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I've been tasked with finding an element $d\in \mathbb Q[x]$ such that $$(d)=(x^4-x^2+4x, x^3-x+3).$$ To do this, I'm confused at what exactly I'm meant to do to find this element. First I did the algorithm on the two polynomials: $$x^4-x^2+4x = (x^3-x+3)(x)+(x)\\ x^3-x+3=(x)(x^2-1)+3\\ x=(3)\left(\frac 1 3 x\right)+(0)$$

Does this mean that my (d) is just 1? Or have I misunderstood something, because in this case, the last nonzero remainder must be 1. And if I use backwards substitution, I get that $$\left(-\frac 1 3 x^2+\frac 1 3\right)\left(x^4-x^2+4x)+\right(\frac 1 3-\frac 1 3 x + \frac 1 3)\left(x^3-x+3\right)=1 $$ which seems to check out.

Have I misunderstood what is meant with finding $d$? This feels weirdly easy.

YamahaJacoby
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    You got it right. – Christoph Jan 20 '21 at 18:23
  • @Christoph Neat, thank you very much. I'm curious, using this, can I find if $((x^4-x^2+4x, x^3-x+3)$ is a principal ideal of $\mathbb Z[i]$? I think we had a theorem about something similar to this, I'm not sure how it went anymore. – YamahaJacoby Jan 20 '21 at 18:26
  • Not sure what you mean: you can consider $(x^4-x^2+4x,x^3-x+3)$ as an ideal in $\mathbb Q[x]$, where it is equal to $\mathbb Q$, or as an ideal in $\mathbb Z[x]$, where it is equal to $(3)=3\mathbb Z[x]$. There is no $x$ in $\mathbb Z[i]$. Anyhow, this should be a new question. – Christoph Jan 20 '21 at 18:31
  • @Christoph You are right, I meant to say $\mathbb Z[x]$. I will post a further thread for clarification once the timer is up. Thank you :) – YamahaJacoby Jan 20 '21 at 18:47
  • No need to ask, it's already proved here in many places, e.g. here. – Bill Dubuque Jan 20 '21 at 18:51

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