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Show that $p(x) = x^3 + 9x + 6$ is irreducible in $\mathbb Q[x]$. Let $\alpha$ be a root of $p(x)$. Find the inverse of $1 + \alpha$ in $\mathbb Q[x]$.

So as far as the irreducibility is conccerned we can use the Einseinstein criterion (p=3). But how can we find the inverse of $1 + \alpha$?

Thanks

2 Answers 2

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Presumably, you want the inverse of $1+\alpha$ in ${\bf Q}(\alpha)$, not in ${\bf Q}[x]$.

Divide $p(x)$ by $1+x$; $$p(x)=(1+x)q(x)+r$$ Then substitute $x=\alpha$, and maneuver a bit.

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    This is just the Extended Euclidean algorithm - see my answer.2012-11-09
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    I'd say it's just the Division Theorem.2012-11-09
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    No, it's the special case of the Extended Euclidean Algorithm when it terminates in a *single step*, i.e. when it requires just one application of the Division Algorithm. To invert elements of *higher* degree generally requires *multiple* applications of the Division Algorithm i.e. the full Extended Euclidean Algorithm. Said structurally it amounts to the fact that Euclidean domains are Bezout domains.2012-11-09
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    @Bill, you are making a distinction between using the Division Theorem and using the Division Theorem just once? A distinction without a difference, I think.2012-11-09
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    The Division Algorithm can be applied in myriad ways. One particular way is known as the Extended Euclidean Algorithm. It has it's own name for a very good reason.2012-11-09
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Hint $\ $ Just as for integers, the extended Euclidean algorithm works, and terminates in one step since $\rm\:x+1\:$ has degree $1,\:$ i.e. $\rm\: a\:p + b\:(1\!+\!x) = 1\:$ $\rm\Rightarrow\: (1\!+\!x)^{-1}\!\equiv b\pmod p,\:$ i.e. in $\rm\:\Bbb Q[x]/(p).$