Showing Difference of Relatively Prime Polynomials is Irreducible

[slamminsammya]

Homework Statement

Let K be a field, and f,g are relatively prime in K[x]. Show that f-yg is irreducible in K(y)[x].

Homework Equations

There exist polynomials a,b\in K[x] such that af+bg=u where u\in K. We also have the Euclidean algorithm for polynomials.

The Attempt at a Solution

Assuming towards a contradiction that f-yg were reducible, we have f-yg=hk where h,k\in K(y)[x] are not units. Then by the relative primacy condition we also have af+bg=1, so that multiplying both sides by hk yields hk(af+bg)=hk=f-yg, but this is a contradiction since f-yg is certainly not in our original ring of polynomials (assuming that y\notin K), but the left hand side is most certainly in the original ring. The problem is I don't feel confident at all in this argument. I am having trouble conceptualizing what f-yg is.

 

[foxjwill]

...multiplying both sides by hk yields hk(af+bg)=hk=f-yg, but this is a contradiction since f-yg is certainly not in our original ring of polynomials (assuming that y\notin K),

Correct.

but the left hand side is most certainly in the original ring.

This is incorrect. Each part of the equation hk(af+bg)=hk=f-yg was derived by directly applying noncontradictory definitions (namely (i) af+bg := 1 and (ii) hk := f-yg), so you won't be able to get a contradiction without doing something else.

 

[foxjwill]

The problem is I don't feel confident at all in this argument. I am having trouble conceptualizing what f-yg is.

Think of y as a constant (which is what it is). It might help to use a different letter, say \alpha, instead of y for the time being so you don't accidentally forget it's not a variable.