Irreducibility of a polynomial with complex numbers I'm currently trying to learn more about the irreducibility of several polynomials. Here's one for example:
$Y^2-(X-a_1)(X-a_2)\cdots(X-a_n)$ $\in$ $\mathbb C[X,Y]$ with $n \ge 1$ and $a_1, ...,a_n$ $\in$ $\mathbb C$ being pairwise distinct.
I solved the problem using Eisenstein. However, my book tells me to try solving it with a different methode (as an exercise), but I can't find a way that works.
So what would be a more efficient way to show its irreducibility?
 A: Your claim as stated is false. I will elaborate on it though. If your polynomial ($Y^2-f(X)$) were reducible then it would be reducible as an element of $\mathbb{C}(X)[Y]$. (This is the rings of polynomials in one indeterminate $Y$ and cofficents coming from the field of rational functions in $X$:$\mathbb{C(X)}$). $Y^2-f(X)$ as an element of $\mathbb{C}(X)[Y]$ will be a degree 2 polynomial and therefore reducible iff it has a root in the base field $\mathbb{C}(X)$. You can show easily for example that such a root can not exist if the degree of $f(X)$ as an element of $\mathbb{C}[X]$ is odd.

I am modifying my answer after you edited your question. Now that you put the condition that $a_i$ are distinct, we can prove that your polynomial is irreducible. Let $f(x)=(x-a_1)...(x-a_n)$. If $Y^2-f(X)$ had a root in $\mathbb{C}(X)$, then there would exist two polynomials $A(X),B(X)$ in $\mathbb{C}[X]$ such that $(\frac{B(X)}{A(X)})^2-f(X)=0$, thus $B(X)^2=f(X)A(X)^2$. once you write $A,B$ as a product of monomials ($\mathbb{C}$ is algebraically closed), you will see immediately that it is impossible for such two polynomials $A,B$ to exist because of parity conditions coming from the distinctness of $a_1,...,a_n$. So your polynomial is irreducible.
