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I've tried a lot with the Euclidean Algorithm, but I still can't figure it out. Do you know how I can use the Euclidean Algorithm to find the inverse of $\overline{x^2 -x}$ in $(\mathbb{R} [x]/(x^4 + 1))^*$?

Thanks in advance!

user890995
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    The goal is to produce a Bezout's identity for the two polynomials. This is, polynomials $a,b$ such that $a(x)(x^4+1)+b(x)(x^2-x)=1$. When that is done the class of $b$ is the inverse. Deeper in the link you get the algorithm. – plop Jun 23 '21 at 18:48
  • @plop Thanks for your answer, but I do understand that. However, I'm not exactly sure how to determine this $a, b$, so I'm asking for help here. – user890995 Jun 23 '21 at 19:18
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    When you run Euclid's algorithm you will get two equations $x^4+1=(x^2-x)(x^2+x+1)+(x+1)$ and $x^2-x=(x+1)(x-2)+2$. Well three, but we don't need the very last one $x+1=2(x/2+1/2)+0$. From $x^2-x=(x+1)(x-2)+2$ you can express $1$ in terms of $x+1$ and $x^2-x$ as $1=\frac{1}{2}(x^2-x)-(x/2-1)(x+1)$. Now you can use the first equation to replace the $x+1$ in this one and get $x^2-x$ and $x^4+1$ in its place. – plop Jun 23 '21 at 19:32
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    As you see, Euclid's give you a bunch of equations, in which the last one has essentially the GCD, which should be $1$ for the inverse to exist. Then we go along the equations backwards, substituting one into another, removing the occurrence of the previous remainder in the equation. At the end we get $1$ expressed in terms of the first two polynomials. – plop Jun 23 '21 at 19:36
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    Generally the easiest way is to use the extended Euclidean algorithm in forward form (augmented matrix) as described in my posts in the linked dupes. The common backward form is notoriously error-prone and should be avoided unless one is infallible (e.g. a machine) – Bill Dubuque Jun 23 '21 at 20:21
  • @BillDubuque, I hadn't known that machines were infallible! This is great! :) – paul garrett Jun 23 '21 at 20:47
  • The algorithm is the same as the "forward" computation. The "backward" substitution doesn't need to be avoided. Dogmas like that are absurd. In fact, the backward substitution makes it easier to explain why you get an expression for $1$ in terms of the polynomials, as the $1$ on one side constitutes a loop invariant. Different presentations of the same object can serve different purposes. – plop Jun 23 '21 at 22:22
  • @Plop If I understand correctly you get something like $1 = a(x^4 + 1) + b(x^2 - x)$ right? In that case what is exactly the inverse of $\overline{x^2-x}$? – user890995 Jun 26 '21 at 11:01

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