Proposition 6 (Lagrange interpolation formula)

— Let $K$ be a commutative field, $\alpha_1, \alpha_2, \ldots, a,$ distinct elements of $K$ and $\beta_1, \beta_2, \ldots, \beta_n$ any elements of $K$. For $i = 1, 2, \ldots, n$ we put

where $U(i)$ is the set of integers $j$ such that $j \neq i$ and $1 \leq j \leq n$. Then $\beta_1 f_1 + \ldots + \beta_n f_n$ is the unique element $f$ of $K[X]$ such that $\deg f < n$ and $f(\alpha_i) = \beta_i$ for $1 \leq i \leq n$.

The uniqueness off follows from the Cor. to Th. 1. Let $f = \beta_1 f_1 + \ldots + \beta_n f_n$, then since $f_i$ has degree $n - 1$, we have $\deg f < n$. On the other hand, $f_i(\alpha_j) = 0$ for $j \neq i$ and $f_i(\alpha_i) = 1$, hence $f(\alpha_i) = \beta_i$ for $1 \leq i \leq n$.