# Ring without maximal ideal

• Dec 5th 2009, 05:35 AM
EinStone
Ring without maximal ideal
Hey there,

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

$\sum_{n=1}^{\infty}a_{n}x^{\alpha_n}$ where $a_n \in \mathbb{R}$ and $0 < \alpha_1 < \alpha_2 < \cdots$ are real numbers so that $\lim_{x\to\infty}\alpha_n = \infty$.

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 AM
Shanks
If it has I as a maximal ideal, then R/I is a field.
• Dec 5th 2009, 09:36 AM
EinStone
• Dec 5th 2009, 10:36 AM
NonCommAlg
Quote:

Originally Posted by EinStone
Hey there,

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

$\sum_{n=1}^{\infty}a_{n}x^{\alpha_n}$ where $a_n \in \mathbb{R}$ and $0 < \alpha_1 < \alpha_2 < \cdots$ are real numbers so that $\lim_{x\to\infty}\alpha_n = \infty$.

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

for $f=\sum_{n=1}^{\infty} a_n x^{\alpha_n} \in R$ with $0 < \alpha_1 < \alpha_2 < \cdots$ and $a_1 \neq 0$ define $\deg f = \alpha_1.$ clearly $\deg (f \cdot g)=\deg f + \deg g$ and so $R$ is an integral domain (without the identity element!)

now suppose that $I$ is a maximal ideal of $R.$ let $S=\mathbb{R} + R$ and choose $g \in R - I.$ let $J=Sg.$ see that $J$ is an ideal of $R.$ since $\forall h \in J: \ \deg h \geq \deg g,$ we have $J \neq R.$

so $J$ is a "proper ideal" of $R.$ it's clear that $I \neq J$ because $g \in J - I.$ let $f \in I.$ if $f = 0,$ then clearly $f \in J.$ so we'll assume that $f \neq 0.$ suppose that $\deg g > \deg f.$ then $\frac{g}{f} \in R$
and therefore $g \in Rf \subseteq I,$ which is a false result. so we must have $\deg g \leq \deg f$ and so $\frac{f}{g} \in S.$ thus $f \in Sg = J.$ hence $I \subset J \subset R,$ which contradicts maximality of $I. \ \Box$
• Dec 5th 2009, 10:57 AM
EinStone
I got 2 questions:
1. How do you know that $f, g, h$ have their $a_1 \neq 0$ (why is degree defined?)

2. I dont understand the notation, what does $S = \mathbb{R} + R$ and what does $Sg$ or $Rf$ mean?
• Dec 5th 2009, 06:55 PM
NonCommAlg
Quote:

Originally Posted by EinStone
I got 2 questions:
1. How do you know that $f, g, h$ have their $a_1 \neq 0$ (why is degree defined?)

if $a_j$ is the first non-zero element in the sequence $\{a_1, a_2, \cdots \},$ then $\alpha_j$ (and not $a_j$) is called the degree of $f =\sum a_n x^{\alpha_n}.$ of course we may always "assume" that $a_1 \neq 0$ and so $\deg f = \alpha_1.$

Quote:

2. I dont understand the notation, what does $S = \mathbb{R} + R$ and what does $Sg$ or $Rf$ mean?
the notations are standard and i don't know how you didn't understand them?? $\mathbb{R}$ is the set of real numbers and $S=\mathbb{R} + R = \{r + f: \ \ r \in \mathbb{R}, \ f \in R \}.$

also $Sg = \{sg: \ s \in S \}$ and $Rf = \{hf: \ h \in R \}.$
• Dec 6th 2009, 07:29 AM
EinStone
Ok thanks for clarification.

Quote:

Originally Posted by NonCommAlg
the degree of $f =\sum a_n x^{\alpha_n} \in R$ is the first non-zero element in the sequence $\{\alpha_1, \alpha_2, \cdots \}.$

Here $\{\alpha_1, \alpha_2, \cdots \}$ you mean $\{a_1, a_2, \cdots \}$ right?

Ok and another question:
If we allow $\alpha_1 = 0$ 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?
• Dec 6th 2009, 08:48 AM
NonCommAlg
Quote:

Originally Posted by EinStone

Here $\{\alpha_1, \alpha_2, \cdots \}$ you mean $\{a_1, a_2, \cdots \}$ right?

that was a terrible typo i made. i just fixed it and you should take a look at it again.

Quote:

If we allow $\alpha_1 = 0$ 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?
well, if you allow $\alpha_1=0,$ then you'll exactly get the ring $S=\mathbb{R}+R,$ which i already mentioned. in this case $R$ (your original ring) would be the unique maximal ideal of this new ring $S.$

first see that $R$ is indeed an ideal of $S.$ now let $I$ be any ideal of $S$ not contained in $R.$ so $I$ has an element of the form $f=a_1 + \sum_{j=2}^{\infty}a_jx^{\alpha_j}, \ \ a_1 \neq 0, \ 0 < \alpha_2 < \alpha_3 < \cdots.$ but then $f$
would have an inverse in $S$ and, since $I$ is an ideal, we get $1=f \cdot \frac{1}{f} \in I.$ thus $I=S.$ so every proper ideal of $S$ is contained in $R$ and hence $R$ is the unique maximal ideal of $S.$
• Dec 6th 2009, 09:01 AM
EinStone
Ok good I get this. To see that R is an ideal in S, it is sufficient to show that $r*s \in R$ if $r \in R$ and $s \in S$ right?