Irreducibility of a polynomial with algebraically independent coefficients I am learning some kind of field theory. Let $\mathbb{Q}'$ be the smallest subfield in $\mathbb{C}$ containing all roots of unity. Recently I read a book on Galois theory and met the following statement:Let $n\in\mathbb{N}$, complex numbers $a_0,a_1,\ldots, a_n$ are algebraically independent (over $\mathbb{Q}$, I guess). Consider a polynomial $p$ over $K_0=\mathbb{Q}'(a_0,a_1,\ldots, a_n)$, $p(x)=x^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0$. Then $p$ is irreducible over $K_0$ and its roots are pairwise different.
After that the author writes "it is sufficient to prove our statement for $f(x)=x^n+a_0$". Further he gives a nice proof of this reduction. Could you please explain me the idea of the reduction? I mean why it is sufficient to prove the case when $a_1=\cdots=a_{n-1}=0$? How does it prove the statement?
 A: Let me answer this question. Well $p(x)=r(x,a_0,\cdots,a_{n-1})\in \mathbb Q’[x,a_0,\cdots,a_{n-1}]$. Suppose we have proved that $x^n+a_0$ is irreducible over $K_0$ and then its roots are distinct. 
Notice that $R=\mathbb Q’[a_0,\cdots,a_{n-1}]$ is a UFD, and $p(x)\in R[x]$. Assume that $p(x)$ is reducible: $p(x)=f(x)g(x)$ with monic polynomials $f(x),g(x)\in R[x]=\mathbb Q’[a_0,\cdots,a_{n-1}][x]$. Specializing $a_i$ to $0$ for each $i\geq 1$, we obtain a nontrivial factorization for $x^n+a_0$. This is absurd. Since the characteristic of $K_0$ is $0$ and $p(x)$ is irreducible, its roots are pairwise distinct.
Now we show that $p(x)=x^n+a_0$ is irreducible. Notice that the filed $\mathbb Q’$ contains $n-$th root of unity $\omega$. Let $b$ be a root of $p(x)$. Then $p(x)=\prod^n_{i=0}(x-\omega^ib)$. If $p(x)=f(x)g(x)$ with $f,g\in \mathbb Q’[a_0][x]$ monic, then they are of the form $(x-\omega^{i_1}b)\cdots(x-\omega^{i_s}b)$. Let $S=\mathbb Q’[a_0]$.Then we know $\omega^{i_1+\cdots+i_s}b^s\in S$. Hence $ b^s\in S$ and $b^{n-s}\in S$. Therefore $b^s=u(a_0)$ and $b^{n-s}=v(a_0)$ with $a_0=-b^n=-u(a_0)v(a_0)$. So one of $u,v$ are constant. We may assume $b^s\in \mathbb Q’$. Then $a_0=-b^n$ implies $a_0^s=(-b^s)^n$. But $a_0$ is transcendental over $\mathbb Q’$.
