Hey there,

LetRbe the ring consisting of the power series of the form

where and are real numbers so that .

I want to show that R is an integral domain without a maximal ideal. But I don't see how to start...

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- Dec 5th 2009, 05:35 AMEinStoneRing without maximal ideal
Hey there,

Let*R*be the ring consisting of the power series of the form

where and are real numbers so that .

I want to show that R is an integral domain without a maximal ideal. But I don't see how to start...

- Dec 5th 2009, 07:44 AMShanks
If it has I as a maximal ideal, then R/I is a field.

- Dec 5th 2009, 09:36 AMEinStone
And where is the contradiction?

- Dec 5th 2009, 10:36 AMNonCommAlg
for with and define clearly and so is an integral domain (without the identity element!)

now suppose that is a__maximal__ideal of let and choose let see that is an ideal of since we have

so is a "proper ideal" of it's clear that because let if then clearly so we'll assume that suppose that then

and therefore which is a false result. so we must have and so thus hence which contradicts maximality of - Dec 5th 2009, 10:57 AMEinStone
I got 2 questions:

1. How do you know that have their (why is degree defined?)

2. I dont understand the notation, what does and what does or mean? - Dec 5th 2009, 06:55 PMNonCommAlg
if is the first non-zero element in the sequence then (and not ) is called the degree of of course we may always "assume" that and so

Quote:

2. I dont understand the notation, what does and what does or mean?

also and - Dec 6th 2009, 07:29 AMEinStone
- Dec 6th 2009, 08:48 AMNonCommAlg
that was a terrible typo i made. i just fixed it and you should take a look at it again.

Quote:

If we allow then R has an identity and therefore a maximal ideal. Prove that in this case the maximal ideal is unique and identify it.

I cant find the maximal ideal, anyone has an idea?

first see that is indeed an ideal of now let be any ideal of not contained in so has an element of the form but then

would have an inverse in and, since is an ideal, we get thus so every proper ideal of is contained in and hence is the unique maximal ideal of - Dec 6th 2009, 09:01 AMEinStone
Ok good I get this. To see that R is an ideal in S, it is sufficient to show that if and right?