I've recently read in a book a proof that used the following fact:

Since $(X^m-1, X^n+1)=1$ in $\mathbb{Q}[X],$ there exist $u,v \in \mathbb{Z}[X]$ such that $$(X^m-1)u + (X^n+1)v = 2$$ hence $(b^m-1)u(b) + (b^n+1)v(b) = 2 \quad (*)$

For context, $b$ is a random element of a finite ring $A$ with unity.

My question is the following: how were they able to get to $(*)?$ Are the rings $A[X]$ and $\mathbb{Q}[X]$ somehow connected? How is $u(b)$ even defined, as $u \in \mathbb{Z}[X]?$

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    $\begingroup$ By evaluating at $b$ (more technically by using the universal property of $\Bbb Z[X])$. Polynomial equations in $\Bbb Z[X]$ (such as Bezout equations) can be considered as universal laws (identities) of commutative rings, e.g. see GCD in a PID persists in extension domains. $\endgroup$ – Bill Dubuque Apr 4 at 12:56
  • $\begingroup$ Thank you! But in this example the ring is not necessary commutative. Is this because only one element of the ring is involved in the polynomial equation? (i.e. $b$) $\endgroup$ – AndrewC Apr 4 at 13:06
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    $\begingroup$ Hint: the proof of the Bezout equation in $\Bbb Z[X]$ uses only ring laws plus the hypothesis that $X$ commutes with all elements in the coefficient ring $\Bbb Z$, so the Bezout equation will persist in the image ring if the latter commutativity constraint persists. $\endgroup$ – Bill Dubuque Apr 4 at 13:12
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    $\begingroup$ See also this question on the role that commutativity plays in the proof that polynomial evaluation is a ring homomorphism. $\endgroup$ – Bill Dubuque Apr 4 at 13:18
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    $\begingroup$ See also this answer for further examples of such universal equations and related ideas (the matrix examples there should shed light on the question in your first comment) $\endgroup$ – Bill Dubuque Apr 4 at 13:25

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