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

Question

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?

Answer 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.

Answer 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 $K$

Answer 3

The following is an answer to your question:

Theorem

If $K$ be a finite field of characteristic $p$,where $p$ is a prime,then $K$ contains a subfield isomorphic to $\mathbb F_p$.

Proof: Let $\varphi:\mathbb Z \to K$ be defined by $\varphi(1)=1$ which is the unique ring homomorphism from $\mathbb Z$ to $K$.

Now,$\varphi(\mathbb Z)$ is a subring of $K$,but $K$ being a field is also an integral domain,so any subring of $K$ is also an integral domain.Now by first isomorphism theorem $\mathbb Z/Ker(\varphi)\simeq \varphi(\mathbb Z)$ which is an integral domain,so $Ker(\varphi)$ is a prime ideal of $\mathbb Z$.Also $Ker(\varphi)$ is non-zero as $\mathbb Z$ is finite and $K$ is infinite.So,$Ker(\varphi)=(q)$ for some prime $q$.Thus $\varphi(\mathbb Z)\simeq \mathbb {Z/qZ}=\mathbb F_q$.But observe that $\varphi(q)=0\implies q1=0\implies p\mid q$.So,$q=p$.Hence $K$ contains a copy of $\mathbb F_p$.