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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

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2 Answers 2

up vote 2 down vote accepted

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. –  Bill Dubuque Nov 9 '12 at 0:52
    
I'd say it's just the Division Theorem. –  Gerry Myerson Nov 9 '12 at 0:54
    
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. –  Bill Dubuque Nov 9 '12 at 1:32
    
@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. –  Gerry Myerson Nov 9 '12 at 1:57
    
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. –  Bill Dubuque Nov 9 '12 at 2:04

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 $\rm\:\Bbb Q[x]/(p).$

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