Is there a unique homomorphism extension for noncommutative rings in this proof?

I read the following in my book:

Let $R$ and $S$ be commutative rings, $\eta$ a homomorphism of $R$ into $S$, $u$ an element of $S$. Let $R[x]$ be the ring of polynomials over $R$ in the indeterminate $x$. Then $\eta$ has one and only one extension to a homomorphism $\eta_u$ of $R[x]$ into $S$ mapping $x$ into $u$. I'm assuming here that $x$ commutes with all of $R$.

Proof: If $A=a_0+a_1x+\cdots+a_nx^n$ then we simply put $$\eta_u(A)=a'_0+a'_1u+\cdots +a'_nu^n$$ where, in general, $a'=\eta(a)$. If $B=b_1+b_1x+\cdots+b_mx^m$, then $AB=p_0+p_1x+\cdots+p_{n+m}x^{n+m}$ where $p_i=\sum_{j+k=i}a_jb_k$. Then $$\eta_u(AB)=p'_0+p'_1u+\cdots+p'_{n+m}u^{n+m}$$ and $$p'_i=\sum_{j+k=i}a'_jb'_k$$ since $\eta$ is a ring homomorphism. On the other hand, \begin{align*} \eta_u(A)\eta_u(B) &= (a'_0+a'_1u+\cdots+a'_nu^n)(b'_0+b'_1+\cdots+b'_mu^m)\\ &=p'_0+p'_1+\cdots+p'_{n+m}u^{n+m}=\eta_u(AB). \end{align*} Still easier is the verification that $\eta_u(A+B)=\eta_u(A)+\eta_u(B)$. NOw we have for $a\in R$ that $\eta_u(a)=a'=\eta(a)$, so $\eta_u$ is an extension of $\eta$. Also $\eta_u(1)=\eta(1)=1$, and $\eta_u(x)=u$. Hence $\eta_u$ is a homomorphism of $R[x]$ which extends $\eta$ and maps $x$ into $u$. Since $R[x]$ is generated by $R$ and $x$ it is the only homomorphism having this property.

My question is, does this proof work equally well if $R$ and $S$ are noncommutative? I don't think commutativity is used anywhere, but I'm wary I may be missing something subtle, else the commutativity assumption in the theorem statement seems unnecessary. If it doesn't work, what must be modified?

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As an aside, noncommutative polynomial rings are strange things. Polynomials over the real Hamilton Quaternions $\mathbb{H}$, for instance, cannot be homomorphically identified as a subring of the functions on $\mathbb{H}.$ –  user17794 Jul 8 '12 at 22:31
Hint: $\rm\ a\,x = x\,a\:\Rightarrow\: \eta(a)\,u = u\,\eta(a),\:$ so it works iff $\rm\:\eta(R)\:$ is central in $\rm\,S.\,$ See here for more. –  Bill Dubuque Jul 8 '12 at 23:23

It depends on what you mean by $R[x]$.
If $x$ commutes with every element of $R$, then $u$ must commute with every element in the image of $\eta$ (you use commutativity in the last computation which is on its own line).
If $x$ is not assumed to commute with every element of $R$, then the corresponding notion of polynomial is more complicated than sums of the form $\sum a_i x^i$, since you also need elements of the form $x^i a_i, x^i a_{ij} x^j, ...$. The corresponding object is usually written $R \langle x \rangle$.
Thank you. I want to assume $x$ commutes with every element of $R$. Does this mean the theorem only works when $\eta(R)\subset C_S(u)$ in $S$? $C_S(u)$ being the centralizer of $u$ in $S$. –  Adelaide Dokras Jul 8 '12 at 22:33