Question

a) Show that the map $\varphi :F\left[x\right]\to F\left[x\right]$given by $\varphi \left(f\left(x\right)\right)=f\left(x+1_F\right)$is an isomorphism such that $\varphi \left(a\right)=a$for every $a\in F$.

b) Use Exercise 31 to show that $f\left(x\right)$is irreducible in $F\left(x\right)$if and only if $f\left(x+1_P\right)$is.

Short answer

It is proved that $f\left(x\right)$is irreducible if and only if $f\left(x+1_P\right)$is irreducible

Step-by-step solution

Statement of Exercise 31

Exercise 31states that if $\varphi :F\left[x\right]\to F\left[x\right]$is an isomorphism in which $\varphi \left(a\right)=-a$for all$a\in F$ , then $f\left(x\right)$will be irreducible in $F\left[x\right]$for role="math" localid="1649886956712" $\varphi \left(f\left(x\right)\right)$.

Show that f(x) is irreducible in F[x]if and only f(x+1P)if is irreducible

The map $\varphi$meets the condition in exercise 31; namely, it will be an isomorphism, which is the identity of constant polynomials. As a result, the conclusion holds, and $f\left(x\right)$will be irreducible if $f\left(x+1_P\right)$is irreducible.

Hence, it is proved that$f\left(x\right)$ is irreducible if and only if $f\left(x+1_P\right)$is irreducible.