An Artinian integral domain is a field

[mathmari]

Hey!

I want to show that an Artinian integral domain is a field.

Let $R$ be an Artinian integral domain and $a\in R$ with $a\neq 0$.
Can we take then the sequence $(a)\supseteq (a)^2\supseteq (a^3)\supseteq \dots $ ? (Wondering)

Since $R$ is an Artinian integral domain we have that $\exists k\in R$ such that $(a^k)=(a^{k+1})$, i.e., $a^k\in (a^k)=(a^{k+1})\Rightarrow a^k=a^{k+1}m$ for some $m\in R$.
We have that $a^k=a^{k+1}m\Rightarrow a^k=a^k\cdot a\cdot m \Rightarrow a^k\cdot 1=a^k\cdot (a\cdot m)$.
Since $a\neq 0$, we have that $a^n\neq 0$ since $R$ is an integral domain, right? (Wondering)
Do we conclude from the last equation that $1=a\cdot m$ ? (Wondering)

 

[Deveno]

In an integral domain, if:

$ar = as$ for $a \neq 0$, then we have:

$a(r - s) = 0$, and since $a \neq 0$, we must have $r - s = 0$, that is: $r = s$.

So, yes (we can "cancel" non-zero terms on each side of an equation).

 

[mathmari]

In an integral domain, if:

$ar = as$ for $a \neq 0$, then we have:

$a(r - s) = 0$, and since $a \neq 0$, we must have $r - s = 0$, that is: $r = s$.

So, yes (we can "cancel" non-zero terms on each side of an equation).

Ah ok... I see... (Nod)

Let $R$ be an Artinian integral domain and $a\in R$ with $a\neq 0$.
Can we take then the sequence $(a)\supseteq (a)^2\supseteq (a^3)\supseteq \dots $ ? (Wondering)

Can we just take this sequence? (Wondering)

 

[Deveno]

Ah ok... I see... (Nod)


Can we just take this sequence? (Wondering)

Sure, and then your derivation of:

$a^k \cdot 1_R = a^k (am)$, shows that any $a \neq 0$ is a unit.

 

[mathmari]

Thanks a lot! (flower)