Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the fraction field of the formal power series ring over a field in finitely many variables $K[[X_1,\dots,X_n]]$? Is there a nice description for this field? When $n=1$, I know this is the formal Laurent series ring over $K$.

share|cite|improve this question
This is a good question. In order to get something similar to the case $n=1$ it would be interesting to know if a formal power series in several indeterminates can be written as a product between a polynomial from $(X_1,\dots,X_n)$ and an invertible power series. (I don't know if this is true or not.) – user26857 Feb 14 '13 at 17:01
@YACP You're asking if $K[[X_1,\dots,X_n]] = K[[X_1,\dots, X_n]]^{\times} K[X_1,\dots,X_n]$? – JSchlather Feb 14 '13 at 18:04
And this is false for $n\ge 2$. Consider $y^2-x^2(1+x)$ which is irreducible in $\mathbb C[x,y]$. It is the product of $y-x\sqrt{1+x}$ and of $y+x\sqrt{1+x}$ as power series. If the latter are polynomials $P, Q\in (x,y)\mathbb C[x,y]$ up to units, then $y^2-x^2(1+x)$ and $PQ$ generate the same ideal in $\mathbb C[[x,y]]$. By faithfull flatness, both ideals would be equal in $\mathbb C[x,y]$, which is impossible – user18119 Feb 14 '13 at 21:55

There’s a really big difference between the one-variable power series ring, together with its fraction field, and a many-variable power series ring. Namely that the one-variable ring has only one irreducible element. This is what lets us have a nice description of the fraction field of $k[[x]]$. A many-variable ring, even $k[[x,y]]$, has infinitely many irreducibles unrelated by unit factors. Much uglier, from this viewpoint, than the one-variable case.

share|cite|improve this answer
Is there an infinite class of irreducible elements in $k[[x,y]]$ that are easy to describe? – JSchlather Feb 14 '13 at 19:38
@Jacob, this is so far from my competence that I hesitate to answer. Although there are polynomials in $k[x,y]$ that split in $k[[x,y]]$, I wonder whether every associate class of irreducibles in the power series ring has an irreducible polynomial in it. – Lubin Feb 15 '13 at 18:28
@Jacob: sorry, my “I wonder whether” was completely stupid. Clearly, $y−f_1(x)$ and $y−f_2(x)$, where $f_1\ne f_2$, are two irreducible power series in $k[[x,y]]$ whose ratio is not a unit of the power-series ring. – Lubin Feb 15 '13 at 19:30
Dear Lubin, how can you convince us that power series are far from you competence ? – user18119 Feb 15 '13 at 22:14

Your Answer


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.