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I am revising for my exams and don't understand how to do the following question, any hints would be very helpful!

Find a minimal polynomial of $\alpha$ when $\alpha$ is an irrational number satisfying $\alpha^3 + 3\alpha^2-2=0$.

I have the definition of a minimal polynomial of $\alpha$ over the rational numbers and what it means for $\alpha$ to be algebraic over the rational numbers.

Marc van Leeuwen
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Koala
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  • Over the rational numbers you mean? Over $\mathbb{R}$ it would be not really nice to factor this expression... – Dirk Apr 11 '17 at 13:05
  • Yes, sorry will edit my question – Koala Apr 11 '17 at 13:07
  • Either the cubic is irreducible over $\Bbb Q$ or it is not. If it is, then it is the minimal polynomial. If it is not, then it factors as $(x - r) q(x)$ for some rational $r$, so that $q(\alpha) = 0$; since $\alpha$ is irrational, $q$ is irreducible. – Travis Willse Apr 11 '17 at 13:07
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    Factorizing gives $(\alpha+1)(\alpha^2+2\alpha-2)$ so does this mean that since $\alpha$ is irrational the polynomial $\alpha^2+2\alpha-2$ is irreducible and so is the minimal polynomial of $\alpha$? – Koala Apr 11 '17 at 16:15

1 Answers1

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The polynomial $X^3 + 3X^2 - 2$ factors as $$X^3 + 3X^2 - 2 = (X + 1)(X^2 + 2X - 2).$$ The quadratic factor is irreducible (over ${\mathbb Q}$) because it has no roots (in ${\mathbb Q}$).

Since $\alpha$ is a root of $X^3 + 3X^2 - 2$, it is a root of either $X + 1$ or of $X^2 - 2X - 2$. Because $\alpha$ is irrational, it is not equal to $-1$, so it is a root of $X^2 + 2X - 2$.

So, $X^2 + 2X - 2$ is an irrediducible, monic polynomial with $\alpha$ as a root, so it is the minimal polynomial of $\alpha$.

Magdiragdag
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