Find all ring morphisms from $\Bbb Z^n$ to $\Bbb Z$.

Question

I tested the case where $n=1$ and I found that there exist just one homomorphism which the idendity.

Answer 1

A ring homomorphism is a linear map over $\Bbb Z$ so it is given by multiplication by a $1\times n$ integer matrix $A$. Being a ring homomorphism means that for every two vectors $u,v$ of dim $n$ we have $A(u\circ v)= (Au)(Av)$ where $\circ$ is the coordinate-wise product of vectors.

Let $A=(a_1,...,a_n)$. Take $u=v=e_i$ (the $i$-th basic vector. Then $a_i=a_ia_i$. So each coordinate of $A$ is $ 0 $ or $1$. Then take $u=v=(1,1,...,1)^T$. Then $(\sum a_i)^2=(\sum a_i)$. So among coordinates of $A$ there are at most $1$ coordinate $1$ and the rest are $0$. There are $n+1$ such vectors $A$ and all of them work.

Answer: all homomorphisms are either $0$ (maps everything to $0$) or $v\mapsto e_i^T v$, $i=1,...,n$.