Generic methods to check irreducibility of polynomials in $K[[X,Y]]$

Question

I usually find it difficult to check irreducibility of polynomials in $K[[X,Y]]$ ($K$ algebraically closed). Does anyone know about generic methods that can be used ? And especially of ones that can be applied to polynomials like $XY-(X+Y)(X^2+Y^2)$ (I found it reducible by using Hensel's lemma after a suitable change of variable, if I did not make mistake).

Answer 1

I have a paper giving an irreducibility test (and factoring "algorithm", in some sense of the word) for formal power series over a PID, which should eventually appear in Trans. of the AMS. In particular it applies to $K[[X]][[Y]]$.

See Theorem 1 of http://arxiv.org/abs/1107.4860

In essence: Write any polynomial $f$ in $K[[X,Y]]$ as a polynomial in $Y$ with coefficients in $K[[X]]$, and let $f_0$ and $f_1$ be the coefficients of $Y^0$ and $Y^1$, respectively. Then $f$ is irreducible in $K[[X,Y]]$ if and only if either (1) $f_0 = 0$ and $f_1$ is a unit in $R[[X]]$, or (2) $f$ has a unique factor in $K[[X]][Y]$ with constant term not a unit in $K[[X]]$. (This works, by the way, even if $f$ is in $K[[X]][Y]$.)

Added on 4/3/14: The criterion above is equivalent to the following. A polynomial $f \in K[[Y]][X]$ is irreducible in $K[[X,Y]] = K[[Y]][[X]]$ if and only if one of the following conditions holds.

(1) $f_0 = 0$ and $f_{1,0} \neq 0$.

(2) $f_0 \neq 0$ and $f_{0,0} = 0$, and if $f = gh$ with $g,h \in k[[Y]][X]$, then either $g_{0,0} \neq 0$ or $h_{0,0} \neq 0$.

Here, $f_{i,j}$ denotes the coefficient of $X^i Y^j$ in $f$, and $f_i$ denotes the coefficient of $X^i$ in $f$ as a polynomial in $K[[Y]][X]$. To test whether (2) holds or not in a given situation, you need to be able to factor polynomials over the complete DVR $K[[Y]]$. To do so you can use Hensel's lemma, Newton polygons, and other methods for factoring polynomials over local fields.

Answer 2

Your argument is correct, and we can generalize it.

Take the lowest-degree terms of your polynomial. In your case, this is $XY$. Factor this into linear terms. If any distinct linear terms show up, you can use this to give a factorization. So the polynomial is always reducible in this case.

Write the lowest-order terms as $l_1^{e_1} l_2^{e_2} \dots l_n ^{e_n}$ where $l_1,\dots,l_n$ are distinct linear terms. To check Hensel lifting for $(l_1^{e_1} +\alpha_1) ( l_2^{e_2} + \alpha_2) \dots (l_n^{e_n} + \alpha_n) = f(X,Y)$ we need to check that the ideal generated by $df/d\alpha_1, df/d\alpha_2 , \dots df/d\alpha_n$ includes all polynomials of degree $> \sum_{i=1}^n e_i$. Indeed it does, because $df/d\alpha_i$ gets everything up to a multiple of $l_i^{e_i}$, so together they get everything up to a multiple of $l_1^{e_1}l_2^{e_2} \dots l_n^{e_n}$, but they can all get multiples of $l_1^{e_1}l_2^{e_2} \dots l_n^{e_n}$.

If there is only a single linear factor, it is more subtle whether the polynomial is irreducible. You can change variables such that that factor is $X$ or $Y$, and then look at the Newton polygon. I think if that is indeterminate you can change the "degree" of the two variables so that there are multiple terms of lowest degree, and try to factor that and apply Hensel's lemma again in a similar way. I"m not sure whether this always tells you whether it is irreducible.