# Why is the unique ring homomorphism $\mathbb Z[x] \to S$, where $x \mapsto s$, irrelevant of $S$ being commutative?

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?

• 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 $n\in\mathbb{Z}$. – Arturo Magidin Jun 10 at 20:24
• 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. – Daniel Schepler Jun 10 at 20:35

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, $$\iota(a_2) s^2 \iota(b_1) s = \iota(a_2) \iota(b_1) s^2 s = \iota(a_2 b_1) s^3$$.
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 $$X ↦ s$$.