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Find the minimal polynomial of $t^2+t$ over $\mathbb{Q}$ where t satisfies $t^3-3t^2-3=0$.

Okay, so I was working on this one for awhile today with my buddy and we couldn't figure it out, haha. We got creative with this and tried a lot of stuff but couldn't figure it out, so i'm betting somebody here makes it look really easy like you always do.

One thing we did was try plugging $t^2+t$ into $x^3-3t^2-3$ and trying to look for clues or even make it zero. Another route I took was dividing $\frac{x^3-3x^2-3}{x-t}$ and yeah of course it divides without remainder but yeah I don't know haha I need a new perspective

Math is hard
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    I think it is easier to find the minimal polynomial $\mu(x)$ of $\left(t+\dfrac12\right)^2$ over $\mathbb{Q}$ first. Note that, if $t$ is a root of $f(x):=x^3-3x^2-2$, then $s:=t+\dfrac{1}{2}$ is a root of $g(x):=f\left(x-\frac12\right)$. It is not a difficult task to find the cubic polynomial $h(x)$ with roots $s^2$, knowing $g(x)$. Most likely, $h(x)$ is irreducible over $\mathbb{Q}$, and so we may set $\mu(x):=h(x)$. Now, $\mu\left(x+\dfrac{1}{4}\right)$ is then the minimal polynomial of $\left(t+\dfrac12\right)^2-\dfrac14=t^2+t$ over $\mathbb{Q}$. – Batominovski Oct 20 '18 at 22:27
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    Just noticed a typo. In my comment above, $f(x)$ should be $x^3-3x^2-3$. – Batominovski Oct 20 '18 at 22:47
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    Here's a very similar question. – Viktor Vaughn Oct 21 '18 at 00:40

2 Answers2

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A more conceptual method:

If $t$ is a root of $t^3-3t^2-3=0$, then $t$ is an eigenvalue of the matrix $$A = \begin{pmatrix}0 & 0 & 3 \\ 1 & 0 & 0 \\ 0& 1 & 3\end{pmatrix}$$ hence $t^2+2t$ is an eigenvalue of the matrix $$A^2+2A = \left( \begin{array}{ccc} 0 & 3 & 15 \\ 2 & 0 & 3 \\ 1 & 5 & 15 \\ \end{array} \right)$$ computing characteristic polynomial shows $t^2+2t$ satisfies $x^3-15 x^2-36 x-69 = 0$. It is indeed the minimal polynomial because $[\mathbb{Q}(t^2+2t):\mathbb{Q}] = 3$.

However, in a computational point of view, this method is quite expensive when dimension of matrix is large.

pisco
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Let $\alpha=t^2+t$. Then $$\begin{align}\alpha^2&=t^4+2t^3+t\\&=(t+2)t^3+t\\&=(t+2)(3t^2+3)+t\\&=3t^3+6t^2+4t\\&=3\cdot(3t^2+3)+6t^2+4t\\&=15t^2+4t+9\end{align}$$ and $$\begin{align}\alpha\cdot(\alpha^2-15\alpha)&=(t^2+t)\cdot(-11t+9)\\ &=-11t^3-2t^2+9t\\&=-11(3t^2+3)-2t^2+9t\\&=-35t^2+9t-33\end{align} $$ so that $$\alpha^3= \alpha\cdot(\alpha^2-15\alpha)+15\alpha^2=190t^2+69t+102.$$ Now find a linear dependency between $$\begin{matrix}\alpha^3=&190t^2&+69t&+102\\ \alpha^2=&15t^2&+4t&+9\\ \alpha^1=&t^2&+t\\ \alpha^0=&&&1\end{matrix} $$

Hagen von Eitzen
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