Question

Let $D:R\left[x\right]\to R\left[x\right]$be the derivative map defined by $D(a_0+a_{1}x+a_2{x}^2+....+a_n{x}^n)=a_1+2a_{2}x+3a_3{x}^2+...+n{a}_n{x}^{n-1}$.

Is D a homomorphism of rings? An isomorphism?

Short answer

No, D is not a homomorphism because $D\left(x^2\right)\ne D{\left(x\right)}^2$.

Step-by-step solution

Given part 

It is given that $\phi :R\left[x\right]\to R\left[x\right]$is the derivative map defined by,

$D(a_0+a_{1}x+a_2{x}^2+....+a_n{x}^n)=a_1+2a_{2}x+3a_3{x}^2+...+n{a}_n{x}^{n-1}$

Proof part

D will be a homomorphism if $D\left(x^2\right)=D{\left(x\right)}^2$.

D is not a homomorphism because D does not hold for the product as follows:

$D\left(x^2\right)=2x$

And,

$D\left(x^2\right)=1^2=1$

It is known that the derivative of all the constant polynomial is 0. Hence, it is not injective. Also, it cannot be an isomorphism.