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.