+\sqrt{0}=A}\) and

\(\displaystyle{1\in A\implies (\exists\,a\in M)\,(\exists\,x\in\sqrt{0}), 1=x+a}\).

But \(\displaystyle{x^m=0}\) for some \(\displaystyle{m\in\mathbb{N}}\) and then,

\(\displaystyle{0=x^m=(1-a)^m=1+\sum_{i=0}^{n-1}\binom{n}{i}(-a)^{n-i}\implies 1\in M\implies M=A}\), a contradiction.

So, \(\displaystyle{\sqrt{0}\subseteq M}\) for every maximal ideal \(\displaystyle{M}\) and

\(\displaystyle{\sqrt{0}\subseteq J(A)}\). Finally, \(\displaystyle{J(A)=\sqrt{0}}\).

[Tsakanikas Nickos]

Thank you very much, Vaggelis, for your elegant solutions.

I would like to note that the lemma that you proved says the following: An Artinian domain is, in fact, a field. Using this lemma, we could give an even shorter proof (essentially the same as yours) of "Application 1":

Let $ \mathfrak{p} $ be a prime ideal of $A$. Then the quotient $A / \mathfrak{p}$ is an artinian ring and an integral domain, thus a field.