polynomial rings

**dori1123** · February 27th 2009, 08:22 PM

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?

**kalagota** · February 28th 2009, 12:01 AM

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 $R=0$ .

**dori1123** · February 28th 2009, 12:35 PM

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...

**ThePerfectHacker** · February 28th 2009, 03:29 PM

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.

**kalagota** · February 28th 2009, 05:04 PM

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

**dori1123** · February 28th 2009, 05:31 PM

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.

**ThePerfectHacker** · February 28th 2009, 07:12 PM

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.

**dori1123** · February 28th 2009, 08:26 PM

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?

**ThePerfectHacker** · February 28th 2009, 08:44 PM

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 $f(a) = 0$ .

**dori1123** · February 28th 2009, 09:42 PM

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.

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