# Proof Verification: $\text{Hom}(\mathbb Z[x],S)=S$ (as rings)

How to prove $$\text{Hom}(\mathbb Z[x],S)=S$$ (as rings), where S is any ring?

My attempt: took an element $$b$$ in $$S$$, defined a map , $$b: \mathbb Z[x]\to S$$ which maps $$f(x)$$ to $$f(b)$$. Clearly $$b$$ is a ring homomorphism, hence we proved one side inclusion. for other side, take $$b$$ in $$\text{Hom}(\mathbb Z[x],S)$$ with the same mapping, since $$b$$ is a ring homomorphism from $$\mathbb Z[x]$$ to $$S$$, clearly $$f(b)$$ is in $$S$$. Since $$f(x)$$ is in $$\mathbb Z[x]$$, surely $$b$$ is in $$S$$.

Is my proof valid?

Also in case to show the generalised result for $$n$$ variables, can I use induction?

• What is $S$?$\hspace{1pt}$ Jul 7 '20 at 18:17
• S is any ring. Forgot to mention. Sorry! Jul 7 '20 at 18:19
• Then this claim cannot be true by the existence of non-commutative rings. Jul 7 '20 at 18:20
• @ElliotG The term as rings in the statement should refer to the $\operatorname{Hom}$, i.e. we consider ring homomorphisms from $\Bbb Z[x]$ to $S$. The equality sign should then refer to a bijection, i.e. isomorphism in the category $\bf{Set}$. Jul 7 '20 at 18:21
• @ElliotG Could you please elaborate on it a bit? I am new to study of rings. Jul 7 '20 at 18:28

As I understand, you give a map let's call it $$\phi : S \rightarrow Hom(\mathbb{Z}[x], S)$$ which takes $$b \in S$$ to the map $$\phi_b: \mathbb{Z}[x] \rightarrow S$$ which sends a polynomial $$p \in \mathbb{Z}[x]$$ to $$\phi_b(p) = p(b)$$. Note $$S$$ has an identity element, call it $$e$$, and $$\phi_b(1) = e$$. To check this map is well defined, for each $$b \in S$$ we should check $$\phi_b$$ is indeed a ring homomorphism. So we would show things like $$\phi_b(p\cdot q) = (p\cdot q)(b) = p(b)\cdot q(b) = \phi_b(p) \cdot \phi_b(q)$$.
To show that this map is indeed a bijection we can construct an inverse. Let's make a map $$\psi : Hom(\mathbb Z[x], S) \rightarrow S$$. We want $$\psi(\phi_b) = b$$ so a natural choice for $$\psi$$ is a map sending any homomorphism $$f : \mathbb Z[x] \rightarrow S$$ to $$\psi(f) = f(x)$$ where $$x \in \mathbb Z[x]$$ is a polynomial.
We have $$\psi(\phi_b) = \phi_b(x) = b$$ as desired. We just need to show that $$\phi_{\psi(f)} = f$$ for any $$f \in Hom(\mathbb Z[x], S)$$. So take any $$p \in \mathbb Z[x]$$, we have $$\phi_{\psi(f)}(p) = p(\psi(f)) = p(f(x)) = f(p)$$. The last equality $$p(f(x)) = f(p)$$ should be justified carefully! And this is the part which (I think) will break when you try to generalise this to $$n$$ variables.
Addendum for non-unital rings. If $$S$$ doesn't have an identity then $$Hom(\mathbb Z, S)$$ may be trivial when $$S$$ is non-trivial. E.g. Take $$S = 2\mathbb Z$$ a non-unital ring contained in $$\mathbb Z$$, then take $$f \in Hom(\mathbb Z[x], S)$$. We have $$f(1) = 2n$$ for some $$n \in \mathbb Z$$. Since $$f$$ is a homomorphism of rings, $$2n = f(1) = f(1^2) = f(1)^2 = 4n^2$$ hence $$n = 0$$. So $$Hom(\mathbb Z[x], S)$$ is the zero ring but $$S$$ is not the zero ring. To avoid these pathologies we have assumed $$S$$ has an identity element.
• Which part of the argument relies on $1 \in S$? Jul 8 '20 at 12:14
• When you say $p(b)$, which element of $S$ is that if $p(x)=x+1$ and $1\notin S$? Jul 8 '20 at 15:05