Understanding resultant Let us  denote by  $P_k$  the vector space polynomials over $\mathbb{K}$ of the degree $< k$.
Take  $f, g \in \mathbb{K}[x]$, $deg(f) = n,$ $deg(g) = m$  and define the linear map 
$$T : P_m \oplus P_n \to P_{n+m}, \; T(a,b) = af + bg.$$
I understand that $T$ is surjective iff $det T \neq 0$ iff $(f,g) \neq 1$ iff $f ,g$ have a common roots in some extension  of $\mathbb{K}$. But how do I prove that $$detT = \prod (\alpha_i - \beta_i)  $$ 
where the product is taken by all the pairs of roots ($f(\alpha_i) = 0, g(\beta_i) = 0.$
I guess that I need to compute the degree of both as polynomials in roots by it is not obvious how to do that.
 A: The easy way to show this is to let $\mathbb K = \mathbb Z[\alpha_1,\ldots,\alpha_n,\beta_1,\ldots,\beta_m]$ - that is, to consider the free commutative ring upon $n+m$ variables - and to let $f=\prod_{i}(x-\alpha_i)$ and $g=\prod_j(x-\beta_j)$ be two elements of $\mathbb K[x]$, then to construct the linear map $T$ from $P_m\oplus P_n\rightarrow P_{n+m}$ given by $T(a,b)=af+bg$ as described in the question. The determinant of this map will be some element of $\mathbb K$. Since everything here is within the language of rings, we can then get the general result by choosing whatever ring $R$ we like to work in and then mapping the results via the unique homomorphism $\mathbb K\rightarrow R$ taking the desired values on each $\alpha_i$ and $\beta_j$.
First, observe that if $\pi : \mathbb K \rightarrow R$ is any morphism such that $\pi(\alpha_i) = \pi(\beta_j)$ for any pair $i,j$, then $\pi(\det T) = \det \pi(T) = 0$ where $\pi(T)$ is the application of $\pi$ to each element of the matrix for $T$. This is by the results you note, that $T$ fails to be injective if two roots coincide - and this does apply to all rings, since if $\alpha_i = \beta_j$ then
$$g\cdot \prod_{i'\neq i}(x-\alpha_{i'}) = f \cdot \prod_{j'\neq j}(x-\beta_{j'})$$
witnesses the failure of injectivity.
This applies, in particular, to the quotient maps $\pi_{i,j} : \mathbb K \rightarrow \mathbb K / \langle \alpha_i - \beta_j\rangle$. By the prior argument, $\det T$ belongs to the kernel of each of these maps - thus is in $\bigcap_{i,j}\langle \alpha_i -\beta_j \rangle$. You can then proceed as you wish to see that this means that $\prod_{i,j}(\alpha_i-\beta_j)$ divides $\det T$ (for instance, you can use that $\mathbb K$ is a UFD and that these terms are coprime).
You can then note that in the matrix $T$, the values $\alpha_i$ each appear $m$ times and the values $\beta_j$ each appear $n$ times. In the product $\prod_{i,j}(\alpha_i-\beta_j)$, each variable already has the listed maximum degree - and $\det T$ is a multiple of this, therefore must be an integer multiple of this product. You can verify that this coefficient is $1$ by choosing any example that is already known to you - for instance setting every $\alpha_i$ to $0$ and every $\beta_j$ to $1$. 
Thus
$$\det T = \prod_{i,j}(\alpha_i-\beta_j)$$
where this is a statement inside of $\mathbb K$ that can be mapped via homomorphism to any ring. Note that this only applies to monic polynomials (it is otherwise false, although it is not hard to work out the factors arising from the leading coefficients).
You can also work out this fact by considering multiplication by $(x-y)$ as a linear map $\mathbb F[x,y]/(f(x),g(y))$ and working in the basis $\{x^ay^b : a < \deg f, b < \deg g\}$. The characteristic polynomial of this map has roots at the differences $\alpha_i-\beta_j$ counted with appropriate multiplicity - which is a more natural result, but working out the connection between these two linear maps is very very laborious, although it gives a bit more insight since the $(x-y)$ term representing the difference between roots actually makes an appearance.
