1

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
  • 23,582

2 Answers2

3

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
  • 179,216
  • This is just the Extended Euclidean algorithm - see my answer. – Bill Dubuque Nov 09 '12 at 00:52
  • 2
    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
  • 1
    @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
-1

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
  • 272,048