# Let $\alpha$ be a root of $X^3+X^2-2X+1\in\mathbb{Q}[X]$.

Here is a question in my homework.

Let $\alpha$ be a root of $X^3+X^2-2X+1\in\mathbb{Q}[X]$. Express $(1-\alpha^2)^{-1}$ as a $\mathbb{Q}$-linear combination of $1$, $\alpha$ and $\alpha^2$. Justify the assertion that the cubic is irreducible over $\mathbb{Q}$, using Gauss' Lemma.

This is the first question of my homework so I kind of expect a fast solution. But I couldn't do it the first part.

For the last part, I showed that this is irreducible over $\mathbb{Z}$ because it is cubic and has no solution in $\mathbb{Z}$ and applied Gauss' lemma. If someone kindly shows me how to do the first part, can you please also say why this relates to the second part?

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Just in case you might think I have typed the question wrong, I did not. I have triple checked. – Chuwei Zhang Oct 21 '12 at 21:12
I've changed algebra tag to abstract-algebra, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Oct 22 '12 at 13:43

Hint: You are looking for $a,b,c\in\mathbb Q$ such that $$(a+b\alpha+c\alpha^2)(1-\alpha^2 )=1.$$ You obtain a polynomial of degree $4$ in $\alpha$ on the left side. Do polynomial division to reduce this to degree $<3$. (The first step is that The leading term is $-c\alpha^4$, hence you can add $c\alpha f(\alpha)$ to obtain degree $3$). Then you obtain somethng $???+???\alpha+???\alpha^2=1+0\alpha+0\alpha^2$, i.e three equations in three unknowns.

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Rationalizing denominators works. Since $\rm\:1/(1-a^2) = -1/((a-1)(a+1))\:$ we can compute $\rm\:1/(a\pm 1)\:$ then multiply the results. Let $\rm\:b = a-1.\:$ Then $\rm\:0 = a^3\!+a^2\!-2a+1 = b^3\! + 4b^2\! + 3b+1\:$ therefore $\rm\:b(b^2+4b+3) = -1,\:$ hence

$$\rm \frac{1}{b}\, =\, \frac{b^2+4b+3}{b(b^2+4b+3)}\, =\, \frac{b^2+4b+3}{-1}$$

Similarly with $\rm\:c = a+1\:$ we have $\rm\: c(c^2-2c-1) = -3\:$ hence

$$\rm \frac{1}c\, =\, \frac{c^2-2c-1}{c(c^2-2c-1)}\, =\, \frac{c^2-2c-1}{-3}$$

Multiplying these, substituting $\rm\:b = a-1,\ c = a+1\:,\:$ then reducing mod $\rm\:a^3\!+a^2\!-2a+1\:$ yields

$$\rm \frac{-1}{a^2-1}\, =\, \frac{-1}{bc}\, = \,\frac{a^2+3a+1}{3}$$

Alternatively you could use the extended Euclidean algorithm to compute the Bezout identity, i.e. with $\rm\:f = 1-a^2,\:$ and $\rm\:g = a^3\!+a^2\!-2a+1,\:$ since $\rm\:gcd(f,g) = 1\:$ we obtain $\rm\:h\,f+k\,g = 1\:$ hence $\rm\:mod\ g\!:\ hf\equiv 1\:\Rightarrow\:h\equiv f^{-1}.\:$ Or, solve for $\rm\:h\:$ by undetermined coefficients, as in Hagen's answer.

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