Answer to Question #350805 in Abstract Algebra for MIS

Question

Let K be a field and f : Z → K the homomorphism of

integers into K.

a) Show that the kernel of f is a prime ideal. If f is an embedding,

then we say that K has characteristic zero.

b) If kerf f= {0}, show that kerf is generated by a prime number

p. In this case we say that K has characteristic p.

Answer 1

Recall that kernel of any ring homomorphism is an ideal. Hence $Ker(f)$ is an ideal of $\mathbb{Z}$. Then the ideal $Ker(f)$ is a prime ideal if and only if $\mathbb{Z}/Ker(f)$ is an integral domain. Indeed if $xKer(f),yKer(f)$ are in $\mathbb{Z}/Ker(f)$ and $xKer(f)yKer(f)=Ker(f)$. Then $xy\in Ker(f)$, then $f(xy)=f(x)f(y)$. But $K$ is a field either $f(x)=0$ or $f(y)=0$. i.e. either $x\in Ker(f)$ of $y\in Ker(f)$. Hence $\mathbb{Z}/Ker(f)$ is an integral domain.

On the other hand if $Ker(f)=\{0\}$, then $f(n)\ne 0$ for any $n\ne0$. In particular $0\ne f(1)=f(1)f(1)$ and $f(1)$ is an element of the field implies multiplying from right by its inverse we have $f(1)$ is the multiplicative identity of the field $K$. Hence for any integer n and any non-zero element $a\in K$ we have $na=f(n)\ne 0$ as product of two non-zero element in a field is non-zero. Hence characteristic of the field $K$ is zero.

(b) On the other hand since $\mathbb{Z}/Ker(f)$ is an integral domain when $Ker(f)\ne0$, the ideal $Ker(f)$ must be a prime ideal. But the only prime ideals of the ring $\mathbb{Z}$ are the ones generated by prime numbers. Hence $\mathbb{Z}/Ker(f)\cong\mathbb{Z}_p$. Hence $K$ has a smallest subfield isomorphic to $\mathbb{Z}_p$. It follows that $K$ has characteristic $p$ for the prime $p$.