Exercise A0.V.1.06

Proposition

Let $I$ be an ideal of $R[x]$ and let $A$ be the subset of $R$ containing $0$ and the leading coefficients of all polynomials in $I.$ Then $A$ is an ideal of $R.$

Proof

Take arbitrary $a, b \in A.$

If $a,b$ and $a - b$ are all non-zero, we have polynomials $a', b' \in I$ with the corresponding leading coefficients. Since $I$ is an ideal, we can multiply $a'$ or $b'$ by $x$ until their degrees match, and this gives us polynomials $\hat{a}, \hat{b}$ whose degrees are equal and whose leading coefficients are $a$ and $b$ respectively. The leading coefficient of $\hat{a} - \hat{b} \in I$ is $a - b.$

If $a - b = 0$ then this is already contained in $A.$ If $b = 0,$ then $a - b = a \in A.$ If $a = 0,$ then $a - b = -b$ which is the leading coefficient of $-b'.$

Thus $A$ is a subgroup of the additive group of $R.$

Take any $r \in R.$ If $ra = 0,$ then $ra \in A.$ Otherwise, the leading coefficient of $ra'$ is $ra.$

Thus $A$ satisfies the absorption requirement. Hence, $A$ is an ideal.

$\blacksquare$