# Thread: Factoring given a root

1. ## Factoring given a root

Suppose that $\beta$ is a zero of $f(x)=x^4+x+1$ in some field extension $E$ of $\mathbb{Z}_2$. Write $f(x)$ as a product of linear factors in $E[x]$.

Since $\beta$ is a zero, $x-\beta$ is a factor of $f(x)$. My plan was to simply divide $x-\beta$ into $f(x)$ to obtain a cubic and then find a root of the cubic, etc. until I obtained all the linear factors. Is there a reason that $x-\beta$ shouldn't divide $f(x)$? I don't see one but it is not dividing it when I try. Does the fact that $\beta$ lies in $E$ and not $\mathbb{Z}_2$ affect this?

2. ## Re: Factoring given a root

Originally Posted by AlexP
Suppose that $\beta$ is a zero of $f(x)=x^4+x+1$ in some field extension $E$ of $\mathbb{Z}_2$. Write $f(x)$ as a product of linear factors in $E[x]$.
$\beta$ satisfies $\beta^4+\beta+1=0$ in $E$ . But $(\beta^2)^4+\beta^2+1=(\beta^4+\beta+1)^2=0$ so, $\beta^2$ is another root of $f(x)$ in $E$ . Now, prove that $1+\beta$ and $1+\beta^2$ are also roots of $f(x)$ in $E$ . That is, $f(x)=(x-\beta)(x-\beta^2)(x-1-\beta)(x-1-\beta^2)$ .

3. ## Re: Factoring given a root

I got it. Since we're in characteristic 2 we have $f(a)=f(a+1)$ for arbitrary $a$ (I did work out the details).

Question about something I'm not clear on though... My book (Contemporary Abstract Algebra, Gallian, 5th ed.) says "A field $E$ is an extension field of a field $F$ if $F \subseteq E$ and the operations of $F$ are those of $E$ restricted to $F$." That wording bothers me... Is that saying that E inherits the operations of F (and thus has the same characteristic?)? That's not what it seems to be saying to me, but it's what seems to make sense.

4. ## Re: Factoring given a root

Originally Posted by AlexP
I got it. Since we're in characteristic 2 we have $f(a)=f(a+1)$ for arbitrary $a$ (I did work out the details).

Question about something I'm not clear on though... My book (Contemporary Abstract Algebra, Gallian, 5th ed.) says "A field $E$ is an extension field of a field $F$ if $F \subseteq E$ and the operations of $F$ are those of $E$ restricted to $F$." That wording bothers me... Is that saying that E inherits the operations of F (and thus has the same characteristic?)? That's not what it seems to be saying to me, but it's what seems to make sense.
More generally, $F$ is an extension of $E$ if there is an embedding $\sigma:E\hookrightarrow F$. In particular, since $\sigma$ is an injective group morphism we know that $\text{char}(F)=|1_F|=|\sigma(1_E)|=|1_E|=\text{cha r}(E)$.

5. ## Re: Factoring given a root

Got it. Thanks to both of you.