Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If we let $R=\mathbb{Z}[x]$ and $D=\mathbb{Z}[[x]]$. We say that $z\in D$ is rational if there is $g\in R$, $g\ne 0$ such that $zg\in R$. Let $S$ be the set of all rational elements in $D$. Then $S$ is a subring of $D$.

So, in this case, we may assume that $zg=f\in R$ with $f,g$ have no common factors. By some computations, we get that $g(0)=\pm 1$.

So, is it still true for the finitely many variables case and also, infinitely many variables?

In particular, if we let $P$ be the kernel of the map from $R$ to $\mathbb{Z}$ mapping all the variables to 0, fixing the constant term, then $P=(x_1,x_2,\cdots)$ is a prime ideal in $R$. Let $S^*$ be the set of invertible elements in $S$, so, may we have

$S^*\cap R=\pm 1 + P$ ???

For the case $n=1$ variable above, I can do since $\mathbb{Q}[x]$ is PID. But in general, I dont know.

Would some one give me some ideas or suggested sources for reading?

Thanks in advance.

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

For any domain $R$ an element of $R[[X]]$ is invertible if and only if the constant term is invertible in $R$.

Applying this repeatedly, one gets that an element of $\mathbb{Z}[[X_1,\dots,X_n]]$ is invertible (in this domain) if and only if its constant term is invertible in $\mathbb{Z}$ thar is it is in $\pm 1 + (X_1,\dots,X_n)$.

So, regarding your 'in particular' with finitely many variables, you have for $S$ the intersection of $\mathbb{Q}(X_1,\dots,X_n)$ and $\mathbb{Z}[[X_1,\dots,X_n]]$ that is the elements are a quotient of two rational (or equivalently integral) polynomials that are an integral power series, as in the question, that $S^{\ast}$ is contained in the set of power series with constant coeffiecient $\pm 1$ as$S^{\ast} \subset \mathbb{Z}[[X_1,\dots,X_n]]^{\ast}$. Yet, not each power series with constant coefficient $\pm 1$ is an element of $S^{\ast}$ for example as $\mathbb{Q}(X_1,\dots,X_n)$ is countable while there are uncountably many power series with constant coefficient $\pm 1$.

Thus, using the notation of the question, $S^{\ast} \subset \pm 1 + (X_1,\dots,X_n)$ yet the inclusion is strict, and this is not an equality.

The case of infinitely many variables: since you consider quotients of polynomials one can reduce to only considering the substructure where only the (finitely many) variables occuring in the polynomials are present (for each quotient individually).

Alternatively, we mainly need that the constant term of an invertible power series (finitely or infinitely many variable) is invertible in the base domain. This follows just by noting that the constant term of the product is the product of the constant terms, so if the product is $1$ it/they have to be invertible. Thus also showing the inclusion. That it is strict follows by restricting to a substructure with finitely many variables (or a suitable adaption of the cardinality argument).

One thing I am now somehow not completely sure about, though I think so, (but in any case it is not needed here) is whether in infinitely many variables also the invertability of the constant term is sufficient to imply that the power series is invertible.

share|improve this answer
    
Thanks, I see for the case finitely many variables. So you think it is also true for the case infinitely many variables? Could you be more precise for the inclusion. Thanks alot. –  kakalotte Jun 24 '11 at 10:48
    
Kakalotte, you are welcome. I expanded my answer along the lines your requested. –  quid Jun 24 '11 at 13:05
    
Thank quid. Could you please give me some sources to look for the rationality of power series, Dirichlet series . . . Thanks. –  kakalotte Oct 19 '11 at 19:53
    
If I rememeber correctly, the book of Gilmer 'Multiplicative Ideal Theory' contains material for power series; yet this might not be the most easily accessible book. And, my memory is vague so I do not vouch for this information. Regarding Dirichlet series I cannot offer a good reference at moment, sorry. Yet, adding some details and motivation, you could always ask a new question. –  quid Oct 19 '11 at 21:32
    
Thank Quid alot.^^ –  kakalotte Oct 22 '11 at 20:03
add comment

Edit: As pointed out by unknown (google) and Andreas Blass it's not true that the power series ring is in general a direct limit of it's subrings in finitely many variables. So what I have written below only holds in the subring $\varinjlim_J R[[x_j|j \in J]]$ where $J$ runs through the finite subsets of $I$.


Let me work out the case of an infinite index set $I$. Let $R$ be a ring with unit. Then $$U := R[[x_i| i \in I]]$$ can be seen as the union of the "finite" $R[[x_{i_1}, \dots, x_{i_n}]]$ with $i_1, \dots, i_n \in I$.

Now suppose $f\in U$ is invertible. So there is $g \in U$ such that $fg=1$. Moreover we can find a large enough $n$ such that $f,g \in R[[x_{i_1}, \dots, x_{i_n}]]$. From the finite case it follows that

$f \in R^{\times} + (x_{i_1}, \dots, x_{i_n}) \le R^{\times} + (x_i | i \in I) =: P$.

Conversely suppose $f \in P$. Then $f \in R^{\times} + (x_{i_1}, \dots, x_{i_n})$ for some $n$. Thus there is $g \in R[[x_{i_1}, \dots, x_{i_n}]] \le U$ such that $fg=1$. So $f$ is invertible in $U$, showing $U^{\times} = P$.

Most properties of $U$ can be derived in this way.

share|improve this answer
    
Perhaps I am confused, but in contrast to the situation for polynomials, I do not think that the power series ring over infinitely many variables is the union of the ones with finitely many variables. –  quid Jun 24 '11 at 13:08
    
Really ? If the index set is countable there are monomorphisms $R[[x_1]] \to R[[x_1,x_2]] \to ...$ that should constitute the direct limit as a union. But I will check this in detail. What are your doubts ? –  Ralph Jun 24 '11 at 13:16
    
@Ralph: I think unknown(google) is referring to the possibility that a power series, having infinitely many terms, could have terms involving more and more variables. Such a series would not be in the image of any of the rings in your direct system. –  Andreas Blass Jun 24 '11 at 13:26
    
@Andreas: OK, I see. Thanks. –  Ralph Jun 24 '11 at 13:29
    
Yes, Andreas Blass, this is what I meant. For completeness, I reproduce the comment I typed before I saw yours. "Something like $\sum_i X_i$, so one power series containing infinitely many variables. But perhaps there are even both definition around for 'the' power series ring in countably many variables." –  quid Jun 24 '11 at 13:34
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.