Field of Prime Characteristic has Unique Prime Subfield

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Theorem

Let $F$ be a field whose characteristic is $p$.

Then there exists a unique $P \subseteq F$ such that:

$(1): \quad P$ is a subfield of $F$
$(2): \quad P \cong \Z_p$.

That is, $P \cong \Z_p$ is a unique minimal subfield of $F$, and all other subfields of $F$ contain $P$.


This field $P$ is called the prime subfield of $F$.


Proof

Let $\left({F, +, \times}\right)$ be a field such that $\operatorname{Char} \left({F}\right) = p$ whose unity is $1_F$.

Let $P$ be a prime subfield of $F$.

From Intersection of Subfields, this has been show to exist.


We can consistently define a mapping $\phi: \Z_p \to F$ by:

$\forall n \in \Z_p: \phi \left({\left[\!\left[{n}\right]\!\right]_p}\right) = n \cdot 1_F$


Suppose $a, b \in \left[\!\left[{n}\right]\!\right]_p$.

Then $a = n + k_1 p, b = n + k_2 p$

So:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \phi \left({a}\right)\) \(=\) \(\displaystyle \phi \left({n + k_1 p}\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left({n + k_1 p}\right) \cdot 1_F\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle n \cdot 1_F + k_1 p \cdot 1_F\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle n \cdot 1_F\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    

and similarly for $b$, showing that $\phi$ is well-defined.


Let $C_a, C_b \in \Z_p$.

Let $a \in C_a, b \in C_b$ such that $a = a' + k_a p, b = b' + k_b p$.

Then:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \phi \left({C_a}\right) + \phi \left({C_b}\right)\) \(=\) \(\displaystyle \phi \left({a' + k_a p}\right) + \phi \left({b' + k_b p}\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left({a' + k_a p}\right) \cdot 1_F + \left({b' + k_b p}\right) \cdot 1_F\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left({a' + k_a p + b' + k_b p}\right) \cdot 1_F\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Integral Multiple Distributes over Ring Addition          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left({a' + b'}\right) \cdot 1_F + \left({\left({k_a p + k_b p}\right) \cdot 1_F}\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left({a' + b'}\right) \cdot 1_F\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \phi \left({C_a +_p C_b}\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    

Similarly for $\phi \left({C_a}\right) \times \phi \left({C_b}\right)$.


So $\phi$ is a ring homomorphism.

From Homomorphism from Field Either Monomorphism or Zero Homomorphism, it follows that $\phi$ is a ring monomorphism.


Thus it follows that $P = \operatorname{Im} \left({\phi}\right)$ is a subfield of $F$ such that $P \cong \Z_p$.


  • Let $K$ be a subfield of $F$, and $P = \operatorname{Im} \left({\phi}\right)$ as defined above.

We know that $1_F \in K$.

It follows that $1_F \in K \implies P \subseteq K$.

Thus $K$ contains a subfield $P$ such that $P$ is isomorphic to $\Z_p$.


  • The uniqueness of $P$ follows from the fact that if $P_1$ and $P_2$ are both minimal subfields of $F$, then $P_1 \subseteq P_2$ and $P_2 \subseteq P_1$, thus $P_1 = P_2$.

$\blacksquare$

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