Let $R$ be a ring (assumed to be commutative and with unit). What is the center of the category of $R$-algebras, i.e. $\text{Z}(R\text{-Alg})$? This is a commutative monoid. See this recent question for the definition of the center. Actually it is easy to reformulate this using the free $R$-algebra on one generator, which classifies elements in $R$-algebras:

$\text{Z}(R\text{-Alg})$ consists of the polynomials $p \in R[t]$ such that $p(0)=0, p(1)=1$ and in $R[t_1,t_2]$ we have $p(t_1 + t_2) = p(t_1) + p(t_2)$ and $p(t_1 * t_2) = p(t_1) * p(t_2)$. The commutative monoid structure on the center corresponds to $(p_1,p_2) \mapsto p_1(p_2)$. You can also view the center as the endomorphism monoid of the bialgebra $R[t]$.

If we write $p = a_n t^n + ... + a_1 t$ with $a_n \neq 0$, then from $p(t_1 t_2) = p(t_1) p(t_2)$ we get that $a_n$ annihilates every $a_i$ with i < n and that every $a_i$ is idempotent. Thus if $R$ has only trivial idempotents, it follows $p = t^n$ and we have to ask when $(t_1 + t_2)^n = t_1^n + t_2^n$ in order to restrict $n$. Besides, even if $R$ has idempotents, $p(t_1 + t_2) = p(t_1) + p(t_2)$ implies that $(t_1 + t_2)^n = t_1^n + t_2^n$. For $R = \mathbb{Z}$, it is easy to see $n=1$ and thus the center is trivial. For $R = \mathbb{F}_p$, it is easy to see that $n$ is a power of $p$, so that the center is $\cong \mathbb{N}$, generated by the Frobenius. This means that the Frobenius is the universal endomorphism of the rings of characteristic $p$. But what about other rings $R$?

Let $R$ be a ring (assumed to be commutative and with unit). What is the center of the category of $R$-algebras, i.e. $\text{Z}(R\text{-Alg})$? This is a commutative monoid. See this recent question for the definition of the center. Actually it is easy to reformulate this using the free $R$-algebra on one generator, which classifies elements in $R$-algebras:

$\text{Z}(R\text{-Alg})$ consists of the polynomials $p \in R[t]$ such that $p(0)=0, p(1)=1$ and in $R[t_1,t_2]$ we have $p(t_1 + t_2) = p(t_1) + p(t_2)$ and $p(t_1 * t_2) = p(t_1) * p(t_2)$. The commutative monoid structure on the center corresponds to $(p_1,p_2) \mapsto p_1(p_2)$. You can also view the center as the endomorphism monoid of the bialgebra $R[t]$.

If we write $p = a_n t^n + ... + a_1 t$ with $a_n \neq 0$, then from $p(t_1 t_2) = p(t_1) p(t_2)$ we get that $a_n$ annihilates every $a_i$ with i < n and that every $a_i$ is idempotent. Thus if $R$ has only trivial idempotents, it follows $p = t^n$ and we have to ask when $(t_1 + t_2)^n = t_1^n + t_2^n$ in order to restrict $n$. Besides, even if $R$ has idempotents, $p(t_1 + t_2) = p(t_1) + p(t_2)$ implies that $(t_1 + t_2)^n = t_1^n + t_2^n$. For $R = \mathbb{Z}$, it is easy to see $n=1$ and thus the center is trivial. For $R = \mathbb{F}_p$, it is easy to see that $n$ is a power of $p$, so that the center is $\cong \mathbb{N}$, generated by the Frobenius. This means that the Frobenius is the universal endomorphism of the rings of characteristic $p$. But what about other rings $R$?

Let $R$ be a ring (assumed to be commutative and with unit). What is the center of the category of $R$-algebras, i.e. $\text{Z}(R\text{-Alg})$? This is a commutative monoid. See this recent question for the definition of the center. Actually it is easy to reformulate this using the free $R$-algebra on one generator, which classifies elements in $R$-algebras:

$\text{Z}(R\text{-Alg})$ consists of the polynomials $p \in R[t]$ such that $p(0)=0, p(1)=1$ and in $R[t_1,t_2]$ we have $p(t_1 + t_2) = p(t_1) + p(t_2)$ and $p(t_1 * t_2) = p(t_1) * p(t_2)$. The commutative monoid structure on the center corresponds to $(p_1,p_2) \mapsto p_1(p_2)$. You can also view the center as the endomorphism monoid of the bialgebra $R[t]$.

It is straight forward to show that $\text{Z}(\mathbb{Z}\text{-Alg}) = 1$ and $\text{Z}(\mathbb{F}_p\text{-Alg}) \cong \mathbb{N}$, generated by the Frobenius $x \mapsto x^p$. The latter says that the Frobenius (and of course its powers) are the only "universal" endomorphisms of rings of characteristic $p$. What about other rings?

Let $R$ be a ring (assumed to be commutative and with unit). What is the center of the category of $R$-algebras, i.e. $\text{Z}(R\text{-Alg})$? This is a commutative monoid. See this recent question for the definition of the center. Actually it is easy to reformulate this using the free $R$-algebra on one generator, which classifies elements in $R$-algebras:

$\text{Z}(R\text{-Alg})$ consists of the polynomials $p \in R[t]$ such that $p(0)=0, p(1)=1$ and in $R[t_1,t_2]$ we have $p(t_1 + t_2) = p(t_1) + p(t_2)$ and $p(t_1 * t_2) = p(t_1) * p(t_2)$. The commutative monoid structure on the center corresponds to $(p_1,p_2) \mapsto p_1(p_2)$. You can also view the center as the endomorphism monoid of the bialgebra $R[t]$.

If we write $p = a_n t^n + ... + a_1 t$ with $a_n \neq 0$, then from $p(t_1 t_2) = p(t_1) p(t_2)$ we get that $a_n$ annihilates every $a_i$ with i < n and that every $a_i$ is idempotent. Thus if $R$ has only trivial idempotents, it follows $p = t^n$ and we have to ask when $(t_1 + t_2)^n = t_1^n + t_2^n$ in order to restrict $n$. Besides, even if $R$ has idempotents, $p(t_1 + t_2) = p(t_1) + p(t_2)$ implies that $(t_1 + t_2)^n = t_1^n + t_2^n$. For $R = \mathbb{Z}$, it is easy to see $n=1$ and thus the center is trivial. For $R = \mathbb{F}_p$, it is easy to see that $n$ is a power of $p$, so that the center is $\cong \mathbb{N}$, generated by the Frobenius. This means that the Frobenius is the universal endomorphism of the rings of characteristic $p$. But what about other rings $R$?