TITLE: Invariance of $\mathbb{Z}[x]$ under a self-equivalence of the category of commutative rings with 1
QUESTION [26 upvotes]: Let $\mbox{Rings}$ be the category of commutative rings with $1$.  
Is there an equivalence of categories $F: \mbox{Rings} \to \mbox{Rings}$ such that
$$F(\mathbb{Z}[x])\not\cong \mathbb{Z}[x]?$$

REPLY [12 votes]: Here is a new, short proof, that the category $\text{CRing}$ of commutative rings is rigid:
Lemma: Let $R,S$ be commutative  rings, such that $\text{CAlg}(R), \text{CAlg}(S)$ are equivalent as categories. Then $R,S$ are isomorphic.
Proof: $R$ is the initial object of $\text{CAlg}(R)$ and the comma category $\text{CAlg}(R) / R$ is (via unitalization) equivalent the category $\text{CAlg}'(R)$ of commutative $R$-algebras which are not assumed to be unital. Now $0$ is a zero object of $\text{CAlg}'(R)$, thus null morphisms are defined. A morphism $\alpha : A \times A \to A$ in $\text{CAlg}'(R)$ with $\alpha (\text{id}_A,0) = \text{id}_A, \alpha (0,\text{id}_A) = \text{id}_A$ exists if and only if $(x,y) \mapsto x+y$ is multiplicative, i.e. the multiplication on $A$ is trivial, that is $A$ is just a $R$-module. Thus we have reconstructed $\text{Mod}(R)$ categorically from $\text{CAlg}(R)$. Finally, we can reconstruct $R$ as the center of $\text{Mod}(R)$. $\checkmark$
Theorem: Every self-equivalence of $\text{CRing}$ is isomorphic to the identity.
Proof: Let $F : \text{CRing} \to \text{CRing}$ be an equivalence. Now $\text{CAlg}(R)$ is just the comma category $R \backslash \text{CRing}$, so we get an equivalence $\text{CAlg}(R) \cong \text{CAlg}(F(R))$. The lemma implies $F(R) \cong R$, in particular, $F(\mathbb{Z}[x]) \cong \mathbb{Z}[x]$ as commutative rings. But then we get also an isomorphism of corings and then $F \cong \text{id}$, as explained in my previous proof. $\checkmark$
Remark, however, that this proof is not applicable for automorphisms of $\text{CAlg}(R)$.