# Prime elements in $\mathbb{Q}[[X,Y,Z]]$ whose status as an infinite series is unchanged by arbitrary multiplication

Let's suppose $R$ is the ring $\mathbb{Q}[[X,Y,Z]]$. I'm interested in finding power series $f(x,y,z) \in R \setminus \mathbb{Q}[X,Y,Z]$ which are, first of all, prime elements in $R$, but also satisfy the following condition: if $g(x,y,z)$ is any nonzero formal power series, then $fg$ is still in $R \setminus \mathbb{Q}[X,Y,Z]$. In other words I'd like to find examples of prime infinite series which are "stable," in the sense that their product with other (nonzero) power series always results in another infinite series.

The problem is a lot of power series can be multiplied by ordinary polynomials such that the result is no longer an infinite power series. Something like $e^x$ is kind of stable though, since if $f(x)$ were a polynomial, if $f(x)e^x$ were no longer an infinite series then we could evaluate this at $0$, implying $e$ is the root of a polynomial in $\mathbb{Q}$.

However, $e^x$ is neither prime (it's a unit) and the fact that it's a unit means there does exist an infinite series (namely its inverse) such that $e^x(e^x)^{-1}$ is no longer an infinite series.

I was thinking of using something like $y + e^x - 1$, which is $y + x + x^2/2 + x^3/3! + ...$, clearly a nonunit. I have no idea whether this element is prime or not though. Perhaps I could relate the transcendentalism of $e$ to $y + e^x - 1$ being stable as an infinite series, as I did in the last paragraph.

If this works, then perhaps I could work in other transcendental numbers, for example $\sum\limits_{n \geq 1} 10^{-n!}$, so that $y + \sum\limits_{n \geq 1}10^{-n!}x^n$ is prime and stable. I really don't know if this is a fruitful approach though.

-
Just an algebraic comment. Take $S=\mathbb Q[X,Y,Z]\setminus \{0\}$. This is a multiplicative system of $R$. What can we say about the ring of fractions $S^{-1}R$? I think an answer to this question could help for solving your problem. (For instance, if this ring is a field, then there is no such $f\neq 0$.) –  user26857 Feb 13 '13 at 23:53
Well, $S^{-1}R$ is an integral domain containing the quotient field $qf(\mathbb{Q}[X,Y,Z])$. It definitely isn't a field, since not everything in $R$ has an inverse, much less one in $S$. I suppose if I could find any principal prime ideals of $S^{-1}R$, they may correspond to the principal prime ideals I was looking for in $R$, namely those which miss $S$. Is this where you were going with this? –  D_S Feb 14 '13 at 1:17
I guess you want to say that $S^{-1}R$ is contained in the field of fractions of $R$. Could you give me (with proof) a concrete example of non-invertible element in $S^{-1}R$? (It's obvious that I'm trying to use the correspondence between prime ideals of $R$ and those of $S^{-1}R$.) –  user26857 Feb 14 '13 at 1:27
Okay, you don't need to say anymore. I think I've got it. Thank you! –  D_S Feb 14 '13 at 5:56
@Shankman: $z:=y+x+x^2/2+...$ is prime because $\mathbb Q[[x,y]]/(z)=\mathbb Q[[x]]$ is integral. –  user18119 Feb 14 '13 at 22:26

Take a power series $t(X)\in X\mathbb Q[[X]]$ which is transcendental over $\mathbb Q[X]$ (this exists by comparing the cardinalities of powers series and algebraic power series). Consider $f(X,Y,Z)=Y-t(X)$. Suppose $$f(X,Y, Z)g(X,Y,Z)=P(X,Y,Z)\in \mathbb Q[X,Y,Z].$$ Write $P=P_0(X,Y)+P_1(X,Y)Z+...$ and substitute $Y$ with $t(X)$ in the above equality: $$0 = P(X, t(X), Z)=P_0(X, t(X))+P_1(X, t(X))Z+...$$ This implies that $P_i(X,Y)=0$, hence $P=0$ and $g=0$.
Edit The element $f(X, Y, Z)$ is prime because $\mathbb Q[[X, Y, Z]]/(f)=\mathbb Q[[X, Z]]$ is an integral domain.
To answer YACP's question in the comments: the localization $S^{-1}Q[[X, Y, Z]]$ has Krull dimension $2$: as the maximal ideal of $\mathbb Q[[X, Y, Z]]$ mets $S$, this localization has dimension at most $2$. Let $s(X)\in X\mathbb Q[[X]]$ be transcendental over $\mathbb Q[X, t(X)]$ (cardinality argument: the transcendental degree of $\mathbb Q[[X]]$ over $\mathbb Q[X]$ is infinite and even uncountable). Consider the ideal $\mathfrak p$ generated by $Y-t(X), Z-s(X)$. It is prime because the quotient $\mathbb Q[[X, Y, Z]]/\mathfrak p=\mathbb Q[[X]]$ is integral. The same argument above shows that $\mathfrak p\cap S=\emptyset$. Therefore $\mathfrak p$ defines a prime ideal of height $2$ in the localization.
Yes, but OP want $f$ to be a prime element, a concern that you haven't addressed. –  Gerry Myerson Feb 14 '13 at 23:52