Question

Show by example that this statement is false:If $f\left(x\right)\in \mathrm{\mathbb{Z}}\left[x\right]$and there is no prime $p$ satisfying the hypotheses of Theorem 4.24, then $\mathit{f}\left(x\right)$is reducible in $\mathrm{\mathbb{Q}}\left[x\right]$.

Short answer

It is proved that $f\left(x\right)$is irreducible in$\mathrm{\mathbb{Q}}\left[x\right]$ .

Step-by-step solution

 Step 1: To show there exists no prime   satisfying hypotheses of the theorem

Let us see the definitions of Reducible polynomial and Irreducible polynomial.

A nonconstant polynomial $f\left(x\right)\in F\left[x\right]$, where F is a field, is irreducible if its only divisors are its associates and the nonzero constant polynomials(units).

A nonconstant polynomial that is not irreducible is called a reducible polynomial.

Consider a polynomial $f\left(x\right)=x^2+1$.

Comparing this with the polynomial $f\left(x\right)=a_0+a_{1}x+a_2{x}^2$, we get:

$a_0=1$and $a_2=1$( where $a_1=0$).

From Theorem 4.24, it follows that:

There is no P that divides $a_2=1$and $p^2$ that divides $a_0=1$.

Thus, no such a prime P that satisfies the hypotheses of Theorem 4.24.

To prove fx is irreducible in  ℚx

Now, the polynomial $f\left(x\right)$ is a quadratic polynomial without a root in $\mathrm{\mathbb{Q}}$since we have:

$$\begin{gathered} f\left(x\right)=x^2+1 \\ \Rightarrow x^2=-1 \\ \Rightarrow x=\sqrt{-1} \end{gathered}$$

$\therefore x=\sqrt{-1}\notin \mathrm{\mathbb{Q}}$

Hence, $f\left(x\right)$ is an irreducible polynomial.

Thus, $f\left(x\right)=x^2+1$ is an example of a polynomial that shows that is $f\left(x\right)$irreducible in $\mathrm{\mathbb{Q}}\left[x\right]$.

Therefore, the statement given in the question is false.