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 $K$.
Fraction field of the formal power series ring in finitely many variables
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commutative-algebra
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0This 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.) – 2013-02-14
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0@YACP You're asking if $K[[X_1,\dots,X_n]] = K[[X_1,\dots, X_n]]^{\times} K[X_1,\dots,X_n]$? – 2013-02-14
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2And 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 – 2013-02-14
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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.
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0Is there an infinite class of irreducible elements in $k[[x,y]]$ that are easy to describe? – 2013-02-14
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0@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. – 2013-02-15
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0@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. – 2013-02-15
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0Dear Lubin, how can you convince us that power series are far from you competence ? – 2013-02-15