Question

If I is an ideal in R, prove that $\mathit{I}\left[x\right]$ (polynomials with coefficients in I) is an ideal in the polynomial ring$\mathit{R}\left[x\right]$ .

Short answer

It is proved that $I\left[x\right]$ is an ideal in$R\left[x\right]$.

Step-by-step solution

Theorem statement

Let R be a ring and Iand S are non-empty subsets of the ring, then I is said to be ideal if and only if it has the following two properties:

1. If $a,b\in I$, then$a-b\in I$.

2. If $r\in R$and$a\in I$ , then$ra\in I$ and $ar\in I$.

Satisfied property (i) of theorem

It is given that $0_R\in I$ because I is an ideal, and for $0_R\in I\left[x\right]$so $I\left[x\right]$is non-empty.

Now, assume that $f\left(x\right),g\left(x\right)\in I\left[x\right]$ then, there exists non-negative integer m and $a_0,a_1,.......a_n$then, we have:

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

Similarly, there exists non-negative integer n and $b_0,b_1.....b_n$, then, we have:

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

Now, assume that $m\ge n$. Then, $b_{n+1}=b_{n+2}=.....=b_m=0_R$, now find the difference between $f\left(x\right)$and $g\left(x\right)$.

$$\begin{gathered} f\left(x\right)-g\left(x\right)=\left(a_0+a_{1}x+......a_m{x}^m\right)-\left(b_0+b_{1}x+......b_n{x}^n\right) \\ =\left(a_0-b_0\right)-\left(a_1-b_1\right)x+\left(a_m-b_m\right)x^m \end{gathered}$$

As, $a_i-b_i\in I$,$i=0,1,....,m$ because I is an ideal, it can be concluded that $f\left(x\right)-g\left(x\right)\in I\left[x\right]$.

Similarly, for $m<n,f\left(x\right)-g\left(x\right)\in I\left[x\right]$.

Hence, property (i) of the theorem is stated in step 1.

Satisfied property (ii) of theorem

Let $h\left(x\right)\in R\left[x\right]$then,

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

For any non-negative integer k and $c_0,c_1+......+c_k\in R$we have,

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

Where$d_I=a_0{c}_I+a_1{c}_{I-1}+........+a_{I-1}{c}_1+a_I{c}_0$ for all $i=0,1,......,m$and $j=0,1,......k$we have $a_i\in I$and $c_j\in R$, so$a_i{c}_i\in I$ because I is closed under addition, then it is clear that al d’ is are from I for all $I=0,1,.......,m+k$, which implies that$f\left(x\right)h\left(x\right)\in I\left[x\right]$ .

Similarly, we have $h\left(x\right)f\left(x\right)\in I\left[x\right]$

Thus, property (ii) is also satisfied for the theorem stated in step 1.

Hence, proved that $I\left[x\right]$is an ideal in $R\left[x\right]$.