Every polynomial with constant term 1 can be factorized using degree one polynomials of the for (1-ax)

Question

I've seen a statement along the lines of:

If $K$ is an algebraically closed field, then every polynomial $P$ with $P(0) = 1$ in $K[x]$ can be expressed as $\Pi_{1 \leq i \leq n} (1 - \lambda_i x)$.

What is a short proof of this?

Answer 1

I assume you know that if $\rho_1, \ldots, \rho_n$ is a complete list of the roots of $P$ (with multiple roots appearing the appropriate number of times), then: $$ P(x) = c \prod_{i = 1}^n (\rho_i - x) \tag*{(*)} $$ for some $c \in K$. Now if $P(0) = 1$, no $\rho_i$ is zero and also: $$ c \prod_{i = 1}^n \rho_i = P(0) = 1 $$ So we may divide the iterated product in (*) above by $\prod_{i=1}^n\rho_i$ (by dividing the $i$-th term by $\rho_i$) to get $$ P(x) = \left(c \prod_{i=1}^n\rho_i\right) \prod_{i = 1}^n (1 - \lambda_i x) = \prod_{i = 1}^n (1 - \lambda_i x) $$ where $\lambda_i = 1/\rho_i$.

Here we use the fact that $K$ is algebraically closed to ensure that $K$ contains a complete list of the roots of $P$. The argument works for any $K$ that contains all of the roots of $P$.