Show that if $f(x) \in \mathbb Z \left[x\right]$ then $f(x)$ factors into a product of two polynomials $\mathbb Q \left[x\right]$ is equivalent to

Question

I want to show that, if $f(x) \in \mathbb Z \left[x\right]$ then $f(x)$ factors into a product of two polynomials of lower degrees $r$ and $s$ in $\mathbb Q \left[x\right]$ if and only if it has such a factorization with polynomials of the same degrees $r$ y $s$ en $\mathbb Z \left[x\right]$

For this I think that i can use two things:

1. The product of two primitive polynomials is primitive
2. The content of $f(x),g(x)\in \mathbb Z \left[x\right]$ satisfies cont$(f(x)g(x))=$ cont$(f(x))$ cont$(g(x))$

My attemp is

For the implication, we can factor out the content to write: $$ f(x)=c_{f}f^{\ast}(x), G(x)=c_{g}g^{\ast}(x), H(x)=c_{h}h^{\ast}(x)$$

where $c_{f}$ is an element of $\mathbb Q $, and $f^{\ast}(x)$ is a primitive polynomial in $\mathbb Z \left[x\right]$

Then we have:

$$c_{f}f^{\ast}(x)= (c_{g}c_{h})g^{\ast}(x)h^{\ast}(x)$$

By the property mentioned, the product $g^{\ast}(x)h^{\ast}(x)$ is primitive. Thus gives $c_{f} = c_{g}c_{h}$ or $c_{f} = c_{g}c_{h}$. In either case, $c_{g}c_{h}$.is an integer, so we let $g(x) = (c_{g}c_{h})g^{\ast}(x)$ and $h(x) =h^{\ast}(x)$ and that is all.

So for the other direction, really i need use that cont$(f(x)g(x))=$ cont$(f(x))$ cont$(g(x))$? I am not clear about the procedure and do not want to use ideals in the demo process, any suggestions? Any help is welcome!

Answer 1

Conversely, assume that $f(x)$ has such a factorization with polynomials of the same degrees $r$ and $s$ in $\mathbb Z \left[x\right]$, say $f(x)=g(x)h(x)$ with $\deg(g)=r$ and $\deg(h)=s$. Then, we can write $g(x)=c_g \tilde{g}(x)$ and $h(x)=c_h \tilde{h}(x)$, where $\tilde{g}(x)$ and $\tilde{h}(x)$ are primitive polynomials in $\mathbb Z \left[x\right]$.

Now, let $c$ be the content of $f(x)$, so that $f(x)=c \tilde{f}(x)$ with $\tilde{f}(x)$ primitive in $\mathbb Z \left[x\right]$. We have $c=c_g c_h$ (by property 2) and $\tilde{f}(x)=\tilde{g}(x) \tilde{h}(x)$ (since $\tilde{g}(x)$ and $\tilde{h}(x)$ are primitive and have the same degrees as $g(x)$ and $h(x)$ respectively).

Since $c_g$ and $c_h$ are integers, their product is also an integer, so $c$ is an integer. Moreover, since $\tilde{g}(x)$ and $\tilde{h}(x)$ are primitive, their product $\tilde{f}(x)=\tilde{g}(x) \tilde{h}(x)$ is also primitive (by property 1). Therefore, we have expressed $f(x)$ as the product of two primitive polynomials in $\mathbb Z \left[x\right]$, namely $c_g \tilde{g}(x)$ and $c_h \tilde{h}(x)$, both of degree $r$ and $s$ respectively, which implies that $f(x)$ factors into two polynomials of lower degrees $r$ and $s$