Proposition 12

Let $K$ be a commutative field and $f, g$ two elements of $K[X]$. For every polynomial $d$ in $K[X]$ the following properties are equivalent:

(i) The polynomial $d$ divides $f$ and $g$ and every polynomial which divides both $f$ and $g$ divides $d$.

(ii) The polynomial $d$ divides $f$ and $g$ and there exist two polynomials $u$ and $v$ such that $d = u f + v g$.

(iii) The relation $(d) = (f) + (g)$ holds between ideals in $K[X]$.

The polynomial $d$ is determined up to multiplication by a non-zero element of $K$ by these properties. If $f$ and $g$ are not both zero, then $d \neq 0$ and the degree of $d$ majorizes the degree of every polynomial dividing both $f$ and $g$.

When $f$ and $g$ are zero, each of the properties (i) to (iii) is satisfied only for $d = 0$, hence they are then equivalent. Henceforth we assume that $f, g$ are not both 0 and we denote by $a$ the ideal $(f) + (g)$ of $K[X]$.

We remark that for any polynomials $u$ and $v$ in $K[X]$ the properties $(u) \supset (v)$ and « $u$ divides $v$ » are equivalent. The assertion (ii) is thus equivalent to « $(d) \supset (f)$ and $(d) \supset (g)$ and $d \in (f) + (g)$ », that is (iii). It is clear that (ii) implies (i). Finally suppose that (i) holds; we have $(d) \supset (f)$ and $(d) \supset (g)$, whence $(d) \supset a$; on the other hand, by Prop. 11 (IV, p. 12) there exists a polynomial $d_1$ such that $a = (d_1)$; since $d_1$ divides both $f$ and $g$, it divides $d$ by hypothesis, whence $(d) \subset a$, and finally we have $(d) = a$, that is, (iii).

The other assertions of Prop. 12 are immediate consequences of Prop. 11 applied to the ideal $a = (f) + (g)$.