The kernel of the unique homomorphism $\varphi:\mathbb Z\to K$ is a prime ideal.

I am a graduate student of Mathematics.In the book M. Artin Algebra I found a statement:

Let $K$ be a finite field.Then the kernel of the unique ring homomorphism from $\mathbb Z$ to $K$ is a prime ideal.

My question is:

$1.$ Why is the homomorphism from $\mathbb Z$ to $K$ unique and what is it?(I think it is unique because a ring homomorphism with domain ring $\mathbb Z$ is determined by its value at $1$ now since it is a ring homomorphism so $1$ must be mapped to $1\in K$.)

$2.$ Why is the kernel a prime ideal?

Can someone help me to find answers to these questions?

Solution 1:

The integers are initial among unital rings: for any unital ring $R$, there is one and only one (unital) ring homomorphism $f\colon \mathbb{Z}\to R$, namely the one sending $1$ to $1_R$, the unity of $R$.

(Even if you do not require ring morphisms to be unital, when $R$ is a field the image of $1$ must satisfy $f(1)^2=f(1)$, hence is either $1_R$ or $0_R$. In the latter case, you get the zero morphism.)

Since the image of $\mathbb{Z}$ under the unique nontrivial morphism is contained in a field, it is an integral domain. The isomorphism theorems tell you the image is isomorphic to $\mathbb{Z}/\mathrm{ker}(f)$, and it is a standard theorem that an ideal $I$ of a commutative unital ring $R$ is prime if and only if $R/I$ is an integral domain. So the kernel is a prime ideal.

Solution 2:

1. Any (commutative) ring homomorphism sends $0$ to $0$ and $1$ to $1$ (the most common convention), so there can only one map from $\mathbb{Z}$ to any ring (hence field) $A$. Explicitly, it sends $n \in \mathbb{Z}$ to itself (the sum of $n$ 1's in $A$)

2. The kernel of this unique map is the set of elements that are sent to $0$, which is precisely $(p)$, where $p$ is the characteristics of the field $K$