Question

Show that the map $\theta :R\left[x\right]\to R$that sends each polynomial$f\left(x\right)$ to its constant term is a surjective homomorphism.

Short answer

It can be proved that$\theta$ is a surjective homomorphism.

Step-by-step solution

Definition of homomorphism

In order to prove$\theta$ is a homomorphism, we need to prove that

$$\begin{gathered} \theta \left(a+b\right)=\theta \left(a\right)+\theta \left(b\right) \\ \theta \left(a·b\right)=\theta \left(a\right)·\theta \left(b\right) \end{gathered}$$

Proving θ is homomorphism

Let $f\left(x\right),g\left(x\right)\in \mathrm{ℝ}\left[x\right]$.

Let the constant term of $f\left(x\right),g\left(x\right)$be, respectively, $a,b$

$\begin{array}{rcl}\theta \left(f\left(x\right)+g\left(x\right)\right)& =& a+b\\ & =& \theta \left(f\left(x\right)\right)+\theta \left(g\left(x\right)\right)\end{array}$

$\begin{array}{rcl}\theta \left(f\left(x\right)·g\left(x\right)\right)& =& a·b\\ & =& \theta \left(f\left(x\right)\right)·\theta \left(g\left(x\right)\right)\end{array}$

Hence,$\theta$ is a homomorphism.

Proving θ is surjective

It is surjective since every $c\in \mathrm{ℝ}$.

There is $c\in \mathrm{ℝ}\left[x\right]$and$\theta \left(c\right)=c$
.

So,$\theta$ is surjective.

Conclusion

Hence,$\theta$ is a surjective homomorphism.