# Ring without maximal ideal

• Dec 5th 2009, 04:35 AM
EinStone
Ring without maximal ideal
Hey there,

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

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

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, 06:44 AM
Shanks
If it has I as a maximal ideal, then R/I is a field.
• Dec 5th 2009, 08:36 AM
EinStone
• Dec 5th 2009, 09:36 AM
NonCommAlg
Quote:

Originally Posted by EinStone
Hey there,

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

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

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

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

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

so $\displaystyle J$ is a "proper ideal" of $\displaystyle R.$ it's clear that $\displaystyle I \neq J$ because $\displaystyle g \in J - I.$ let $\displaystyle f \in I.$ if $\displaystyle f = 0,$ then clearly $\displaystyle f \in J.$ so we'll assume that $\displaystyle f \neq 0.$ suppose that $\displaystyle \deg g > \deg f.$ then $\displaystyle \frac{g}{f} \in R$
and therefore $\displaystyle g \in Rf \subseteq I,$ which is a false result. so we must have $\displaystyle \deg g \leq \deg f$ and so $\displaystyle \frac{f}{g} \in S.$ thus $\displaystyle f \in Sg = J.$ hence $\displaystyle I \subset J \subset R,$ which contradicts maximality of $\displaystyle I. \ \Box$
• Dec 5th 2009, 09:57 AM
EinStone
I got 2 questions:
1. How do you know that $\displaystyle f, g, h$ have their $\displaystyle a_1 \neq 0$ (why is degree defined?)

2. I dont understand the notation, what does $\displaystyle S = \mathbb{R} + R$ and what does $\displaystyle Sg$ or $\displaystyle Rf$ mean?
• Dec 5th 2009, 05:55 PM
NonCommAlg
Quote:

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

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

Quote:

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

also $\displaystyle Sg = \{sg: \ s \in S \}$ and $\displaystyle Rf = \{hf: \ h \in R \}.$
• Dec 6th 2009, 06:29 AM
EinStone
Ok thanks for clarification.

Quote:

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

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

Ok and another question:
If we allow $\displaystyle \alpha_1 = 0$ 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, 07:48 AM
NonCommAlg
Quote:

Originally Posted by EinStone

Here $\displaystyle \{\alpha_1, \alpha_2, \cdots \}$ you mean $\displaystyle \{a_1, a_2, \cdots \}$ 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 $\displaystyle \alpha_1 = 0$ 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 $\displaystyle \alpha_1=0,$ then you'll exactly get the ring $\displaystyle S=\mathbb{R}+R,$ which i already mentioned. in this case $\displaystyle R$ (your original ring) would be the unique maximal ideal of this new ring $\displaystyle S.$

first see that $\displaystyle R$ is indeed an ideal of $\displaystyle S.$ now let $\displaystyle I$ be any ideal of $\displaystyle S$ not contained in $\displaystyle R.$ so $\displaystyle I$ has an element of the form $\displaystyle f=a_1 + \sum_{j=2}^{\infty}a_jx^{\alpha_j}, \ \ a_1 \neq 0, \ 0 < \alpha_2 < \alpha_3 < \cdots.$ but then $\displaystyle f$
would have an inverse in $\displaystyle S$ and, since $\displaystyle I$ is an ideal, we get $\displaystyle 1=f \cdot \frac{1}{f} \in I.$ thus $\displaystyle I=S.$ so every proper ideal of $\displaystyle S$ is contained in $\displaystyle R$ and hence $\displaystyle R$ is the unique maximal ideal of $\displaystyle S.$
• Dec 6th 2009, 08:01 AM
EinStone
Ok good I get this. To see that R is an ideal in S, it is sufficient to show that $\displaystyle r*s \in R$ if $\displaystyle r \in R$ and $\displaystyle s \in S$ right?