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  1. MHB Master
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    #1
    Hey!!

    Let $f = x^4−2x^2−1 \in \mathbb{Q}[x]$.

    We have that $f(x+1)=(x+1)^4-2(x+1)^2-1=x^4+4x^3+6x^2+4x+1-2(x^2+2x+1)-1=x^4+4x^3+4x^2-2$

    We have that $p=2$ divides all the coefficients $4,4,-2$ and $p^2=4$ does not divide the constant term $-2$.

    So, the polynomial $f(x+1)$ is Eisenstein. This means that $f(x+1)$ is irreducible, or not?

    Knowing that, can we conclude that $f(x)$ is irreducible as follows:

    Suppose that $f(x)$ is reducible, then there are non-constant polynomials $g(x),h(x)\in \mathbb{Q}[x]$ such that $f(x)=g(x)h(x)$. Then $f(x+1)=g(x+1)h(x+1)$. Since the polynomials are non-constant that would mean that $f(x+1)$ is reducible, a contradiction.
    So, $f(x)$ is irreducible in $\mathbb{Q}[x]$.

    Is this correct?

    Let $\rho\in \mathbb{C}$ be a root of $f$. I want to find a basis and the degree of the extension $\mathbb{Q}[\sqrt{2},\rho]$.

    We have that $\text{Irr}(\rho, \mathbb{Q})=f$ and $\text{Irr}(\sqrt{2},\mathbb{Q}[\rho])=x^2-2$, or not?

    Therefore, the degree of the extension $\mathbb{Q}[\sqrt{2},\rho]$ is $\deg f \cdot \deg (x^2-2)=4\cdot 2=8$.

    Is this correct?

    How could we find a basis of the extension?

  2. MHB Journeyman
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    #2
    Quote Originally Posted by mathmari View Post
    Hey!!

    Let $f = x^4−2x^2−1 \in \mathbb{Q}[x]$.

    We have that $f(x+1)=(x+1)^4-2(x+1)^2-1=x^4+4x^3+6x^2+4x+1-2(x^2+2x+1)-1=x^4+4x^3+4x^2-2$

    We have that $p=2$ divides all the coefficients $4,4,-2$ and $p^2=4$ does not divide the constant term $-2$.

    So, the polynomial $f(x+1)$ is Eisenstein. This means that $f(x+1)$ is irreducible, or not?

    Knowing that, can we conclude that $f(x)$ is irreducible as follows:

    Suppose that $f(x)$ is reducible, then there are non-constant polynomials $g(x),h(x)\in \mathbb{Q}[x]$ such that $f(x)=g(x)h(x)$. Then $f(x+1)=g(x+1)h(x+1)$. Since the polynomials are non-constant that would mean that $f(x+1)$ is reducible, a contradiction.
    So, $f(x)$ is irreducible in $\mathbb{Q}[x]$.

    Is this correct?
    This is correst. Basically what we are doing is this: Consider the ring map $R[x]\to R[x]$ which sends $x$ to $x+1$. This is a ring isomorphism. Thus $f(x)$ in $R[x]$ is irreducible iff and only if its image, which is $f(x+1)$, is irreducible.

    Quote Originally Posted by mathmari View Post
    Let $\rho\in \mathbb{C}$ be a root of $f$. I want to find a basis and the degree of the extension $\mathbb{Q}[\sqrt{2},\rho]$.

    We have that $\text{Irr}(\rho, \mathbb{Q})=f$ and $\text{Irr}(\sqrt{2},\mathbb{Q}[\rho])=x^2-2$, or not?
    It is not clear why $x^2-2$ is irreducible over $\mathbf Q(\rho)$. This must be true but it requires a proof. And I am unable to see it myself.

  3. MHB Master
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    #3 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    It is not clear why $x^2-2$ is irreducible over $\mathbf Q(\rho)$. This must be true but it requires a proof. And I am unable to see it myself.
    So that $x^2-2$ is irreducible over $\mathbf Q(\rho)$, the root $\pm \sqrt{2}$ does not have to be an element of $\mathbf Q(\rho)$, right?
    So we have to know that the real part of $\rho$ is not equal to $\pm \sqrt{2}$, or not?

    Let $\rho=\sqrt{2}+bi$ then $f(\rho)=0 \Rightarrow (\sqrt{2}+bi)^4−2(\sqrt{2}+bi)^2−1=0 \\ \Rightarrow -1-10 b^2+b^4+i (4 \sqrt{2} b-4 \sqrt{2} b^3)=0 \\ \Rightarrow -1-10 b^2+b^4=0 \land 4 \sqrt{2} b-4 \sqrt{2} b^3=0 \\ \Rightarrow (b=\pm \sqrt{5+\sqrt{26}} )\land (b=0 \lor b=\pm 1)$
    a contradiction

    Let $\rho=-\sqrt{2}+bi$ then $f(\rho)=0 \Rightarrow (-\sqrt{2}+bi)^4−2(-\sqrt{2}+bi)^2−1=0 \\ \Rightarrow -1-10 b^2+b^4+i (-4 \sqrt{2} b+4 \sqrt{2} b^3)=0 \\ \Rightarrow -1-10 b^2+b^4=0 \land -4 \sqrt{2} b+4 \sqrt{2} b^3=0 \\ \Rightarrow (b=\pm \sqrt{5+\sqrt{26}} )\land (b=0 \lor b=\pm 1)$
    a contradiction

    Therefore, $\sqrt{2}\notin \mathbf Q(\rho)$.

    So, $x^2-2$ is irreducible over $\mathbf Q(\rho)$.

    Is this correct? Could I improve something?

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    #4
    Quote Originally Posted by mathmari View Post
    So that $x^2-2$ is irreducible over $\mathbf Q(\rho)$, the root $\pm \sqrt{2}$ does not have to be an element of $\mathbf Q(\rho)$, right?
    So we have to know that the real part of $\rho$ is not equal to $\pm \sqrt{2}$, or not?
    The first statement is correct because a degree 2 polynomial is irreducible over $\mathbb Q(\rho)$ if and only if it does not have a root in $\mathbf Q(\rho)$. I am unable to see the logic behind your second statement. Can you elaborate on that?

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    #5 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    I am unable to see the logic behind your second statement. Can you elaborate on that?
    What I said is wrong. I thought that we would have that $\overline{\rho}\in \mathbb{Q}(\rho)$, which is not true, is it?

    So, we have to show that $\rho$ cannot be equal to $\sqrt{2}$ so that $\sqrt{2}\notin \mathbb{Q}(\rho)$, right?

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    #6
    Quote Originally Posted by mathmari View Post
    What I said is wrong. I thought that we would have that $\overline{\rho}\in \mathbb{Q}(\rho)$, which is not true, is it?

    So, we have to show that $\rho$ cannot be equal to $\sqrt{2}$ so that $\sqrt{2}\notin \mathbb{Q}(\rho)$, right?
    Are you saying that $\rho\neq \sqrt{2}$ implies $\sqrt{2}\notin \mathbf Q(\rho)$?

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    #7 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    Are you saying that $\rho\neq \sqrt{2}$ implies $\sqrt{2}\notin \mathbf Q(\rho)$?
    Yes. Is this wrong?

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    #8
    Quote Originally Posted by mathmari View Post
    Yes. Is this wrong?
    Yes. This implication is incorrect. Try supplying a proof. You'll see why.

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    #9 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    Try supplying a proof. You'll see why.
    Should I prove that it doesn't hold that $\sqrt{2}\in \mathbb{Q}(\rho)\Rightarrow \rho=\sqrt{2}$ ?


    Quote Originally Posted by caffeinemachine View Post
    It is not clear why $x^2-2$ is irreducible over $\mathbf Q(\rho)$. This must be true but it requires a proof. And I am unable to see it myself.
    So, is there an other way to find the degree of the extension?

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    #10
    Quote Originally Posted by mathmari View Post
    Should I prove that it doesn't hold that $\sqrt{2}\in \mathbb{Q}(\rho)\Rightarrow \rho=\sqrt{2}$ ?
    Yes. Try showing that if $F(\alpha, \beta)$ is an algebraic extension of a field $F$, then it doesn't in general hold that $\beta\in F(\alpha)$ implies $\beta=\alpha$.


    Quote Originally Posted by mathmari View Post
    So, is there an other way to find the degree of the extension?
    One way I can see how to show that $\sqrt{2}\notin \mathbf Q(\rho)$ is by using the artifice of trace of an extension. Are you familiar with it?

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