1. Factoring given a root

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

Since $\displaystyle \beta$ is a zero, $\displaystyle x-\beta$ is a factor of $\displaystyle f(x)$. My plan was to simply divide $\displaystyle x-\beta$ into $\displaystyle f(x)$ 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 $\displaystyle x-\beta$ shouldn't divide $\displaystyle f(x)$? I don't see one but it is not dividing it when I try. Does the fact that $\displaystyle \beta$ lies in $\displaystyle E$ and not $\displaystyle \mathbb{Z}_2$ affect this?

2. Re: Factoring given a root

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

3. Re: Factoring given a root

I got it. Since we're in characteristic 2 we have $\displaystyle f(a)=f(a+1)$ for arbitrary $\displaystyle a$ (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 $\displaystyle E$ is an extension field of a field $\displaystyle F$ if $\displaystyle F \subseteq E$ and the operations of $\displaystyle F$ are those of $\displaystyle E$ restricted to $\displaystyle F$." 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 $\displaystyle f(a)=f(a+1)$ for arbitrary $\displaystyle a$ (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 $\displaystyle E$ is an extension field of a field $\displaystyle F$ if $\displaystyle F \subseteq E$ and the operations of $\displaystyle F$ are those of $\displaystyle E$ restricted to $\displaystyle F$." 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, $\displaystyle F$ is an extension of $\displaystyle E$ if there is an embedding $\displaystyle \sigma:E\hookrightarrow F$. In particular, since $\displaystyle \sigma$ is an injective group morphism we know that $\displaystyle \text{char}(F)=|1_F|=|\sigma(1_E)|=|1_E|=\text{cha r}(E)$.

5. Re: Factoring given a root

Got it. Thanks to both of you.