MHB Why can we show that an other polynomial is irreducible?

[evinda]

Hello! :)
I am looking at the exercise:

Prove that $f(x)=10x^4-18x^3+4x^2+7x+16 \in \mathbb{Z}[x]$ is irreducible in $\mathbb{Q}[x]$.

According to my notes,a way to do this is the following:

We know that $\forall m>1 \exists $ ring homomorphism $\widetilde{ \phi }: \mathbb{Z}[x] \to \mathbb{Z}_m[X]$. We choose a prime $m$,that does not divide the coefficient of the highest degree of the terms.A logic choice is $m=3$.So, if $f(x)=g(x) \cdot h(x), g(x),h(x) \in \mathbb{Z}[x]$,then $\widetilde{ \phi }(f(x))=\widetilde{ \phi }(g(x)) \cdot \widetilde{ \phi }(h(x))$.

$\widetilde{ \phi }(f(x))=x^4+x^2+x+1= \text{ the product of two non-zero polynomial of } \mathbb{Z}_3[x]$

We have $2$ cases:

• $x^4+x^2+x+1=\text{ polynomial of degree } 1 \cdot \text{ polynomial of degree } 3$
• $x^4+x^2+x+1= (ax^2+bx+c) \cdot (a'x^2+b'x+c')$

First case:

Is means that there is a root in $\mathbb{Z}_3$:

$[a]x+, [a] \neq [0] \Rightarrow \text{ root }: [a]^{-1}$.-> Why is $[a]^{-1}$ the root and not $[-b][a]^{-1}$ ?So,we have to check if one of $0,1,2 \in \mathbb{Z}_3$ is a root and we conclude that none of these numbers are roots,so the first case is rejected.

Second case:

$$\left\{\begin{matrix}
x^4: & 1=aa' \\
x^3: & 0=ab'+ba' \\
x^2: & 1=ac'+a'c+bb' \\
x: & 1=bc'+cb' \\
\text{ constant: } & 1=cc'
\end{matrix}\right.$$

$a=a'=1 \text{ or } 2$ and $c=c'=1 \text{ or } 2 $

We get the relations $b+b'=0$ & $c(b+b')=1$ that can't be true.

So,the second case is also rejected.
-> But...why if we have shown that $x^4+x^2+x+1$ is irreducible in $\mathbb{Z}_3 $,have we proven that $f(x)=10x^4-18x^3+4x^2+7x+16 \in \mathbb{Z}[x]$ is irreducible in $\mathbb{Z}$ ,and so also in $\mathbb{Q}$?

 

[Opalg]

$[a]x+, [a] \neq [0] \Rightarrow \text{ root }: [a]^{-1}$.-> Why is $[a]^{-1}$ the root and not $[-b][a]^{-1}$ ?

You are right, it should be $[-b][a]^{-1}$ (but that does not affect the overall argument).

-> But...why if we have shown that $x^4+x^2+x+1$ is irreducible in $\mathbb{Z}_3 $,have we proven that $f(x)=10x^4-18x^3+4x^2+7x+16 \in \mathbb{Z}[x]$ is irreducible in $\mathbb{Z}$ ,and so also in $\mathbb{Q}$?

If there was a factorisation in $\mathbb{Z}[x]$ then by reducing all the coefficients mod 3 you would get a factorisation in $\mathbb{Z}_3[x]$. Since you have shown that there is no such factorisation it follows that the polynomial is irreducible over $\mathbb{Z}$, and hence (by Gauss's lemma) also over $\mathbb{Q}$.

 

[evinda]

You are right, it should be $[-b][a]^{-1}$ (but that does not affect the overall argument).If there was a factorisation in $\mathbb{Z}[x]$ then by reducing all the coefficients mod 3 you would get a factorisation in $\mathbb{Z}_3[x]$. Since you have shown that there is no such factorisation it follows that the polynomial is irreducible over $\mathbb{Z}$, and hence (by Gauss's lemma) also over $\mathbb{Q}$.

Ok,thanks! :)