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Math Help - Radical extension

  1. #1
    ynj
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    Radical extension

    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..
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    Quote Originally Posted by ynj View Post
    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.
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  3. #3
    ynj
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    Quote Originally Posted by ThePerfectHacker View Post
    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...
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    Quote Originally Posted by ynj View Post
    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.
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    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.
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    Quote Originally Posted by NonCommAlg View Post
    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.
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  7. #7
    ynj
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    Quote Originally Posted by ThePerfectHacker View Post
    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?
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    Quote Originally Posted by ynj View Post
    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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