Lemma 2

Let R be a ring and P a monic polynomial in R[X]. There exists a ring R' containing R such that in the polynomial ring R'[X] the polynomial P is a product of monic polynomials of degree 1.

We proceed by induction on the degree $n$ of P, the lemma being obvious for $n = 0$ and $n = 1$. Suppose therefore that $n > 1$. Let $a$ be the ideal of $R[X]$ generated by P and let $f$ be the canonical homomorphism of $R[X]$ onto $B = R[X]/a$. Since P is monic, for every polynomial $Q \in R[X]$, $\deg(PQ) = \deg(P) + \deg(Q)$, whence $a \cap R = 0$; the restriction off to R is therefore injective. Identifying R with the subring $f(R)$ of B by means of $f$ and writing $b = f(X)$, we see that $b$ is a root of P in B, P being considered as a polynomial in $B[X]$. Then there exists a monic polynomial Q in $B[X]$ of degree $n - 1$ such that $P(X) = (X - b)Q(X)$ (Algebra, Chapter IV, § 1, no. 4, Proposition 5). By the induction hypothesis there exists a ring $R' \supseteq B$ such that in $R'[X]$ the polynomial Q is a product of monic polynomials of degree 1; clearly in $R'[X]$ P is then a product of monic polynomials of degree 1.