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From Aluffi, Algebra: Chapter $0$

If $s$ is any element of a ring $S$, then there is a unique ring homomorphism $\mathbb Z[x] \to S$ sending $x$ to $s$ and ‘extending’ the unique ring homomorphism $\iota : \mathbb Z \to S$. In this case the commutativity of $S$ is immaterial (why?).

The unique homomorphism $f: \mathbb Z[x] \to S$, where is $x \mapsto s$, is defined by mapping $f(a_0+a_1x+\cdots+a_nx^n)=\iota(a_0)+\iota(a_1)s+\cdots +\iota(a_n)s^n$ where $\iota: \mathbb Z \to S$ is the unique ring homomorphism.


Consider $(a_0 +a_2x^2)$ and $(b_1x+b_2x^2)$. Then $(a_0 +a_2x^2)(b_1x+b_2x^2)=a_0b_1x+a_0b_2x^2+a_2b_1 x^3+a_2b_2x^4$.

So, $$f(a_0 +a_2x^2)=\iota(a_0) +\iota(a_2)s^2,$$ $$f(b_1x +b_2x^2)=\iota(b_1)s +\iota(b_2)s^2,$$ $$f(a_0b_1x+a_0b_2x^2+a_2b_1 x^3+a_2b_2x^4)=\iota(a_0b_1)s+\iota(a_0b_2)s^2+\iota(a_2b_1)s^3+\iota(a_2b_2)s^4.$$

However, \begin{align*} f(a_0 +a_2x^2)f(b_1x +b_2x^2)&=(\iota(a_0) +\iota(a_2)s^2)(\iota(b_1)s +\iota(b_2)s^2)\\ &=\iota(a_0)\iota(b_1)s+\iota(a_0)\iota(b_2)s^2+\iota(a_2)s^2\iota(b_1)s+\iota(a_2)s^2\iota(b_2)s^2. \end{align*}

If $S$ does not commute then $f$ is not a ring hom? So, why does it not matter if $S$ is commutative?

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    $\begingroup$ What you are forgetting is that $\iota(\mathbb{Z})$ always lies in the center of $S$, because it is a ring homomorphisms. So $s\iota(n) = \iota(n)s$ for all $n\in\mathbb{Z}$. $\endgroup$ – Arturo Magidin Jun 10 at 20:24
  • $\begingroup$ Another way to look at it is that the free ring on $n$ elements is $\mathbb{Z} \langle t_1, \ldots, t_n \rangle$ which consists of elements of a form like $3 t_2 t_1 t_2^2 t_3^5 t_1 - 2 t_3 + 5$ where the $t_i$ do not commute with each other. It just happens that $\mathbb{Z}\langle t \rangle \simeq \mathbb{Z}[t]$ is commutative - whereas for $n \ge 2$, $\mathbb{Z}\langle t_1, t_2, \ldots, t_n \rangle$ is not commutative. $\endgroup$ – Daniel Schepler Jun 10 at 20:35
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The image of $\iota$ is automatically contained in the center of $S$. For example, $\iota(3) = 1_S + 1_S + 1_S$, so for any $x \in S$, we have $$\iota(3) x = (1_S + 1_S + 1_S) x = 1_S x + 1_S x + 1_S x = x + x + x = \cdots = x \iota(3).$$ (If you have not seen this result before, I will leave the formalization of the proof to you, along with the case of $\iota(n)$ for $n < 0$.)

Thus, for instance, $\iota(a_2) s^2 \iota(b_1) s = \iota(a_2) \iota(b_1) s^2 s = \iota(a_2 b_1) s^3$.

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First: If $S$ is any ring with unit $1 ∈ S$, then by induction from “$1·x = x = x·1$” using distributivity, $ℤ·1$ is always within the center of $S$, so you may regard $S$ as a $ℤ$-Algebra.

If $s ∈ S$ is any element, then $s$ and all its powers commute with each other as well as with all elements of $ℤ·1$. You can show that the smallest $ℤ$-Algebra within $S$ containing $s$ is given by $$\{a_ns^n + … + a_1s + a_0;~a_0,…,a_n ∈ ℤ\}$$ and is hence commutative. Let’s call this algebra $ℤ[s]$.

Now, there is a unique $ℤ$-algebra homomorphism $ℤ[X] → ℤ[s]$ with $X ↦ s$, and an inclusion $ℤ[s] → S$, which is an injective ring homomorphism. Hence, there is always a unique ring homomorphism $ℤ[X] → S$ with $X ↦ s$.

Bottom line: Powers of a fixed element commute with each other, hence stuff generated by one element over commutative structures tend to be commutative themselves.

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