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.  
 A: 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.
