The prime subring and the prime field
Definition (Prime field)
Let \(F\) be a field. We define the prime field of \(F\) as the intersection of all subfields of \(F\).
Definition (Prime subring)
Let \(R\) be a ring. The prime subring (or prime ring) of \(R\) is the intersection of all subrings of \(R\); that is, the smallest subring of \(R\).
Note (The prime ring of a field lives inside its prime field)
If \((F, +, \cdot)\) is a field, then it is a ring; since every subfield of \(F\) is in particular a subring, the prime ring of \(F\) is a subring of the prime field of \(F\).
Example (Well-known prime subrings)
The following examples illustrate the definition:
- \(\mathbb{Z}\) is the prime subring of \(\mathbb{Z}\), and \(\mathbb{Z}_{n}\) is the prime subring of \(\mathbb{Z}_{n}\).
- Let \(R\) be a ring and consider \(M_{n \times n}(R)\). Then \(\{ z \cdot I_{n} :\ z \in \mathbb{Z} \}\) is the prime ring of \(M_{n \times n}(R)\).
- If \(R_{1} \subseteq R_{2}\) are rings (respectively, fields), then \(R_{1}\) and \(R_{2}\) have the same prime ring (respectively, prime field).
Note (The chain of number rings)
Note that \(\mathbb{Z} \subseteq \mathbb{Q} \subseteq \mathbb{R} \subseteq \mathbb{C} \subseteq \mathbb{H}\) (as rings):
- they all have the same prime ring, \(\mathbb{Z}\);
- \(\mathbb{Q}\) is the prime field of \(\mathbb{Q}\), \(\mathbb{R}\) and \(\mathbb{C}\).
The characteristic
Definition (Characteristic of a ring)
If \(R\) is a ring and \(R_{0}\) is its prime ring, we define
\[\operatorname{char}(R) = \begin{cases} |R_{0}| & \text{if } R_{0} \text{ is finite}, \\ 0 & \text{otherwise}. \end{cases}\]Theorem (The prime subring is \(\mathbb{Z}\) or \(\mathbb{Z}_{n}\))
Let \(R\) be a ring.
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The prime subring \(R_{0}\) of \(R\) is isomorphic to \(\mathbb{Z}\) or to \(\mathbb{Z}_{n}\), and in that case
\[\operatorname{char}(R) = \begin{cases} n & \text{if } R_{0} \cong \mathbb{Z}_{n}, \\ 0 & \text{if } R_{0} \cong \mathbb{Z}. \end{cases}\] -
If \(D\) is an integral domain, then \(\operatorname{char}(D)\) is \(0\) or a prime number \(p\).
Proof: Part 1. Let \(R_{0}\) be the prime ring of \(R\) and let \(f_{R} : \mathbb{Z} \longrightarrow R\) be such that \(n \longmapsto n \cdot 1_{R}\), which is a ring homomorphism (in fact, the only one; see the exercise on homomorphisms from \(\mathbb{Z}\)). We claim that \(\operatorname{Im}(f_{R}) = R_{0}\): on the one hand, \(\operatorname{Im}(f_{R})\) is a subring of \(R\); on the other, every subring \(S\) of \(R\) contains \(1_{R}\) and is closed under sums and differences, so it contains \(n \cdot 1_{R}\) for every \(n \in \mathbb{Z}\), that is, \(\operatorname{Im}(f_{R}) \subseteq S\). Then \(\operatorname{Im}(f_{R})\) is the smallest subring of \(R\), that is, \(\operatorname{Im}(f_{R}) = R_{0}\). By the first isomorphism theorem,
\[R_{0} = \operatorname{Im}(f_{R}) \cong \mathbb{Z}/\operatorname{Ker}(f_{R}).\]Note that \(\operatorname{Ker}(f_{R})\) is an ideal of \(\mathbb{Z}\), and since \(\mathbb{Z}\) is a PID, we have that \(\operatorname{Ker}(f_{R}) = (n)\) with \(n \in \mathbb{N}\). If \(n = 0\), then \(\operatorname{Ker}(f_{R}) = \{0\}\), so that \(R_{0} \cong \mathbb{Z}\), which is infinite, and \(\operatorname{char}(R) = 0\). If \(n \neq 0\), then \(R_{0} \cong \mathbb{Z}/(n) \cong \mathbb{Z}_{n}\), which has \(n\) elements, and \(\operatorname{char}(R) = n\).
Part 2. It was proved in the homework; it is left as an exercise.
The Frobenius homomorphism and Fermat’s little theorem
Theorem (Frobenius: raising to the \(p\)-th power is a homomorphism)
Let \(R\) be a commutative ring such that \(\operatorname{char}(R) = p > 0\) with \(p\) prime. Then \(\varphi : R \to R\) such that \(x \mapsto x^{p}\) is a homomorphism. In particular,
\[(x + y)^{p} = x^{p} + y^{p} \quad \text{for all } x, y \in R.\]Proof: Note first that \(\varphi(1) = 1^{p} = 1\), and that, since \(R\) is commutative,
\[\varphi(x \cdot y) = (x y)^{p} = x^{p} y^{p} = \varphi(x) \cdot \varphi(y).\]It remains to prove that \(\varphi(x + y) = \varphi(x) + \varphi(y)\), that is, that \((x + y)^{p} = x^{p} + y^{p}\).
Let \(a, b \in R\). By Newton’s binomial theorem, valid in commutative rings,
\[(a + b)^{p} = \sum_{i=0}^{p} \binom{p}{i} a^{i} b^{p-i},\]where \(\binom{p}{i} a^{i} b^{p-i}\) denotes the corresponding integer multiple. We claim that if \(0 < i < p\), then \(\binom{p}{i}\) is a multiple of \(p\): from the identity
\[i!\,(p - i)!\, \binom{p}{i} = p!\]we see that \(p\) divides the right-hand side; but \(p\) is prime and divides neither \(i!\) nor \((p-i)!\), since all their factors are smaller than \(p\); hence \(p \mid \binom{p}{i}\). Now, since \(\operatorname{char}(R) = p\), we have that \(p \cdot 1_{R} = 0\), and if \(\binom{p}{i} = p \cdot t\) with \(t \in \mathbb{Z}\), then
\[\binom{p}{i} \cdot a^{i} b^{p-i} = \big( \binom{p}{i} \cdot 1_{R} \big)\, a^{i} b^{p-i} = \big( t \cdot (p \cdot 1_{R}) \big)\, a^{i} b^{p-i} = 0.\]That is, in the binomial sum only the terms \(i = 0\) and \(i = p\) survive, so that \((a + b)^{p} = a^{p} + b^{p}\).
Note (Frobenius on finite fields and its iterates)
Three remarks on the previous homomorphism:
- \(\varphi\) is called the Frobenius homomorphism.
- If \(R\) is a finite field, then \(\operatorname{Ker}(\varphi) = \{0\}\) (the kernel is a proper ideal of a field), so that \(\varphi\) is injective. Since \(R\) is finite, \(\varphi\) is also surjective, and hence \(\varphi\) is an automorphism.
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\(\varphi^{n} : R \to R\), \(x \mapsto x^{p^{n}}\), is a homomorphism for every \(n \in \mathbb{N}\), being a composition of homomorphisms. In particular,
\[(a + b)^{p^{n}} = a^{p^{n}} + b^{p^{n}} \quad \text{for all } a, b \in R,\ n \in \mathbb{N}.\]
Theorem (Fermat’s little theorem)
Let \(p\) be prime. Then \(n^{p} \equiv n \pmod{p}\) for every \(n > 0\).
Proof: We proceed by induction on \(n\). The base case \(n = 1\) is trivial, since \(1^{p} = 1\). Now, suppose as inductive hypothesis that \(n^{p} \equiv n \pmod{p}\) for some \(n \in \mathbb{N}\). The ring \(\mathbb{Z}_{p}\) is its own prime ring, so \(\operatorname{char}(\mathbb{Z}_{p}) = p\). Using the Frobenius homomorphism on \(\mathbb{Z}_{p}\) and the inductive hypothesis, we have that
\[(n + 1)^{p} = n^{p} + 1^{p} = n + 1 \pmod{p},\]which proves the inductive step. Conclude the result.