Find minimal polynomial given a certain relation

Question

I have the following relation: $$a^3-a-1=0$$

Now I want to find the minimal polynomial of the following elements:

1) $a^2+a+1$

2) $(a^2+1)^{-1}$

I don't really know how to start, but this is what I tried.

$X^0 = 1 \\ X^1 = a^2+a+1 \\ X^2 = (a^2+a+1)^2 = a^4 + 2a^3 + 3a^2 + 2a + 1 = a(a+1)+2(a+1) + 3a^2 + 2a + 1 = 4a^2 + 5a + 3$

I can do this again, but I need a formula where I got only one $a^3$ since I need that the minimal polynomial is monic. How can I do this?

For the second, I tried to find the inverse of $a^2+1$. I found out that this was $-\frac{2}{5}a^2 + \frac{1}{5}a + \frac{4}{5}$ (correct me if I am wrong), after that I can do the same computations as above, but I still can't find a polynomial where the coefficient of $a^3$ is 1. Can somebody please show me how I can find the minimal polynomial?

Answer 1

Assuming $a$ has minimal polynomial $m_a(x)=x^3-x-1$, e.g. work over $\mathbb{Q}$.

For the first question, do it once more, \begin{align*} X^3&=(a^2+a+1)^3\\ &=(a^3 + 3 a^2 + 7 a + 11) \times (a^3 - a - 1) + (16 a^2 + 21 a + 12)\\ &= 16 a^2 + 21 a + 12 \end{align*} So $1$, $X^1=a^2+a+1$, $X^2=4a^2+5a+3$, $X^3=16a^2+21a+12$ gives us four vectors in a three-dimensional $\mathbb{Q}$-vector space, hence there is a linear dependence which we can calculate by successively eliminating powers of $a$: \begin{align*} X^3-4X^2&=a\\ X^2-4X &= a-1\\ \implies (X^3-4X^2)-(X^2-4X)-1&=0 \end{align*} so $m_{a^2+a+1}(x)=x^3-5x^2+4x-1$, because we also see in this process $1,X,X^2$ is linearly independent.

For the second question, it is easier to calculate the minimal polynomial of $a^2+1$ first, then take its reverse (or reciprocal) polynomial.

Answer 2

Here is an illustration of a technique I sometimes use, which involves naming the expression we are working with and eliminating $a$. You will note that I have avoided using polynomial fractions by multiplying through by what would have been the denominator - this incidentally avoids accidentally dividing by zero. I think you will find it pretty much as efficient as the other suggestions.

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For the first we let $x=a^2+a+1$ and then $$0=a^3-a-1=ax-a^2-2a-1=ax-x-a$$ so $$a(x-1)=x$$

Multiply $a^3-a-1=0$ by $(x-1)^3$ to obtain $$\left(a(x-1)\right)^3-a(x-1)^3-(x-1)^3=x^3-x(x-1)^2-(x-1)^3=0$$

This gives you a monic cubic polynomial in $x$, and the minimal polynomial for $x$ will be a factor of this.

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For the second, we let $y=\frac 1{a^2+1}$ so that $y(a^2+1)=1$ and so $a^2y=1-y$ to use later.

This expression involves only $a^2$ so we want to eliminate odd powers of $a$. This is done by using $a^3-a=1$ (gathering the odd powers on one side and the even powers on the other) and squaring it to get $a^6-2a^4+a^2=1$.

Then we multiply through by $y^3$ to obtain $$a^6y^3-2a^4y^3+a^2y^3=y^3= (1-y)^3-2y(1-y)^2+y^2(1-y)$$ and that gives a cubic in $y$. This time the cubic is not monic, but if you are allowed rational coefficients you can divide through be the leading coefficient to obtain a monic polynomial.