1. ## polynomial rings

Let $f(x)$ be a polynomial in $\mathbb{Z}[x]$. Prove that if $f(x)$ has a root $\alpha \in \mathbb{Z}$, then $f(x)$ has a linear factor in $\mathbb{Z}[x]$.

Suppose $f(x)$ has a root $\alpha \in \mathbb{Z}$, then $f(x)$ is reducible, and then $f(x) = a(x)(x - \alpha)$ for some nonconstant polynomial $a(x) \in \mathbb{Z}[x]$. So $x - \alpha$ is a linear factor in $\mathbb{Z}[x]$.

Is this correct? Did I miss anything important in this proof?

2. Originally Posted by dori1123
Let $f(x)$ be a polynomial in $\mathbb{Z}[x]$. Prove that if $f(x)$ has a root $\alpha \in \mathbb{Z}$, then $f(x)$ has a linear factor in $\mathbb{Z}[x]$.

Suppose $f(x)$ has a root $\alpha \in \mathbb{Z}$, then $f(x)$ is reducible, and then $f(x) = a(x)(x - \alpha)$ for some nonconstant polynomial $a(x) \in \mathbb{Z}[x]$. So $x - \alpha$ is a linear factor in $\mathbb{Z}[x]$.

Is this correct? Did I miss anything important in this proof?
there is no assumption on the degree of $f$ so you cannot say that $a(x)$ is non constant..

well, this is a direct consequence of the division algorithm. since you think that $x-\alpha$ is the linear factor, then try writing $f(x)=g(x)(x-\alpha) + R$ where $R$ is a non zero constant or $R=0$.

3. Originally Posted by kalagota
there is no assumption on the degree of $f$ so you cannot say that $a(x)$ is non constant..

well, this is a direct consequence of the division algorithm. since you think that $x-\alpha$ is the linear factor, then try writing $f(x)=g(x)(x-\alpha) + R$ where $R$ is a non zero constant or $R=0$.
But $\mathbb{Z}[x]$ is not a field, there's no Euclidean Algorithm...

4. Originally Posted by dori1123
But $\mathbb{Z}[x]$ is not a field, there's no Euclidean Algorithm...
The division algorithm does work for non-fields too, $\mathbb{Z}[x]$ is an example.

5. another example is $\mathbb{Z}$ itself..

6. Suppose $f(x) \in \mathbb{Z}[x]$ has a root $\alpha \in \mathbb{Z}$. By the division algorithm, $f(x) = q(x)(x-\alpha)+r$ for some polynomial $q(x) \in \mathbb{Z}[x]$ and for some $r \in \mathbb{Z}$. Since $f(\alpha)=0$, we have $r=0$. So $f(x)=q(x)(x-\alpha)$ and so $x-\alpha$ is a linear factor of $f(x) \in \mathbb{Z}[x]$. Is this correct?

Also, for the converse to be true, if $f(x) \in \mathbb{Z}[x]$ has a linear factor, then the linear factor must be a monic polynomial to guarantee that $f(x)$ has a root in $\mathbb{Z}$. Right? That's the only condition I can think of.

7. Originally Posted by dori1123

Also, for the converse to be true, if $f(x) \in \mathbb{Z}[x]$ has a linear factor, then the linear factor must be a monic polynomial to guarantee that $f(x)$ has a root in $\mathbb{Z}$. Right? That's the only condition I can think of.
If $f(x) = (x-a)g(x)$ then $f(a) = 0$, so $a$ is a zero.

8. Originally Posted by ThePerfectHacker
If $f(x) = (x-a)g(x)$ then $f(a) = 0$, so $a$ is a zero.
I understand that the converse is true if $f(a)=0$ but why is $a$ zero?

9. Originally Posted by dori1123
I understand that the converse is true if $f(a)=0$ but why is $a$ zero?
No, $a$ is not zero, $a$ is a zero of $f$. This means $f(a) = 0$.

10. Originally Posted by ThePerfectHacker
No, $a$ is not zero, $a$ is a zero of $f$. This means $f(a) = 0$.
Oh, you mean a root. Sorry, I got used to say roots.