# Question about polynomials and elements of random rings

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 $$u \in \mathbb{Z}[X]?$$

• 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. – Bill Dubuque Apr 4 at 12:56
• 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. $b$) – AndrewC Apr 4 at 13:06
• 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. – Bill Dubuque Apr 4 at 13:12
• See also this question on the role that commutativity plays in the proof that polynomial evaluation is a ring homomorphism. – Bill Dubuque Apr 4 at 13:18
• 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) – Bill Dubuque Apr 4 at 13:25