Question

Let $\phi :\mathbf{F}\left[x\right]\to \mathbf{F}\left[x\right]$ be an isomorphic such that $\phi \left(\mathbf{a}\right)=\mathbf{a}$ for every $a\in \mathbf{F}$. Prove that $\mathbf{f}\left(\mathbf{x}\right)$ is irreducible in $\mathbf{F}\left[\mathbf{x}\right]$ if and only if $\phi \left(\mathbf{f}\left(\mathbf{x}\right)\right)$ is.

Short answer

It is proved that $f\left(x\right)$ is irreducible in $\mathrm{F}\left[\mathrm{x}\right]$ if and only if $\phi \left(f\left(x\right)\right)$ is irreducible.

Step-by-step solution

Statement of Exercise 23

Exercise 23states that $f\left(x\right),g\left(x\right),h\left(x\right)\in F\left[x\right]$and $r\in F$.

1. When $f\left(x\right)=g\left(x\right)+h\left(x\right)$in $F\left(x\right)$, demonstrate that $f\left(r\right)=g\left(r\right)+h\left(r\right)$ in F.
2. When $f\left(x\right)=g\left(x\right)h\left(x\right)$in $F\left(x\right)$, demonstrate that $f\left(r\right)=g\left(r\right)h\left(r\right)$ in F.

Show that fx  is irreducible in  Fx if and only if φfx  is irreducible

We aim to prove that $f\left(x\right)$ is irreducible if and only if $\phi \left(f\left(x\right)\right)$ is irreducible.

The main goal of the proof is to establish the lemma as follows:

$\mathrm{deg}\left(\phi \left(f\left(x\right)\right)\right)=\mathrm{deg}\left(f\left(x\right)\right)$

(1)

Suppose that with the sake of contradiction, $\mathrm{deg}\left(\phi \left(x\right)\right)=0$.

This entails that for arbitrary $f\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0\in F\left[x\right]$:

$\begin{array}{c}\phi \left(f\left(x\right)\right)=\phi \left(a_n{x}^n+\cdots +a_{1}x+a_0\right)\\ =\sum _{i=0}^n\phi \left(a_i{x}^i\right)\\ =\sum _{i=0}^n\phi \left(a_i\right)\phi \left(x^i\right)\\ =\sum _{i=0}^n{a}_i\phi \left(x^i\right)\\ =\sum _{i=0}^n{a}_i\phi {\left(x\right)}^i\in F\end{array}$

Thus, $\phi$ cannot be isomorphism since all the elements in the image of $\phi$ are of degree 0. This is a contradiction.

As a result, $\mathrm{deg}\left(\phi \left(x\right)\right)\ne 0$.

(2)

Suppose that $\mathrm{deg}\left(\phi \left(x\right)\right)>1$. Since $\phi$ would be an isomorphism, there should be an element $p=\underset{i-0}{\overset{m}{\sum b_i}}{x}^i\in F\left[x\right]$ in which $\phi \left(p\right)=x$, however, this is impossible because $\mathrm{deg}\left(\phi \left(x\right)\right)>1$ and $\phi \left(a\right)=a$ for every $a\in F$ indicates that $\mathrm{deg}\left(\phi \left(p\right)\right)>m$ and clearly $m>1$since otherwise, $\phi \left(p\right)$ is a constant polynomial. This is a contradiction.

As a result, $\mathrm{deg}\left(\phi \left(x\right)\right)=1$.

It is now clear that $\mathrm{deg}\left(\phi \left(f\left(x\right)\right)\right)=\mathrm{deg}\left(f\left(x\right)\right)$.

Show that  φfx is reducible 

Theorem 4.11considers $F$as a field. A non-zero polynomial $f\left(x\right)$ would be reducible in $F\left[x\right]$ such that if $f\left(x\right)$could be expressed as the product of two polynomials of lower degree.

Suppose that $f\left(x\right)$ is irreducible. For the purpose of contradiction, consider that $\phi \left(f\left(x\right)\right)$ is irreducible.

However, according to theorem 4.11, it is known that there is each$p,q\in F\left[x\right]$ with a lower degree than $\phi \left(f\right)$such that,

$\begin{array}{c}\phi \left(f\right)=pq\\ =\phi \left(p_1\right)\phi \left(q_1\right)\\ =\phi \left(p_1{q}_1\right)\end{array}$

Where $p_1$ and $q_1$ are in which $\phi \left(p_1\right)=p$and$\phi \left(q_1\right)=q$ . (It is known that certain elements exist because of surjectivity of $\phi$).

Use injectivity of $\phi$, it may deduce that $f=p_1{q}_1$. Clearly, the degrees of $p_1$ and $q_1$ are lower than the degree of $f$ because isomorphism maintains degrees.

This demonstrates the intended implication.

Consider that $\phi \left(f\left(x\right)\right)$ is irreducible. For the purpose of contradiction, consider that $f\left(x\right)$ is reducible.

However, according to theorem 4.11, it is known that there is some $p,q\in F\left[x\right]$ contain lower degree than $f$ such that $f\left(x\right)=p\left(x\right)q\left(x\right)$.

It implies that:

$\begin{array}{c}\phi \left(f\left(x\right)\right)=\phi \left(p\left(x\right)q\left(x\right)\right)\\ =\phi \left(p\left(x\right)\right)\phi \left(q\left(x\right)\right)\end{array}$

It implies that $\phi \left(x\right)$ could be expressed as a product of two polynomials of lower degree because our isomorphism preserves degrees/

According to the reverse of theorem 4.11, it implies that $\phi \left(f\left(x\right)\right)$ is reducible.

This demonstrates that intended implication.

Hence, it is proved that $f\left(x\right)$ is irreducible in $\mathrm{F}\left[\mathrm{x}\right]$ if and only if $\phi \left(f\left(x\right)\right)$ is irreducible.