Show $y^4 - x^5$ is irreducible in $\mathbb C[[x,y]]$

Question

Is there a 'conceptual' way to see that $f(x,y) = y^4 - x^5$ is not the product of two power series $a(x,y)$ and $b(x,y)$ unless either $a$ or $b$ are invertible?

I guess I am thinking of $\mathbb{C}[[x,y]]$, but how much does that matter?

Answer 1

Although, yes, there are higher-dimensional Newton polytopes, the lowest-dimensional version, Newton polygons, is understandable and useful.

Situation: polynomials $R[x]$ over a principal ideal domain $R$, such as $\mathbb Z$ or $k[x]$ with field $k$. For a prime $\pi$ in $R$, plot the orders of the the coefficients as functions of the power of $x$. The Newton polygon is the bending-upward convex hull of these dots.

The information captured this way is not generally definitive (insofar as there are many "no information" outcomes), but there are some happy cases where the outcome is clear. One extreme case is Eisenstein's criterion (with Gauss' lemma), where the polynomial is monic, all the interior coefficients are divisible by some prime $\pi$ in $R$, and the constant coefficient is divisible by $\pi$, but not by $\pi^2$. This gives a Newton polygon of length $n$ (=degree) and slope $1/n$. The application of the Newton polygon game (yes, proofs are needed, but/and are available) is that such a fraction implies irreducibility (as a corollary that the extension generated by any root is totally ramified locally at $\pi$).

In the case at hand, the base ring is $R=\mathbb C[[x]]$, the formal power series ring, and the polynomial ring is $\mathbb C[[x]][y]$. The prime is the unique non-zero prime ideal in the formal power series ring, generated by $x$. The Newton polygon has a single side, with slope $5/4$. This is in lowest terms, with denominator $4$, so the polynomial is irreducible in $\mathbb C[[x]][y]$.

Then it is certainly irreducible over $\mathbb C[x][y]$, since a factorization there would give one over the formal power series ring. (And the polynomial is monic in $y$.)

The general prescription of interpreting Newton polygons takes a little more preparation, but is worthwhile, in my opinion, since this is one of the few devices we have to easily understand ramification (and/or irreducibility). :)