Proposition 11

— Let $K$ be a commutative field. (i) For every non-zero ideal $a$ of $K[X]$ there exists precisely one monic polynomial $f$ in $K[X]$ such that $a = (f )$. (ii) Let $f_1$ and $f_2$ be in $K[X]$; for $(f_1) = (f_2)$ to hold it is necessary and sufficient that there exist a non-zero element $\lambda$ of $K$ such that $f_2 = \lambda f_1$. Let us prove (ii), the sufficiency of the stated condition being clear. The case where $f_1$ and $f_2$ generate the zero ideal is trivial. Thus assume that the non-zero polynomials $f_1$ and $f_2$ generate the same ideal of $K[X]$. Then there exist polynomials $u_1$ and $u_2$ such that $f_1 = u_1 f_2$ and $f_2 = u_2 f_1$; it follows that u_1 u_2 = 1, whence $\deg u_1 + \deg u_2 = 0$ and so $\deg u_2 = 0$. We have thus shown that $u_2$ is a non-zero element of $K$.

To prove (i), let $f$ be a monic polynomial in $a$ of least possible degree. Given $g$ in $a$, let $u$ and $v$ be the quotient and remainder of the Euclidean division of $g$ by $f$; then $v = g - u f$ belongs to $a$ and we have $\deg v < \deg f$; if $v$ were non-zero, there would be a non-zero element $A$ of $K$ such that $\lambda v$ is monic, and since $\lambda v \in a$, this would contradict the definition off. We thus have $a = (f)$; the uniqueness of the monic polynomial $f$ such that $a = (f)$ now follows from (ii).