# How does ring homomorphisms taking $1$ to $1$ implies a diagram is commutative?

Let $$I = \langle 1 − ξ \rangle, \xi$$ is a primitive $$p$$-th root of unity, $$I$$ is a prime ideal of ring of integers $$\mathbb Z[\xi]$$. Consider the following diagram -

Consider the following diagram -

Given, $$i$$ is the inclusion map, $$f$$ is a isomorphism and $$π$$ and $$σ$$ are the natural quotient maps.

I understand , how both $$σ$$ and $$f ◦ π ◦ i$$ are ring homomorphisms taking $$1$$ to $$1$$.

But I could not follow they coincide on $$\mathbb Z$$ and so the diagram is commutative. Can anyone explain or prove in detail pleae?

A ring morphism $$f: \Bbb{Z} \to R$$ that preserves the unit is unique:
Indeed, if $$n \geq 1$$ then $$f(n) = f(\sum_{i=1}^n 1) = \sum_{i=1}^n f(1) = \sum_{i=1}^n 1_R = n1_R$$ and $$f(0) = 0$$ and similarly $$f(n) = n1_R$$ for $$n \leq -1$$.
We conclude that $$f(n) = n 1_R$$ for all $$n \in \Bbb{Z}$$. In your case you have two ring morphisms preserving the unit with domain $$\Bbb{Z}$$ so they must coincide.
• You missed the point for $n=pq+r, f(n)=r$ Aug 11, 2020 at 9:36