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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

user42912
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2 Answers2

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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.

Gerry Myerson
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  • This is just the Extended Euclidean algorithm - see my answer. – Bill Dubuque Nov 09 '12 at 00:52
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    I'd say it's just the Division Theorem. – Gerry Myerson Nov 09 '12 at 00:54
  • 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 09 '12 at 01:32
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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. – Gerry Myerson Nov 09 '12 at 01:57
  • My point is that students reading only your answer likely will have no idea how to invert other polynomials (of higher degree) by using the *extended Euclidean algorithm* (of which the above is a special (linear) case, i.e. the case when the algorithm terminates in a single step). And, btw, there are many proofs that "use the division algorihm more than once" that have nothing to do with the extended Euclidean algorithm. – Bill Dubuque Apr 29 '23 at 18:41
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As for integers, the extended Euclidean algorithm computes modular inverses, and terminates in one step, since here $\rm\,x+1\,$ has degree $1,\,$ i.e. $\rm\, a\,p + b\,(1\!+\!x) = 1\,$ $\rm\Rightarrow\, (1\!+\!x)^{-1} = b\,$ in $\rm\,\Bbb Q[x]/(p).$

Or $ $ let $\,X=x\!+\!1\,$ and $\, p(x) = p(X\!-\!1) = g(X)\:\!X-c,\ $
therefore $\bmod p(x)\!:\ \ g(X)\:\!X^{\phantom{|^|}}\!\!\!\equiv c\,\Rightarrow\, X^{-1}\equiv g(X)/c\,\ $ (remark: this can be viewed as rationalizing the denominator of $\,\frac{1}X\,$ using the norm $\,c\,$ as a "simpler multiple" of $X$)

Bill Dubuque
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