In proving an irreducible curve has only finitely many singular points, is $f_x \not \equiv 0$?

Question

Artin Algebra Chapter 11

This has been answered on the site, but I want to ask about the solution given by Brian Bi:

The constant polynomials are not considered irreducible, so f is not constant. Therefore ∂f/∂x or ∂f/∂y is nonzero (possibly both). Applying Theorem 11.9.10 to f and one of its nonzero partial derivatives yields the desired result.

My question is about

Therefore ∂f/∂x or ∂f/∂y is nonzero (possibly both).

In Chapter 11.9, we have a definition of irreducible for a complex polynomial of two variables that is part of a sentence, but I'm not sure if the last part is part of the sentence is part of the definition.

Let's assume that the polynomial f is irreducible - that it is not a product of two nonconstant polynomials, and also that it has positive degree in the variable x.

Here is the context:

Now:

• If positive degree in x is part of the definition of irreducible and if $f$ has positive degree in $x$, then it's definitely the case that ∂f/∂x is nonzero while ∂f/∂y is the one that we are not sure is nonzero right?

• If positive degree in x is not part of the definition of irreducible, then how do we say constants are irreducible? In Chapter 11.8, this is explicit.

For reference, the definition of an irreducible polynomial in one variable in the previous section Chapter 11.8 is:

A polynomial with coefficients in a field is called irreducible if it is not constant and if is not the product of two polynomials, neither of which is a constant.

Here is the context:

Answer 1

It looks like you're parsing the last quote wrong. It should be read as

Let's assume [that the polynomial f is irreducible - that it is not a product of two nonconstant polynomials], and also that it has positive degree in the variable x.

In other words, the condition that $f$ has positive degree in $x$ is not part of the definition of "irreducible", but is instead an additional assumption being made in addition to $f$ being irreducible. To be clear, the precise definition of what it means for $f$ to be irreducible is that $f$ is nonconstant and cannot be written as a product of two nonconstant polynomials. This sentence is in no way intended as a redefinition of the term "irreducible"; it just contains a brief reminder for the reader of what it means (which omits the nonconstant condition).

In any case, in problem 9.9, there is no assumption that $f$ has positive degree in $x$. You are right that if we knew it had positive degree in $x$, then $\partial f/\partial x$ would be nonzero.

(Incidentally, Brian Bi's solution is incomplete. Theorem 11.9.10 only applies to polynomials with no nonconstant common factor, and so some argument must be given that $f$ and its nonzero partial derivative satisfy this condition. Here you must use the fact that $f$ is irreducible, not just nonconstant.)