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Can someone please guide me along on this. Not sure where to start.

Let $t=1+\sqrt[3]5+\sqrt[3]5^2$, and let $f(x) \in \mathbb{Q}[x]$ be the minmal polynomial of $t$ over $\mathbb{Q}$.

Question:

  • Find and simplify $f(x)$. (Hint: consider $(t-1)^3$)
  • Use Cardano's formula to express the roots of $f(x)$. (Hint: substitute to get the form $x^3+px+q$.)
user26857
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user551155
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    What have you tried? Where are you stuck? (For instance, you have been given a few solid hints, have you tried following them? What happened?) – Arthur Apr 15 '20 at 17:08
  • $(t-1)^3=t^3-3t^2+3t-1$, so I thought to view t in the form $a+b\sqrt[3]5+csqrt[3]5^2$ and substitute but I'm not sure what that does for me. – user551155 Apr 15 '20 at 17:13
  • What did you get when you computed $(t-1)^3$? – Angina Seng Apr 15 '20 at 17:32
  • basis: ${1, \sqrt[3]5, \sqrt[3]5^2}$, dimension = 3. so $1, \sqrt[3]5, \sqrt[3]5^2$ are linearly independent. – user551155 Apr 15 '20 at 17:32
  • as above: $(t-1)= t^3-3t^2+3t-1$ – user551155 Apr 15 '20 at 17:35
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    You should use the fact that $ t - 1 = \sqrt[3]{5} + \sqrt[3]{5}^2$ and expand that version of $(t-1)^3$, which can then be expressed in terms of $t$ (slightly magically). – Calvin Lin Apr 15 '20 at 17:36

1 Answers1

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  • Following the first hint we have: \begin{gather} (t-1)^3 = (1+ \sqrt[3]{5} + \sqrt[3]{5^2} - 1)^3 = \\ ( \sqrt[3]{5} + \sqrt[3]{5^2})^3= 5 (1+\sqrt[3]{5})^3 = \\ 5(1+5+3\sqrt[3]{5} + 3\sqrt[3]{5^2}) = 5(3+3t) \end{gather} So $t$ is a root of the polynomial: \begin{equation} f(x)=(x-1)^3-5(3+3x)=x^3-3x^2-12x-16 \end{equation} and $f(x)$ is irreducible in $\mathbb Q[x]$ (to show this, you can try rational solutions or noted that $t\notin \mathbb Q$ and the minimal polynomial of $t$ has degree $1$ or $3$)

  • Using the sobstitution $x-1=y$ we obtain: \begin{gather} f(x)=(x-1)^3-15(x-1)-30\\ f(y+1)=g(y)=y^3-15y-30 \end{gather} And now you can use the Cardano's Formula in order to find the roots $y_1,y_2,y_3$ of $g(y)$ and then the roots of $f(x)$: \begin{gather} x_1 = 1+ y_1 = 1 + \sqrt[3]{5} + \sqrt[3]{5^2}\\ x_2=1+y_2 = 1 + \left(-\frac{1}{2}\right)\left[\sqrt[3]{5}(1-i\sqrt 3)+ \sqrt[3]{5^2}(1+i\sqrt 3)\right]\\ x_3=1+y_3 = 1 + \left(-\frac{1}{2}\right)\left[\sqrt[3]{5^2}(1-i\sqrt 3)+ \sqrt[3]{5}(1+i\sqrt 3)\right] \end{gather}

Remark: we have another proof that $f$ is irreducible since $g$ is irreducible by Eisenstein criterion ($p=3$).

Menezio
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    How did you get $f(x)$ after the substitution of $y=x-1$? Did you substitute directly in to $f(x)=x^3-3x^2-12x-16$? Maybe my math is wrong but I keep getting a different result. – user551155 Apr 15 '20 at 18:12
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    @user551155 I did this: $f(x)=(x-1)^3-5(3+3x)=(x-1)^3-15(x+1)=(x-1)^3-15(x-1+2)=(x-1)^3-15(x-1) - 15(2) = (x-1)^3-15(x-1)-30$ – Menezio Apr 15 '20 at 18:19
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    @user551155 Maybe I was not clear in the answer. The sobtitution to do is $x=y+1$ in polynomial $f(x)$. So $f(y+1) = g(y)=y^3-15y-30$ – Menezio Apr 15 '20 at 18:34
  • If I reduce $f(x)=x^3-3x^3-12x-16$ to $\mathbb{Z_3}$ I get $f(x)=x^3-1 \in \mathbb{Z_3}$ and $f(1)=0 \in \mathbb{Z_3}$ indicating $f(x)$ has a root at $x-1$. Does this indicate $f(x)$ is reducible? – user551155 Apr 15 '20 at 19:26
  • No, it's the other implication: If $f(x)$ is reducible in $\mathbb Z[x]$ then it is reducible in $\mathbb Z_p[x]$ Here the proof. The converse statement is false (as you have shown in this example). Here you can find an explcit example. – Menezio Apr 15 '20 at 19:40
  • @menzio I'm confused as to how you came up with $x_2, x_3$? – user551155 Apr 15 '20 at 20:28
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    I see $x_3$ is the conjugate of $x_2$(which makes sense) just not sure how to get $x_2$ – user551155 Apr 15 '20 at 20:31
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    @user551155 Here you can find the Cardano's formula to get the roots of polynomial of degree $3$. In particular see the paragraph "Cardano's Formula". – Menezio Apr 15 '20 at 21:19