Question

Question 6: The function$\mathit{\phi }\mathbf{:}\mathbf{ℝ}\left[x\right]\mathbf{\to }\mathbf{ℝ}$ given by $\mathit{\phi }\left(f\left(x\right)\right)\mathbf{=}\mathit{f}\left(2\right)$is a homomorphism of rings by Exercise 24 of Section4.4(with$\mathit{a}\mathbf{=}\mathbf{2}$ ). Find the kernel of$\mathit{\phi }$ . [Hint: Theorem 4.16.]

Short answer

Kernel of$Ker\left(\phi \right)=\left\{\left(x-2\right)g\left(x:g\in \mathrm{ℝ}\left[x\right]\right)\right\}$

Step-by-step solution

Applying definition of kernels 

Consider $\phi :\mathrm{ℝ}\left[x\right]\to \mathrm{ℝ}$such that $\phi \left(f\left(x\right)\right)=f\left(2\right)$

Kernel $\phi$is the set of all $f\left(x\right)\in \mathrm{ℝ}\left[x\right]$, such that:

$$\begin{gathered} \phi \left(f\left(x\right)\right)=f\left(2\right) \\ =0 \end{gathered}$$

 Step 2: Using theorem 4.16

Here,$f\left(2\right)=0$means all those polynomials whose root is 2.

This implies $x-2$is a factor of $f\left(x\right)$.

Hence, the kernel of $\phi$is the set of polynomials that are multiple of $x-2$.

Conclusion

Hence, kernel of $\phi$is ideal generated by $x-2$.

Thus,$Ker\left(\phi \right)=\left\{\left(x-2\right)g\left(x:g\in \mathrm{ℝ}\left[x\right]\right)\right\}$