Radical extension

**ynj** · September 5th 2009, 06:01 AM

Prove the splitting field of $x^3+x^2+1\in Z_{2}[x]$ is a radical extension...
when it comes to $Z_{2}$ ,I am a little bit confused. Since many theorem is based on $charF=0$ ..

**ThePerfectHacker** · September 5th 2009, 07:31 AM

Originally Posted by ynj

Prove the splitting field of $x^3+x^2+1\in Z_{2}[x]$ is a radical extension...
when it comes to $Z_{2}$ ,I am a little bit confused. Since many theorem is based on $charF=0$ ..

The theorem about determining if something is a radical extension (that is, the polynomial is solvable) is about charachteristic zero fields. However, the definition of radical extensions is more general then that.

Remember, $K/F$ is a radical extension (by definition) iff there exists $a_1,...,a_n\in K$ and $e_1,...,e_n\geq 1$ such that $F(a_1,...,a_n) = K$ , $a_j^{e_j} \in F(a_1,...,a_{j-1})$ .

So construct the splitting field of this and argue that you can write it in the above form.

**ynj** · September 5th 2009, 07:39 AM

Originally Posted by ThePerfectHacker

The theorem about determining if something is a radical extension (that is, the polynomial is solvable) is about charachteristic zero fields. However, the definition of radical extensions is more general then that.

Remember, $K/F$ is a radical extension (by definition) iff there exists $a_1,...,a_n\in K$ and $e_1,...,e_n\geq 1$ such that $F(a_1,...,a_n) = K$ , $a_j^{e_j} \in F(a_1,...,a_{j-1})$ .

So construct the splitting field of this and argue that you can write it in the above form.

Yeah,but I simply know nothing about the structure of the splitting field...

**ThePerfectHacker** · September 5th 2009, 10:50 AM

Originally Posted by ynj

Prove the splitting field of $x^3+x^2+1\in Z_{2}[x]$ is a radical extension...

Notice that $x^3+x^2+1$ is irreducible over $F = \mathbb{Z}_2[x]$ . Therefore, as you know, there exists an extension field $K$ which has $\alpha\in K$ that solves this polynomial. Therefore, $\alpha^3 = \alpha^2 + 1$ . Now $x^3 + x^2 + 1 = (x+\alpha)(x^2 + (\alpha+1)x+\alpha(\alpha+1))$ . You need to ask now whether $x^2+(\alpha+1)x+\alpha(\alpha+1)$ has a zero in $F(\alpha)$ . Sadly, it does not, this can be confirmed by checking $a\alpha^2+b\alpha + c$ where $a,b,c\in \{0,1\}$ . Thus, $F(\alpha)$ is not the splitting field over $x^3+x^2+1$ . However, we know there exists $L/F(\alpha)$ such that there is $\beta \in L$ which solves $x^2+(\alpha+1)x+\alpha(\alpha +1 ) \in F(\alpha)[x]$ . The extension field $F(\alpha,\beta)$ will therefore become the splitting field over $F$ . Now it remains to argue this satisfies the conditions of being a radical extension.

**NonCommAlg** · September 5th 2009, 03:42 PM

this is maybe an easier way: let $p(x)=x^3+x^2+1 \in \mathbb{F}_2[x].$ see that if $p(\alpha)=0,$ then $\alpha^7 = 1 \in \mathbb{F}_2.$ thus the splitting field of $p(x)$ is a radical extension of $\mathbb{F}_2.$

**ThePerfectHacker** · September 5th 2009, 04:09 PM

Originally Posted by NonCommAlg

this is maybe an easier way: let $p(x)=x^3+x^2+1 \in \mathbb{F}_2[x].$ see that if $p(\alpha)=0,$ then $\alpha^7 = 1 \in \mathbb{F}_2.$ thus the splitting field of $p(x)$ is a radical extension of $\mathbb{F}_2.$

But $F(\alpha)$ is not the splitting field.

You need to show that $F(\alpha,\beta)$ satisfies $\beta^{n_1} \in F(\alpha)$ and $\alpha^{n_2} \in F$ .
You have shown that $n_2=7$ works so it still remains to find $n_1$ .

**ynj** · September 5th 2009, 04:19 PM

Originally Posted by ThePerfectHacker

But $F(\alpha)$ is not the splitting field.

You need to show that $F(\alpha,\beta)$ satisfies $\beta^{n_1} \in F(\alpha)$ and $\alpha^{n_2} \in F$ .
You have shown that $n_2=7$ works so it still remains to find $n_1$ .

Yeah.. $F(\alpha,\beta)$ is the splitting field. But we know $\alpha^7\in F,\beta^7\in F\subset F(\alpha)$ . Is that right?

**NonCommAlg** · September 5th 2009, 04:22 PM

Originally Posted by ynj

Yeah.. $F(\alpha,\beta)$ is the splitting field. But we know $\alpha^7\in F,\beta^7\in F \subset F(\alpha)$ . Is that right?

correct!

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