Exercise 7.5.3

Proof. We give another proof of Exercise 2, knowing that $f(x,y)=a(x)-\mathit{yb}(x)$ is irreducible in $F[x,y]$. We must prove that a factorization

implies $A\in F(y)$ or $B\in F(y)$.

(a)

$A$ is expressed by $$A(x,y)=\frac{a_0(y)}{b_0(y)}+\frac{a_1(y)}{b_1(y)}x+\cdots +\frac{a_m(y)}{b_m(y)}{x}^m.$$

If we take $g(y)=b_0(y)\cdots b_m(y)\in F[y]$ the product of the $b_i$ (or the lcm of the $b_i$), then $g(y)\frac{a_i(y)}{b_i(y)}\in F[y]$, thus $A_1=g(y)A\in F[x,y]$. Similarly, there is $h\in F[y]$ such that $B_1=h(y)B\in F[x,y]$.

(b)

Therefore, $g(y)h(y)(a(x)-\mathit{yb}(x))=A_1{B}_1\in F[x,y]$, where $g,h,f,A_1,B_1$ are in $F[x,y]$. As $f$ is irreducible and divides $A_1,B_1$ in the UFD $F[x,y]$, $f$ divides $A_1$ or $f$ divides $B_1$.

(c)

Suppose by example that $f$ divides $A_1$ (the other case is similar): $$A_1=(a(x)-\mathit{yb}(x))A_2,A_2\in F[x,y].$$

Then, dividing the equality in (b) by $a(x)-\mathit{yb}(x)\ne 0$, we obtain

$$g(y)h(y)=A_2{B}_1.$$

The degree of $x$ in $A_2{B}_1$ is zero, thus the degree of $x$ in $B_1$ is also 0, so $B_1\in F[y]$.

(d)

Consequently $B=B_1(y)∕h(y)\in F(y)$. In the other case, we obtain $A\in F(y)$. So $a(x)-\mathit{yb}(x)$ is irreducible in $F(y)[x]$.