Question

Show that the set$I$ of all polynomials with even constant terms is an ideal in$\mathrm{\mathbb{Z}}\left[x\right]$ .

Short answer

It is proved$I$ is an ideal.

Step-by-step solution

Determine Theorem 6.1

Consider the zero polynomial$z\left(x\right)=0$ ; it is in$I$ . Thus,$I$ is non-empty.

Now, suppose $f\left(x\right),g\left(x\right)\in I$, then

$f\left(x\right)=2a_0+a_{1}x+...a_m{x}^m$

Here, some non-negative integers $m$and $a_0,a_1,.....a_m\in \mathrm{\mathbb{Z}}$.

$g\left(x\right)=2b_0+b_{1}x+......+b_n{x}^n$

Some non-negative integer$n$ and$b_0,b_1,.....b_n\in \mathrm{\mathbb{Z}}$ without loss of generality.

Consider that $m\ge n$, then

$b_{n+1}=b_{n+2}=....=b_m=0$, and then I gives

$f\left(x\right)-g\left(x\right)=\left(2a_0-2b_0\right)+\left(a_1-b_1\right)x+\left(a_2-b_2\right)x^2+....+\left(a_m-b_m\right)x^m\in I$.

As $2a_0-2b_0=2\left(a_0-b_0\right)$, it is an even integer and constant term of $f\left(x\right)-g\left(x\right)$

Therefore, condition (i) of Theorem 6.1 is satisfied.

Determine set I is ideal in  ℤx

Consider $h\left(x\right)\in \mathrm{\mathbb{Z}}\left[x\right]$, then there exits negative integer $K$and $c_0,c_1,........c_k\in \mathrm{\mathbb{Z}}$, so that

$h\left(x\right)=c_0+c_{1}x+....+c_k{x}^k$

Now,

$f\left(x\right)h\left(x\right)=d_0+d_{1}x+......+d_{m+k}{x}^{m+k}$

Here,

$d_0=\left(2a_0\right)c_0=2a_0{c}_0$

That is an even integer and a constant term of $f\left(x\right)h\left(x\right)$. Thus,$f\left(x\right)h\left(x\right)\in I\mathrm{\mathbb{Z}}\left[x\right]$ .

is a commutative ring.

$h\left(x\right)f\left(x\right)-f\left(x\right)h\left(x\right)\in I$; thus, condition (ii) of Theorem 6.1 is satisfied.

Therefore, $I$ is an ideal.