Fields

In the posts of this category we discuss fields. In particular, the purpose of this category is to introduce Galois theory.

Prime Fields

Let us first recall the following definition.

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

Usually, a field is written as $\mathbb{K}$. On the other hand, since a field is a ring satisfying certain conditions, by collecting only these we can form a subcategory of $\Ring$, namely $\Field$. 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 $1$. ([Algebraic Structures] §Definition of a Ring, ⁋Definition 3) Thus any subring of a ring $A$ also has a multiplicative identity, and this is precisely equal to $A$’s multiplicative identity $1$. 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 can only be an inclusion or the zero map.

Proposition 2 Let arbitrary fields $\mathbb{K},\mathbb{K}'$ be given, and consider a ring homomorphism $f:\mathbb{K} \rightarrow \mathbb{K}'$. Then $f$ is an injection.

Proof

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

Meanwhile, in that post [Algebraic Structures] §Field of Fractions, we were able to obtain $\mathbb{Q}$ as the total field of fractions of $\mathbb{Z}$, $\Frac(\mathbb{Z})$. On the other hand, the following example shows an example of a field different from $\mathbb{Q}$.

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

To verify this, let an arbitrary nonzero $a+p\mathbb{Z}\in \mathbb{F}_p$ be given. Then since $\gcd(a,p)=1$, from [Ring Theory] §Integral Domains, ⁋Corollary 9,

\[1=ax+bp\]

holds for some $x, b\in \mathbb{Z}$, and it suffices to consider 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 $\mathbb{P}$. Here, $\mathbb{P}$ belongs to the center of $A$ and is contained in every subfield of $A$.

Proof

Let $\mathbb{K}$ be a subfield of $A$. If we let $Z$ be the center of $A$, then $\mathbb{K}':= \mathbb{K} \cap Z$ is also a subfield of $A$.

Consider the unique ring homomorphism from $\mathbb{Z}$ to $A$, $f: \mathbb{Z} \rightarrow A$, 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 [Algebraic Structures] §Field of Fractions, ⁋Proposition 10, either $\mathfrak{p} = (0)$ or, for some prime $p$, $\mathfrak{p} = (p)$.

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

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 $Z$.

Now consider another subfield of $A$, $L$. 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, since every subfield contains $\mathbb{P}$, $\mathbb{P}$ is contained in every subfield of $A$ and is unique as a prime field.

Since we are mostly interested only in the case where the ring $A$ is a field, the hypothesis of the above theorem has little significance. 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 among the subfields of $\mathbb{K}$.

Corollary 6 A field $\mathbb{K}$ is a prime field if and only if $\mathbb{K}$ 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 $\mathbb{P}$. Meanwhile, examining the proof of Theorem 4, the method of obtaining this $\mathbb{P}$ is also given quite explicitly: since $\mathbb{P}$ must contain the multiplicative identity of $A$, $1$, from this

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

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

Definition 7 Suppose a ring $A$ has a subfield. Here, for the kernel of $f:\mathbb{Z} \rightarrow A$ in the proof of Theorem 4, we define its non-negative integer generator $n$ as the characteristic of the ring $A$ and denote it by $\ch(A)$.

That is, if $A$ has a subfield, and thus has 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 Consider a nonzero ring $A$ having a subfield. The following hold:

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

Proof

This is obtained because for any integer $n$ and element $x\in A$, considering, from $\mathbb{Z}$ to $A$, the unique ring homomorphism $f$, the equation

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

holds.

Naturally, the objects we are familiar with are those where $\ch(A)=0$. If $\ch(A)=p$, several interesting results hold. To this end, let us first prove the following lemma.

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

Proof

We prove this by mathematical induction on $i$. In the case $i = 1$, since $\binom{p}{1} = p$, 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 in 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. Meanwhile, $i$ is not a multiple of $p$, so 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 equations

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

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

Proof

The equation for multiplication requires no further proof since $A$ is commutative. In the case of the equation for addition, 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, for all $1 \leq i \leq p-1$, every $i$ satisfies $\binom{p}{i} \equiv 0 \pmod{p}$, 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 in Theorem 10

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

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 of $A$, $S$, we write the image under the $f$-fold composition of the Frobenius endomorphism as $S^{p^f}$.

In addition, we introduce a few notations. For a subring of $A$, $K$, and an arbitrary subset $S\subseteq A$, $K[S]$ means the subring of $A$ generated by $K\cup S$. If $A$ is a field, then by [Algebraic Structures] §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 containing $K\cup S$ in $A$.

Then under 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 $\{a_i\}_{i \in I}$ is a $K$-module generating $K[S]$, then the set $\{a_i^{p^f}\}_{i \in I}$ is a $K^{p^f}$-module generating $K[S]^{p^f}$.

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 $\Frob_p^f(K[S])$. This holds because the image of the subring generated by $K\cup S$ under $\Frob_p^f$ is equal to the subring generated by $\Frob_p^f(K\cup S)$.
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 Rings

Now we define the following.

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

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 $p$.
$\phi : A \rightarrow A^{1/p^\infty}$ is a ring homomorphism and satisfies 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 characteristic $p \ne 0$ commutative ring $A$, the perfect closure of $A$, $(A^{1/p^\infty}, \phi)$, 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 flow. Intuitively, the condition that $\Frob_p$ is bijective means that any element of $A$ has a (unique) $p$-th root. Therefore, the perfect closure of $A$ is obtained by adjoining, for any element of $A$, its $p$-th roots.

To this end, we consider the following directed system:

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

If we consider the direct limit of this directed system,

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

any element of $\varinjlim A$ is represented by a 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 if we write out the multiplication in $\varinjlim A$ explicitly. By definition, in $\varinjlim A$, the multiplication of two elements $[a_i\in A_i]$ and $[a_j\in A_j]$ is, assuming without loss of generality that $j\geq i$, given by the formula

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

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

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

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 $\phi: a\mapsto [a\in A_0]$ being injective is equivalent to $A$ being reduced.

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

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

Proof

If $p=0$, then $q=1$, and hence the given function is the identity function, so there is nothing to prove. If $q \ne 1$, since $q$ is a positive prime, the characteristic of $\mathbb{K}$ 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 clear because $\mathbb{K}$ is a field and therefore has no nilpotent elements.

Definition 17 For a characteristic exponent $q$ field $\mathbb{K}$, 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 must be $1$, or (2) the characteristic exponent must be $q>1$ and it must be 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, and we have already observed in the proof of that proposition that the function $\mathbb{K} \rightarrow \mathbb{K}^{q^f}\subseteq \mathbb{K}$ of Proposition 16 is always injective without any 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 $\Omega_{E/A}$. ([Multilinear Algebra] §Differential Modules, ⁋Example 10) According to this structure, for an arbitrary polynomial $f\in E$, the element $\dd{f}\in\Omega_{E/A}$ is given by the following formula:

\[\dd{f}=\sum_{i\in I} \frac{\partial f}{\partial\x_i}\dd{\x_i}\]

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 $A$ having a subfield and polynomial algebra $E=A[\x_i]_{i\in I}$, let the subset of $E$, $S$, be the set of those satisfying $\dd{f}=0$ among polynomials $f$. 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 $S=E^p$ holds.

K

Fields

Prime Fields

Characteristic

Perfect Rings

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