Local ring

**Arczi1984** · December 5th 2010, 11:17 AM

Hi,

Could You help me with this task?

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

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

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

**tonio** · December 5th 2010, 12:20 PM

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 $a, b \in A$ , from this that $a+b=1$ we have $a$ is an invertible element or $b$ is an invertible element."

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

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

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

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

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

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

Tonio

**Arczi1984** · December 5th 2010, 12:53 PM

I found longer solution to this implication. Your is shorter and easier

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