Question

Let $\mathit{\phi }\mathbf{:}\mathit{C}\mathbf{\to }\mathit{C}$be an isomorphism of rings such that $\mathit{\phi }\left(a\right)\mathbf{=}\mathit{a}$for each $\mathit{a}\mathbf{\in }\mathit{\phi }$. Suppose $\mathit{r}\mathbf{\in }\mathit{C}$is a root of $\mathit{f}\left(x\right)\mathbf{\in }\mathbf{\mathbb{Q}}\left[x\right]$. Prove that $\mathit{\phi }\left(r\right)$is also a root of f (x).

Short answer

$\phi \left(r\right)$is a root of f (x). Hence, it is proved.

Step-by-step solution

Determine φ(r) is also a root of f (x)

Let’s consider that $f=a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_1{x}_1+a_0$,

By using the definition of the polynomial,

$$\begin{gathered} f\left(\phi \left(r\right)\right)=\sum _{i=0}^n{a}_i{\left(\phi \left(r\right)\right)}^i \\ \stackrel{\left(1\right)}{=}\sum _{i=0}^n{a}_i\phi \left(r^i\right) \\ \stackrel{\left(2\right)}{=}\sum _{i=0}^n\phi \left(a_i\right)\phi \left(r^i\right) \\ \stackrel{\left(1\right)}{=}\sum _{i=0}^n\phi \left(a_i{r}^i\right) \\ \stackrel{\left(1\right)}{=}\phi \left(\sum _{i=0}^n{a}_i{r}^i\right) \\ \stackrel{\left(3\right)}{=}\phi \left(0\right) \\ =0 \end{gathered}$$

Here,

$\phi$is a homomorphism.

$\phi \left(a\right)=a$for all$a\in \phi$

r is root of f

Hence, $\phi \left(r\right)$is root of f (x).