D+F 7.3.10

Decide which of the following are ideals in $\mathbb{Z}[x]$

(a) The set of all polynomials whose constant term is a multiple of 3.

$\textit{Answer}$ : This set, call it $J$, is a subring. $0$ is a multiple of 3 i.e. $0 \equiv 0 (mod3)$. So, $0\in{J}$. Also, polynomial subtraction leaves the difference of 2 polynomial's constant terms in the form $3n-3m=3(n-m) \equiv 0 (mod3) \quad n,m,\in{\mathbb{Z}}$.
Multiplication of polynomials is done by continued use of the distributive law. The only purely constant term in the product of 2 polynomials in $\mathbb{Z}[x]$ will be the last term. Let $f(x)$ be a polynomial with constant term $3k, k\in{\mathbb{Z}}$. Let $g(x)\in{\mathbb{Z}[x]}$. Then, the product polynomial $f(x)g(x)=g(x)f(x)$ will have constant term $3\cdot k \cdot n \equiv 0 (mod 3)$. Thus, the set of polynomials with a constant term being a multiple of 3 is closed under left and right multiplication in the ring $\mathbb{Z}[x]$.
$\therefore$ $J$ is an ideal of the ring $\mathbb{Z}[x]$.

(b) The set of all polynomials whose coefficient of $x^2$ is a multiple of 3.

$\textit{Answer}$ : Consider the polynomial $3x^2+x$ which is in the given set. Then, $(3x^2+x)\cdot (3x^2+x)=9x^4+6x^3+x^2$. Thus, this set is not even closed under multiplication by its own elements. So, it is not an ideal of $\mathbb{Z}[x]$.

(c) The set of all polynomials whose constant term, coefficient of $x$ and coefficient of $x^2$ are zero.

$\textit{Answer}$ : This set, J, is a subring since $0\in{J}$ and subtraction does not change degrees of terms.
Multiplication of polynomials only adds degree to terms. Polynomials in $J$ either have degree 0 (the zero element) or degree greater than 2. So, the product of an arbitrary polynomial with one in $J$ must have either degree 0 or have degree greater than 2. Thus, this set is closed under right and left multiplication by polynomials in $\mathbb{Z}[x]$.
$\therefore$ $J$ is an ideal of $\mathbb{Z}[x]$.

(d) $\mathbb{Z}[x^2]$ i.e. the polynomials in which only even powers of $x$ appear.

$\textit{Answer}$ : This set, $J$, is a subring since $0\in{J}$ and subtraction of polynomials does not change the degree of any individual terms. However, $J$ is not an ideal. Consider $x^2\in{J}$. $x^2 \cdot x = x^3 \not \in{J}$.
$\therefore $ $J$ is not an ideal of $\mathbb{Z}[x]$.

(e) the set of polynomials whose coefficients sum to zero.

$\textit{Answer}$ : $p(x)=0\in{J}$. If $f(x),g(x) \in{J}$ then $f(1)-g(1)=(f-g)(1)=0-0=0$. Now let $f(x)\in{J}$ and $h(x)\in{\mathbb{Z}[x]}$. Then, $f(1)h(1)=0 \cdot c = (fh)(1)= 0$ Thus, $fh \in{J}$. Polynomial multiplication is commutative, so $hf\in{J}$. So, $J$ contains the additive identity, is closed under subtraction and is closed under left and right multiplication by arbitrary elements in $\mathbb{Z}[x]$.
$\therefore \quad J$ is an ideal of $\mathbb{Z}[x]$.

(f) the set of polynomials $p(x)$ s.t. $p'(0)=0$, where $p'(x)$ is the usual first derivative of $p(x)$ with respect to $x$.

$\textit{Answer}$ : Let $f(x)=x^2+1 \in{J}$ and $g(x)=x^2+x\in{\mathbb{Z}[x]}$. Then, $$(fg)'(0)=f(0)g'(0)+g(0)f'(0)=1 \cdot 1 + 0 \cdot 0 = 1$$ So, $fg \not \in{J}$.
$\therefore \quad J$ is not an ideal of $\mathbb{Z}[x]$.