Let $f$ be a nonconstant polynomial over a field $F$.
Since there exists a splitting field of $f$ over $F$, $f$ can be decomposed as $f=c\prod_{i=1}^n (X-\alpha_i)$
Hence, it is possible to define $D_f=c^{2n-2} \prod_{1\leq i<j\leq n} (\alpha_i-\alpha_j)^2$ and call it the discriminant of $D_f$.
However, I saw in wikipedia that discriminant of a monic polynomial over a commutative ring is important in algebraic geometry, but how do I define the discriminant in this case?
Let $R$ be a commutative ring and $f\in R[X]$ be nonconstant monic polynomial.
Then, does there always exist a ring extension $R\leq S$ such that $f$ splits over $S$?
Here's an approach using the fact that the discriminant is a symmetric polynomials in the roots.
First, consider the ring $R=\mathbb{Z}[T_1, \ldots, T_n]$, and look at the "universal" monic polynomial $P(X) = \prod_{i=1}^n (X-T_i) \in R[X]$. Its discriminant is $D=\prod_{1\leq i<j\leq n}(T_i-T_j)^2$. This is a symmetric polynomial in the $T_i$, so it is a polynomial in the elementary symmetric polynomials $\sigma_1, \ldots, \sigma_n$. To be clear: the ring homomorphism $$ \mathbb{Z}[S_1, \ldots, S_n] \to \mathbb{Z}[T_1, \ldots, T_n]$$ taking each variable $S_i$ to the elementary symmetric polynomial $\sigma_i$ is injective, and its image is the set of symmetric polynomials in the variables $T_i$. Therefore the polynomial $D$ above has a unique preimage, which we call $\Delta = \Delta(S_1, \ldots, S_n)$.
Again, just to be clear: to get the discriminant of $P$, we simply evaluate $\Delta$ at $(\sigma_1, \ldots, \sigma_n)$. In short, $D = \Delta(\sigma_1, \ldots, \sigma_n)$. Now, the thing to notice is that $$P(X) = X^n + \sum_{i=1}^n (-1)^i \sigma_i X^{n-i},$$ so the discriminant of $P$ is $\Delta$ evaluated in the coefficients of $P$ (up to a sign).
Now, we simply carry this idea over to an arbitrary commutative ring $R'$. If $Q = X^n + \sum_{i=1}^n (-1)^i a_i X^{n-1}\in R'[X]$ is any monic polynomial, simply define its discriminant to be $\Delta(a_1, \ldots, a_n).$
If $Q$ splits over (an extension of) $R'$, then this definition is equivalent to the one given by the formula in your post.
As for the question whether there is always an extension of $R'$ such that a given monic polynomial $P$ splits, I suppose you can do as with fields, that is, take $S=R'[X]/(P)$. This should contain a copy of $R'$ (because $P$ is monic), and if you repeat this process often enough, you should get an extension of $R'$ in which $P$ splits.