Fields

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In this post we discuss fields. In particular, our goal is to introduce Galois theory.

Prime Fields

Let us first recall the following definition.

Definition 1 A commutative division ring is called a field. (§Field of Fractions, ⁋Definition 3)

We usually denote a field by $\mathbb{K}$. Since a field is a ring satisfying certain conditions, we can collect all fields into a subcategory $\Field$ of $\Ring$. That is, $\Field$ is a full subcategory of $\Ring$.

Earlier we defined every object of $\Ring$ to have $1$, and every ring homomorphism to preserve $1$. (§Definition of a Ring, ⁋Definition 3) Thus any subring of a ring $A$ also has a multiplicative identity, which is exactly the multiplicative identity $1$ of $A$. From this we also know that $\mathbb{Z}$ is an initial object of $\Ring$.

The following proposition, though obvious, is quite interesting: it shows that a ring homomorphism between fields is either an inclusion or the zero map.

Proposition 2 Let $\mathbb{K},\mathbb{K}'$ be arbitrary fields, and consider a ring homomorphism $f:\mathbb{K} \rightarrow \mathbb{K}'$. Then $f$ is either the zero map or an inclusion.

Proof

$\ker f$ is an ideal of $\mathbb{K}$. Now see §Field of Fractions, ⁋Proposition 4.

Meanwhile, in the post §Field of Fractions we obtained $\mathbb{Q}$ as the total field of fractions $\Frac(\mathbb{Z})$ of $\mathbb{Z}$. The following example gives a field different from $\mathbb{Q}$.

Example 3 Fix a prime $p$. Then the quotient ring $\mathbb{F}_p := \mathbb{Z}/p\mathbb{Z}$ of $\mathbb{Z}$ is a field.

To verify this, let an arbitrary nonzero $a+p\mathbb{Z}\in \mathbb{F}_p$ be given. Since $\gcd(a,p)=1$, by §Integral Domains, ⁋Corollary 9 there exist $x, b\in \mathbb{Z}$ satisfying

\[1=ax+bp\]

and it suffices to take the remainder of both sides upon division by $p$.

Collectively we call these prime fields. Then the following holds.

Theorem 4 Suppose a ring $A$ has some subfield. Then $A$ has a unique prime subfield $\mathbb{P}$. In this case, $\mathbb{P}$ lies in the center of $A$ and is contained in every subfield of $A$.

Proof

Let $\mathbb{K}$ be a subfield of $A$. Let $C$ be the center of $A$; then $\mathbb{K}':= \mathbb{K} \cap C$ is also a subfield of $A$.

Consider the unique ring homomorphism $f: \mathbb{Z} \to A$ and let its kernel be $\mathfrak{p}$. Then every subring of $A$ contains $\im(f)$, so in particular $\mathbb{K}'$ also contains $\im(f)$. Now since $\mathbb{K}'$ is a field it is an integral domain, and therefore $(0)$ is a prime ideal of $\mathbb{K}'$; hence by §Field of Fractions, ⁋Proposition 9 we have either $\mathfrak{p} = (0)$ or $\mathfrak{p} = (p)$ for some prime $p$.

If $\mathfrak{p}=0$, then $f$ is an embedding of $\mathbb{Z}$ into $\mathbb{K}'$, and thus by §Field of Fractions, ⁋Theorem 1 we obtain $\mathbb{Q}\hookrightarrow\mathbb{K}'$.

If $\mathfrak{p} = (p)$, then by the first isomorphism theorem the homomorphic image of $\mathbb{Z}/p\mathbb{Z} = \mathbb{F}_p$ defines a subfield of $\mathbb{K}'$.

Thus in both cases we see that a prime field $\mathbb{P}$ exists inside $\mathbb{K}'$, and this $\mathbb{P}$ is a subfield of $A$ and lies in the center $C$.

Now consider another subfield $L$ of $A$. Then $\mathbb{P} \cap L$ is a subfield of $A$ and a subset of $\mathbb{P}$, so $\mathbb{P} \cap L = \mathbb{P}$, and therefore $\mathbb{P} \subseteq L$. That is, every subfield contains $\mathbb{P}$, so $\mathbb{P}$ is contained in every subfield of $A$ and is unique as a prime field.

Since we are mostly interested in the case where the ring $A$ is a field, the hypothesis of the above theorem is not very restrictive. For instance, we obtain the following corollaries.

Corollary 5 A field $\mathbb{K}$ has a unique prime subfield $\mathbb{P}$, and $\mathbb{P}$ is the smallest of the subfields of $\mathbb{K}$.

Corollary 6 A necessary and sufficient condition for a field $\mathbb{K}$ to be a prime field is that $\mathbb{K}$ has no subfield other than itself.

Now from this we know that whenever an arbitrary field $\mathbb{K}$ is given, there exist a prime field $\mathbb{P}$ and an injection $\mathbb{P}\hookrightarrow \mathbb{K}$ defined in a unique way. In general we call a ring homomorphism $f:A \rightarrow B$ an extension if it is injective, so this can be rewritten as the following slogan:

Every field is, in a unique way, an extension of some prime field.

Characteristic

Using prime fields we can define the characteristic of a ring. Suppose a ring $A$ has some subfield. Then by Theorem 4 this $A$ has a unique prime subfield $\mathbb{P}$. Moreover, looking at the proof of Theorem 4, the way to obtain this $\mathbb{P}$ is given quite explicitly: since $\mathbb{P}$ must contain the multiplicative identity $1$ of $A$, from this

\[n.x=\underbrace{1+1+\cdots+1}_\text{\scriptsize$n$ times}\]

must also belong to $\mathbb{P}$, and if this ever becomes $0$ then $\mathbb{P}$ is isomorphic to $\mathbb{F}_p$, otherwise $\mathbb{P}$ is $\mathbb{Q}$. Taking this as intuition we define the following.

Definition 7 Suppose a ring $A$ has a subfield. Then we define the positive integer $n$ generating the kernel of $f:\mathbb{Z} \rightarrow A$ in the proof of Theorem 4 as the characteristic of the ring $A$, and write it as $\ch(A)$.

That is, if $A$ has a subfield and hence a prime subfield $\mathbb{P}$, then $\ch A=0$ is equivalent to $\mathbb{P}\cong \mathbb{Q}$, and $\ch A=p$ is equivalent to $\mathbb{P}\cong \mathbb{F}_p$; no other cases exist.

Proposition 8 Let $A$ be a nonzero ring. The following hold:

The characteristic of $A$ is $0$ if and only if for every $n\neq 0$ the map $x\mapsto n.x$ is a bijection from $A$ to $A$.
For a prime $p$, $\ch(A)=p$ is equivalent to $p.x=0$ for all $x\in A$.

Proof

This follows because for any integer $n$ and element $x\in A$, considering the unique ring homomorphism $f$ from $\mathbb{Z}$ to $A$ we have the identity

\[n.x=\underbrace{1+1+\cdots+1}_\text{\scriptsize$n$ times}=f(n)x\]

The objects we are familiar with are of course the case $\ch(A)=0$. If $\ch(A)=p$ then several interesting results hold. To see this we first prove the following lemma.

Lemma 9 Fix a prime $p$. Then for an integer $i$ with $1 \leq i \leq p-1$, the binomial coefficient $\binom{p}{i}$ is a multiple of $p$.

Proof

We prove this by induction on $i$. For $i = 1$ we have $\binom{p}{1} = p$, so there is nothing more to prove. Now assume for $2 \leq i \leq p-1$ that $\binom{p}{i-1}$ is divisible by $p$. Then from the identity

\[i \cdot \binom{p}{i} = (p - i + 1) \cdot \binom{p}{i-1}\]

the right-hand side is a multiple of $p$ by the induction hypothesis. Since $i$ is not a multiple of $p$, for this identity to hold $\binom{p}{i}$ must be a multiple of $p$.

Now the following holds.

Theorem 10 Let $A$ be a commutative ring of characteristic $p>0$. Then the function

\[\Frob_p: A \rightarrow A;\qquad a \mapsto a^p\]

is an endomorphism of $A$. That is, the two identities

\[(a + b)^p = a^p + b^p,\qquad (ab)^p = a^p b^p\]

hold for all $a,b\in A$.

Proof

The multiplicative identity requires no further proof since $A$ is commutative. For the first identity, considering the binomial expansion

\[(a + b)^p = a^p + b^p + \sum_{i = 1}^{p-1} \binom{p}{i}. a^i b^{p-i}\]

by Lemma 9 we have $\binom{p}{i} \equiv 0 \pmod{p}$ for all $i$ with $1 \leq i \leq p-1$, and therefore by Proposition 8

\[\binom{p}{i}. a^i b^{p-i} = 0\]

holds for all $1\leq i\leq p-1$. From this we obtain the desired equality.

Definition 11 When a commutative ring $A$ has characteristic $p$, the function of Theorem 10

\[A \to A,\qquad a \mapsto a^p\]

is called the Frobenius endomorphism.

Then the $f$-fold composition $\Frob_p^f$ of the Frobenius endomorphism is given by the formula

\[a\mapsto a^{p^f}\]

and since this is still an endomorphism of $A$, in particular

\[(a_1+\cdots+a_n)^{p^f}=a_1^{p^f}+\cdots+a_n^{p^f}\]

always holds and each $a_k^{p^f}$ is an element of $A$. More generally, for a subset $S$ of $A$, we write the image of $S$ under the $f$-fold composition of the Frobenius endomorphism as $S^{p^f}$.

We introduce a few more notations. For a subring $K$ of $A$ and an arbitrary subset $S\subset A$, we denote by $K[S]$ the subring of $A$ generated by $K\cup S$. If $A$ is a field, then by §Field of Fractions, ⁋Theorem 1 from $K[S]\hookrightarrow A$ a ring homomorphism

\[\Frac(K[S]) \rightarrow A\]

is induced, and through this we can regard $\Frac(K[S])$ as a subfield of $A$. We agree to write this as $K(S)$. Then $K(S)$ is by definition the smallest subfield of $A$ containing $K\cup S$.

Then in this notation the following holds.

Proposition 12 Fix a commutative ring $A$ of characteristic $p>0$, a subring $K$ of $A$, a subset $S$ of $A$, and a positive integer $f$.

$K[S]^{p^f} = K^{p^f}[S^{p^f}]$ holds. If $A$ is a field then $K(S)^{p^f} = K^{p^f}(S^{p^f})$ also holds.
If the set $\{a_i\}_{i \in I}$ generates $K[S]$ as a $K$-module, then the set $\{a_i^{p^f}\}_{i \in I}$ generates $K[S]^{p^f}$ as a $K^{p^f}$-module.

Proof

The left-hand side of the equality is $\Frob_p^f(K[S])$, so the claim is that the subring of $A$ generated by $K^{p^f}\cup S^{p^f}=\Frob_p^f(K)\cup\Frob_p^f(S)$ equals $\Frob_p^f(K[S])$, which is obvious.
Since $K[S]$ is generated by $K$-linear combinations of the $a_i$, from the binomial expansion and Theorem 10 we know that $K[S]^{p^f}$ is generated by $K^{p^f}$-linear combinations of the respective $a_i^{p^f}$.

Perfect Fields

Now we define the following.

Definition 13 Consider a ring $A$ of characteristic $p \neq 0$ and the Frobenius map $\Frob_p : A \to A$.

If $\Frob_p$ is surjective we call $A$ a semi-perfect ring.
If $\Frob_p$ is bijective we call $A$ a perfect ring.

If $A$ is a perfect ring then $\Frob_p^{-1}$ is also an (auto)morphism from $A$ to $A$, so it is appropriate to write it as $a\mapsto a^{p^{-f}}$ or $a^{1/p^f}$. As before, for a subset $S$ of $A$ we write the image of $S$ under $(\Frob_p^{-1})^f$ as $S^{1/p^f}$.

Definition 14 For a commutative ring $A$ of characteristic $p \ne 0$, a perfect closure is a pair $(A^{1/p^\infty}, \phi)$ satisfying the following two conditions.

$A^{1/p^\infty}$ is a perfect ring of characteristic $p$.
$\phi : A \to A^{1/p^\infty}$ is a ring homomorphism satisfying the following universal property.

Whenever a perfect ring $B$ of characteristic $p$ and a ring homomorphism $f:A \rightarrow B$ are given, there exists a unique ring homomorphism $\hat{f}: A^{1/p^\infty} \rightarrow B$ such that $f=\hat{f}\circ\phi$.

If a perfect closure of $A$ exists then it is uniquely determined by definition. The following theorem shows that a ring of characteristic $p\neq 0$ always has a perfect closure.

Theorem 15 For a commutative ring $A$ of characteristic $p \ne 0$, the perfect closure $(A^{1/p^\infty}, \phi)$ of $A$ exists. Moreover the following hold.

The kernel of $\phi$ is the set of all nilpotent elements of $A$.
For any $x \in A^{1/p^\infty}$, there exists some $n > 0$ such that $x^{p^n} \in \phi(A)$.

Instead of a proof we briefly outline the idea. Intuitively, the condition that $\Frob_p$ is bijective means that every element of $A$ has a (unique) $p$th root. Therefore the perfect closure of $A$ is obtained by adjoining $p$th roots of every element of $A$.

For this we consider the directed system

\[A_0=A\overset{\Frob_p}{\longrightarrow} A_1=A\overset{\Frob_p}{\longrightarrow} A_2=A\rightarrow\cdots\]

Taking the direct limit of this directed system

\[\varinjlim A=\left(\coprod A\right)\bigg/{\sim},\qquad \text{$a_i\sim a_j\iff \Frob_p^{k-i}(a_i)=\Frob_p^{k-j}(a_j)$ for some $k\geq i,j$}\]

an arbitrary element of $\varinjlim A$ is represented by suitable $i$ and $a_i\in A_i$. We denote this by

\[[a_i\in A_i]\]

and define the canonical morphisms by

\[\phi_i: A_i \rightarrow \varinjlim A;\qquad a_i\mapsto [a_i\in A_i]\]

Our claim is that $A^{1/p^\infty}=\varinjlim A$ and $\phi=\phi_0$ satisfy the desired conditions.

This conforms to our intuition when we write out the multiplication in $\varinjlim A$ explicitly. By definition, the product of two elements $[a_i\in A_i]$ and $[a_j\in A_j]$ in $\varinjlim A$ is, assuming without loss of generality that $j\geq i$, given by the formula

\[[\Frob_p^{j-i}(a_j)\in A_j]=[a_i^{p^{j-i}}a_j\in A_j]\]

Our intuition is that $[a_j\in A_j]$ corresponds to $a_j^{1/p^j}$ in $\varinjlim A$, which is to say

\[[a_j\in A_0]=[a_j\in A_j]^{p^j}\]

Then the right-hand side is $[a_j^{p^j}\in A_j]=[\Frob_p^j(a_j)]$, so the above identity holds.

What should be noted, as can be seen from the first result of the theorem, is that $\phi: a\mapsto [a\in A_0]$ is not injective, which is also obvious from a small calculation.

For unified notation it is sometimes convenient to set the case $\ch A=0$ to $1$. For this purpose we define the characteristic exponent of a field $\mathbb{K}$ to be $1$ when $\ch A=0$, and $p$ when $\ch A=p>0$.

Proposition 16 Consider a field $\mathbb{K}$ of characteristic exponent $q$. Then for each $f\geq 0$, the map $x\mapsto x^{q^f}$ induces an isomorphism from $\mathbb{K}$ onto some subfield $\mathbb{K}^{q^f}$ of $\mathbb{K}$.

Proof

If $p=0$ then $q=1$ and therefore the given function is the identity, so there is nothing to prove. If $q \ne 1$ then $q$ is a positive prime, so the characteristic of $K$ is a positive prime $p$ and $q = p$. Then the given function is the $f$-fold composition of the Frobenius endomorphism and is a surjective ring homomorphism from $\mathbb{K}$ onto $\mathbb{K}^{q^f}$. That this is injective is obvious because $\mathbb{K}$ is a field and therefore has no nilpotent elements.

Definition 17 For a field $\mathbb{K}$ of characteristic exponent $q$, if the function of Proposition 16 is an isomorphism from $\mathbb{K}$ onto $\mathbb{K}$ we call $\mathbb{K}$ a perfect field; otherwise we call it an imperfect field.

That is, for a field $\mathbb{K}$ to be perfect, either (1) the characteristic exponent is $1$, or (2) the characteristic exponent is $q$ and it is a perfect ring in the sense of Definition 13.

Then the following holds.

Proposition 18 For a field $\mathbb{K}$, the following hold.

If $\ch(\mathbb{K})=0$ then $\mathbb{K}$ is perfect.
If $\mathbb{K}$ is a finite set then $\mathbb{K}$ is perfect.

Proof

It suffices to show only the second claim; we already observed in the proof of that proposition that the function of Proposition 16, $\mathbb{K} \rightarrow \mathbb{K}^{q^f}\subset \mathbb{K}$, is always injective regardless of conditions on $\mathbb{K}$. Now if $\mathbb{K}$ is a finite set, then by its size we must have $\mathbb{K}^{q^f}=\mathbb{K}$.

Meanwhile, for an arbitrary commutative ring $A$ and its polynomial algebra $E=A[\x_i]_{i\in I}$, we have examined the module of Kähler differentials $\Omega_{E/A}$. (§Differential Modules, ⁋Example 10) According to this structure, for an arbitrary polynomial $f\in E$ the element $df\in\Omega_{E/A}$ is given by the formula

\[df=\sum_{i\in I} \frac{\partial f}{\partial\x_i}\mathop{d\x_i}\]

Therefore, if $A$ has characteristic $p$ then the image of the polynomial $\x_i^p$ in $\Omega_{E/A}$ will be $0$. The following proposition refines this observation more completely.

Proposition 19 For a commutative ring $A$ and polynomial algebra $E=A[\x_i]_{i\in I}$, let $S$ be the subset of $E$ consisting of polynomials $f$ satisfying $df=0$. Then the following hold.

If $\ch(A)=0$ then $S=A$.
If $\ch(A)=p$ then $S=A[\x_i^p]_{i\in I}$, and moreover if $A$ is perfect then $A=E^p$ holds.

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Prime Fields

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