# Local ring

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• Dec 5th 2010, 11:17 AM
Arczi1984
Local ring
Hi,

Could You help me with this task?

"Proof that a commutative ring a is a local ring iff for any $\displaystyle a, b \in A$, from this that $\displaystyle a+b=1$ we have $\displaystyle a$ is an invertible element or $\displaystyle b$ is an invertible element."

Proof. $\displaystyle (\Rightarrow)$ $\displaystyle a,b\in A$ - any element which satysfied equality $\displaystyle a+b=1$ and suppose that $\displaystyle a,b\notin U(A)$, where $\displaystyle U(A)$ is a set of invertible elements of $\displaystyle A$. We have then that $\displaystyle a,b\in \mathfrak{m}$. Since $\displaystyle \mathfrak{m}$ is an ideal, then also $\displaystyle 1=a+b\in \mathfrak{m}$, and so $\displaystyle \mathfrak{m}=A$. This is contradiction, so $\displaystyle a \in U(A)$ or $\displaystyle b \in U(A)$.

$\displaystyle (\Leftarrow)$ - I'm not sure. Could You help me with this implication?
• Dec 5th 2010, 12:20 PM
tonio
Quote:

Originally Posted by Arczi1984
Hi,

Could You help me with this task?

"Proof that a commutative ring a is a local ring iff for any $\displaystyle a, b \in A$, from this that $\displaystyle a+b=1$ we have $\displaystyle a$ is an invertible element or $\displaystyle b$ is an invertible element."

Proof. $\displaystyle (\Rightarrow)$ $\displaystyle a,b\in A$ - any element which satysfied equality $\displaystyle a+b=1$ and suppose that $\displaystyle a,b\notin U(A)$, where $\displaystyle U(A)$ is a set of invertible elements of $\displaystyle A$. We have then that $\displaystyle a,b\in \mathfrak{m}$. Since $\displaystyle \mathfrak{m}$ is an ideal, then also $\displaystyle 1=a+b\in \mathfrak{m}$, and so $\displaystyle \mathfrak{m}=A$. This is contradiction, so $\displaystyle a \in U(A)$ or $\displaystyle b \in U(A)$.

$\displaystyle (\Leftarrow)$ - I'm not sure. Could You help me with this implication?

Let $\displaystyle \displaystyle{S:=A-U(A)$ , and take $\displaystyle x,y\in S$ . If $\displaystyle x+y\notin S$ then

$\displaystyle x+y\in U(A)\Longrightarrow \exists c\in A\,\,s.t.\,\,1=c(x+y)=cx+cy\Longrightarrow cx\in U(A)\,\,or\,\,cy\in U(A)$ ,

but in both cases we get a straightforward contradiction (can you see why?) , so it must be

that $\displaystyle x+y\in S$ and from here $\displaystyle S$ is an ideal, and we're thus done.

Tonio
• Dec 5th 2010, 12:53 PM
Arczi1984
I found longer solution to this implication. Your is shorter and easier ;)