# Fields

In this post, we consider generalizations of the real and complex number systems.

Definition 1. A field is a set $F$ equipped with two binary operations $+:F \times F \to F$ and $\cdot:F \times F \to F$ such that the following properties hold:

(FA1) addition is commutative, viz., $a+b=b+a$ for all $a,b \in F$;
(FA2) multiplication is commutative, viz., $ab=ba$ for all $a,b \in F$;
(FA3) addition is associative, viz., $a+(b+c) = (a+b)+c$ for all $a,b \in F$;
(FA4) multiplication is associative, viz., $a(bc) = (ab)c$ for all $a,b,c \in F$;
(FA5) addition and multiplication are distributive, viz., $a(b+c) = ab+ac$ and $(a+b)c = ac + bc$ for all $a,b,c \in F$;
(FA6) an additive identity $0_F$ exists, so that $a + 0_F = 0_F + a = a$ for all $a \in F$;
(FA7) a multiplicative identity $1_F$ exists, so that $a 1_F = 1_F a = a$ for all $a \in F$;
(FA8) addition is invertible, viz., each $a \in F$ admits $-a \in F$ such that $a + (-a) = (-a) + a = 0_F$;
(FA9) multiplication is invertible, viz., each $a \in F \smallsetminus \{0_F\}$ admits $a^{-1} \in F \smallsetminus \{0_F\}$ such that $aa^{-1} = a^{-1}a = 1_F$.

The set of real numbers $\mathbb{R}$ with usual addition and multiplication is a field. Similarly, the set of complex numbers $\mathbb{C}$ and the set of rational numbers $\mathbb{Q}$ with usual addition and multiplication are also fields.

Note that a field is an ordered quintuplet $(F,+,\cdot,0_F,1_F)$ of a set $F$, two binary operations $+$ and $\cdot$ on $F$, and two identity elements $0_F$ and $1_F$ such that $(F,+,0_F)$ and $F \smallsetminus \{0_F\},\cdot,1_F)$ are abelian groups, and that the dstributive property (FA5) holds.

If there is no danger of confusion, we simply write 0 and 1 for $0_F$ and $1_F$, respectively. Moreover, given $n \in \mathbb{N}$ and $x \in F$, we write $n \cdot x$ to denote the $n$-fold sum $x + \cdots + x$.

Remark 2. $0$ is the unique additive identity, and $1$ is the unique multiplicative identity. Indeed, if $0’$ is another additive identity, then $0 = 0+0′ = 0’$. Similarly, if $1’$ is another multiplicative identity, then $1 = 1 \cdot 1′ = 1’$.

Remark 3. Additive inverses and multiplicative inverses are unique. To this end, we fix $a \in F$. If $b$ and $c$ are additive inverses of $a$, then

Similarly, if $a \neq 0$ and if $b$ and $c$ are multiplicative inverses of $a$, then

Remark 4. $0 \neq 1$. Indeed, (FA7) implies that $1 \cdot 1 = 1$. (FA9), in conjunction with Remark 3, shows that $1$ must be an element of $F \smallsetminus \{0\}$.

Remark 5. $a0 = 0a = 0$ for all $a \in F$. indeed,

by (FA5), and subtracting $a0$ from each side yields $a0 = 0$. Similarly, $0a = 0$.

Remark 6. $-(ab) = (-a)b = a(-b)$ for all $a,b \in F$. Observe that

by (FA5) and Remark 5. Similarly, $ab+(-a)b = 0$ and it follows from Remark 3 that $(-a)b = -(ab)$. An analogous argument shows that $a(-b) = -(ab)$.

Exercise 7. A ring is a set $R$ with two binary operations $+$ and $\cdot$ on $F$ such that (FA1), (FA3), (FA4), (FA5), (FA6), (FA7), and (FA8) hold true. Show that the properties established in Remark 2 through Remark 6 remain valid for rings.

Note that $\mathbb{Q}$, $\mathbb{R}$, amd $\mathbb{C}$ all share the same addition and multiplication operations. In what sense is $\mathbb{Q}$ a “substructure” of $\mathbb{R}$, or $\mathbb{C}$ a “superstructure” of $\mathbb{R}$? We formalize these notions below.

Definition 8. A subfield of a field $F$ is a subset $K$ of $F$ such that

(1) $0_F,1_F \in K$;
(2) $a+b \in K$ and $ab \in K$ whenever $a,b \in K$;
(3) $-a \in K$ and $a^{-1} \in K$ whenever $a,b \in K$.

Definition 9. A field extension of a field $K$ is a field $F$ that contains $K$ as a subfield.

Example 10. $\mathbb{Q}$ has no proper subfield. Indeed, every subfield $F$ of $\mathbb{Q}$ must contain all the integers and their multiplicative inverses. Therefore, $\frac{1}{n} \in F$ for all $n \in \mathbb{Z} \smallsetminus \{0\}$. Moreover, $\frac{m}{n} = \sum_{i=1}^m \frac{1}{n}$ must be contained in $F$, whence $F \supseteq \mathbb{Q}$. It follows that $F = \mathbb{Q}$. $\square$

Example 11. $\mathbb{R}$ is a field extension of $\mathbb{Q}$.

$\mathbb{R}$ is quite a special one, in two respects. Firstly, $\mathbb{R},\leq$ with the usual less-than-or-equal-to ordering $\leq$ is the order-theoretic completion of $\mathbb{Q},\leq$. Indeed, $\mathbb{R},\leq$ is the smallest order-superstructure of $\mathbb{Q},\leq$ such that the least-upper-bound property holds. A typical construction is given by the method of Dedekind cuts.

Moreover, $(\mathbb{R},|\cdot|)$ with the usual absolute-value metric $|\cdot|$ is the Cauchy completion of $\mathbb{Q},|\cdot|$. Indeed, $\mathbb{R},|\cdot|$ is the smallest metric-superstructure of $\mathbb{Q},|\cdot|$ such that all Cauchy sequences converge.

We remark that the method of Cauchy completion works for non-standard metrics on $\mathbb{Q}$ as well. See the Wikipedia article on $p$-adic numbers for an important collection of fields we can obtain from completing certain non-standard metrics on $\mathbb{Q}$. $\square$

Example 12. $\mathbb{C}$ is a field extension of $\mathbb{R}$. In fact, $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$, i.e., the smallest field extension $F$ of $\mathbb{R}$ such that every polynomial with coefficients in $F$ has a root in $F$. That $\mathbb{C}$ is an algebraically closed field is precisely the content of the fundamental theorem of algebra. To see that $\mathbb{C}$ is the smallest such field, we consider the following general fact.

Let $K$ be a field extension of $F$. Since $K$ is closed under addition and $ax \in K$ whenever $a \in F$ and $x \in K$, we see that $K$ is a vector space over $F$. Therefore, we can talk about the $F$-dimension of $F$. For example, the $\mathbb{R}$-dimension of $\mathbb{C}$ is 2. Now, if $K$ is any field extension of $\mathbb{R}$ that is also a subfield of $\mathbb{C}$, then $K$ must be a $\mathbb{R}$-vector subspace of $\mathbb{C}$. Therefore, the $\mathbb{R}$-dimension of $K$ is either 1 or 2. If $\dim_\mathbb{R} K = 1$, then $K = \mathbb{R}$. If $\dim_{\mathbb{R}} K = 2$, then $K = \mathbb{C}$. It follows that $\mathbb{C}$ is the smallest proper field extension of $\mathbb{R}$. This, in particular, shows that $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$. $\square$

Example 13. A field extension $K$ of $\mathbb{Q}$ such that $% $ (see Example 12) is referred to as an algebraic number field. It can be shown that all such extensions of $\mathbb{Q}$-dimension 2, called quadratic fields, are algebraic number fields of the form

where $d$ is a squarefree integer, viz., $d \neq 0$, $d \neq 1$, and the prime factorization of $d$ does not repeat any prime. If $d > 0$, then $\mathbb{Q}[\sqrt{d}]$ is said to be a real quadratic field; if $% $, then $\mathbb{Q}[\sqrt{d}]$ is said to be an imaginary quadratic field. $\square$

Example 14 (Modular arthmetic). So far, every example of a field we have considered was infinite. Let us now construct finite fields. For each $k \in \mathbb{N}$, we define the following equivalence relation on the set of integers $\mathbb{Z}$: $n \sim_k m$ if and only if $n-m$ is divisible by $k$. We denote by $\mathbb{Z}/k\mathbb{Z}$ the quotient set $\mathbb{Z}/\sim_k$ and define the addition and multiplication operations by setting $[n]+[m] = [n+m]$ and $[n] \cdot [m] = [n \cdot m]$ for all $n,m \in \mathbb{Z}$.

Let us check that these operations are well-defined, viz., they do not depend on the choice of representatives. To this end, we assume that $n_1 \sim_k n_2$ and $m_1 \sim_k m_2$. We can then find $a,b \in \mathbb{Z}$ such that $n_1 – n_2 = nk$ and $m_1 – m_2 = bk$. Observe that

and so $n_1 + m_1 \sim_k n_2 + m_2$. Therefore, $[n_1+m_1] = [n_2+m_2]$. Similarly,

and so $n_1m_1 \sim_k n_2m_2$. It follows that $[n_1m_1] = [n_2m_2]$.

Note that modular addition is always invertible, as $[k-n]$ is the additive inverse of $[n]$. Most integers, however, do not have multiplicative inverses, and so invertibility of modular multiplication is a trickier matter.

We claim that $\mathbb{Z}/k\mathbb{Z}$ is not a field if $k$ is not a prime. To see this, we suppose that $p$ is a positive prime number strictly less than $k$ that divides $k$. We can then find $m \in \mathbb{N}$ such that $k = pm$. If $[pn]=[1]$ for some $n \in \mathbb{Z}$, then $pn-1 = kn’ = pmn’$ for some $n’ \in \mathbb{Z}$, so that $p(n – mn’) = 1$. But $p \geq 2$, and so $p$ does not admit a multiplicative inverse in $\mathbb{Z}$. This is absurd, and we conclude that $p$ has no multiplicative inverse in $\mathbb{Z}/k\mathbb{Z}$.

Let us now show that $\mathbb{Z}/k\mathbb{Z}$ is a field whenever $k$ is prime. In order to verify this claim, we shall make use of the Euclidean algorithm. Fix a nonzero integer $n$. The Euclidean algorithm furnishes integers $m_1$ and $m_2$ such that $m_1n + m_2k = \gcd(n,k)$. Since $k$ is assumed to be prime, $\gcd(n,k) = 1$. We now observe that

whence $[m_1]$ is the multiplicative inverse of $[n]$. $\square$

Exercise 15. An integral domain is a ring (see Exercise 7) $R$ such that (FA2) holds (see Definition 1) and that, for all $a,b \in R$, $ab = 0$ implies either $a = 0$ or $b = 0$. Show that $\mathbb{Z}$ is an integral domain but not a field. Show also that $\mathbb{Z}/k\mathbb{Z}$ is an integral domain if and only if it is a field, i.e., when $k$ is prime. If you feel ambitious, prove that every finite integral domain is a field.

Note that $[k] \cdot [1] = [0]$ in $\mathbb{Z}/k\mathbb{Z}$, even though there is no integer $n$ such that $n \cdot 1 = 0$ in $\mathbb{Z}$. This difference is captured by the following notion:

Definition 16. A field $F$ is of characteristic $k$ in case $k$ is the smallest positive integer such that $k \cdot 1_F = 0_F$. If no such positive integer exists, then we say that $F$ is of characteristic zero. We write $\operatorname{char}(F)$ to denote the characteristic of $F$.

Note that $\mathbb{Q}$ is of characteristic zero, and that $\mathbb{Z}/p\mathbb{Z}$ is of characteristic $p$. These are representative of all fields, as far as the computation of characteristics is concerned.

Proposition 17. Every field of positive characteristic must be of prime characteristic.

Proof. Let $k = \operatorname{char}(F)$. We suppose for a contradiction that $k = mn$ for some integers $m,n \geq 2$. Since $% $ and $% $, we can find elements $a$ and $b$ of $F$ such that $ma \neq 0$ and $nb \neq 0$, respectively. Now,

But then

which is absurd. We conclude that $F$ is of prime characteristic. $\square$

Proposition 17, in tandem, with the argument given in Example 14, shows that the subset

of $F$ forms a subfield of $F$. In other words, a field of positive characteristic contains a copy of $\mathbb{Z}/\operatorname{char}(F)\mathbb{Z}$. We formalize this notion below.

Definition 18. A field homomorphism is a function $\varphi:F \to K$ between two fields $F$ and $K$ that satisfies the following properties:

​(1) $\varphi(a+b) = \varphi(a) + \varphi(b)$ for all $a,b \in F$; (2) $\varphi(ab) = \varphi(a)\varphi(b)$ for all $a,b \in F$; (3) $\varphi(0_F) = 0_K$; (4) $\varphi(1_F) = 1_K$.

A field isomorphism is a bijective field homomorphism. We say that $F$ is isomorphic to $K$ if there exists a field isomorphism $\varphi:F \to K$. In this case, we write $F \cong K$.

We shall return to the study of the subfield $P$ at the end of this post. For now, we show that field homomorphisms are structure-preserving.

Proposition 19. The image of a field homomorphism is a subfield of the codomain field.

Proof. Let $\varphi:F \to K$ be a field homomorphism. Evidently, $0_K = \varphi(0_F)$ and $1_K = \varphi(1_F)$ are in $\operatorname{im} \varphi$. We fix $a,b \in \operatorname{im} \varphi$ and find $c,d \in F$ such that $a = \varphi(c)$ and $b = \varphi(d)$. Observe that

Similarly, $ab \in \operatorname{im}\varphi$. We also note that

whence $-a=\varphi(-c) \in \operatorname{im}\varphi$. Similarly, $a^{-1} \in \operatorname{im} \varphi$. We conclude that $\operatorname{im}\varphi$ is a subfield of $K$. $\square$

Unlike group homomorphisms, field homomorphisms are quite rigid.

Proposition 20. Every field homomorphism is injective.

Proof. Let $\varphi:F \to K$ be a field homomorphism. We define the kernel of $\varphi$ to be the set

We first show that $\varphi$ is injective if and only if $\ker \varphi = \{0_F\}$. If $\varphi$ is injective, then the kernel is clearly trivial. Conversely, we suppose that the kernel is trivial and fix $a,b \in F$ such that $\varphi(a) = \varphi(b)$. Note that $\varphi(a-b) = 0_K$, so that $a -b \in \ker \varphi = \{0_F\}$. It follows that $a = b$.

Let us now define an ideal of $F$ to be a subset $I$ of $F$ such that$a \in F$ and $x \in I$ imply $ax \in I$, and that $x,y \in I$ implies $x +y \in I$. We show that $F$ has only two ideals: $\{0_F\}$ and $F$. Indeed, if $x$ is a nonzero element of an ideal $I$, then $1_F = x^{-1}x \in I$, and so $a = a1_F \in I$ for all $a \in F$.

Observe that $\ker \varphi$ is an ideal of $F$. Indeed, if $a \in F$ and $x \in \ker \varphi$, then

so that $ax \in \ker \varphi$. Moreover, if $x,y \in \ker \varphi$, then

and so $x+y \in \ker\varphi$. Now, $\varphi(1_F) = 1_K \neq 0_K$, and so $1_F \notin \ker \varphi$. Since $\ker \varphi$ is an ideal, it follows that $\ker \varphi = \{0_F\}$, whence $\varphi$ is injective. $\square$

As a simple application, we establish the following property of characteristics.

Corollary 21. If there exists a field homomorphism $\varphi:F \to K$, then $\operatorname{char}(F) = \operatorname{char}(K)$.

Proof. By Proposition 20, $\varphi$ is an injective field homomorphism. If $\operatorname{char}(F) = 0$, then $n \cdot 1_F \neq 0_F$ for all $n \geq 1$. Since $\varphi$ is injective,

is never $0_K$ for any $n \geq 1$. We conclude that $\operatorname{char}(K) = 0$.

If $F$ is of positive characteristic, then an analogous reasoning implies that

is nonzero for all $1 \leq n \leq \operatorname{char}(F) – 1$ and is zero when $n = \operatorname{char}(F)$. It follows that $\operatorname{char}(F) = \operatorname{char}(K)$. $\square$

What the above corollary tells us is that it is necessaryto have matching characteristics in order for field homomorphisms $\varphi:F \to K$ to exist. We now show that, if $F$ is either $\mathbb{Q}$ or $\mathbb{Z}/p\mathbb{Z}$, then it is sufficient to have matching characteristics to ensure the existence of field homomorphisms. To lay out the proper context for this result, we introduce the notion of the prime subfield, or the subfield generated by $1$.

Definition 22. Let $F$ be a field and let $S$ be a subset of $F$. The subfield of $F$ generated by $S$ is the smallest subfield of $F$ that contains $S$, viz., the intersection of all subfields of $F$ that contains $S$. If $S = \{1\}$, then the subfield generated by $S$ is referred to as the prime subfield of $F$.

We conclude the post by establishing the promised sufficiency result:

Proposition 23. Let $F$ be a field. If $\operatorname{char}(F) = 0$, then the prime subfield of $F$ is isomorphic to $\mathbb{Q}$. If $\operatorname{char}(F) = p$ for some prime $p$, then the prime subfield of $F$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$.

Since we have shown in Proposition 17 that the characteristic of a field must be either 0 or prime, this covers all possible cases.

Proof. Let $F$ be a field and let $P$ be the prime subfield of $F$. If $\operatorname{char}(F) = 0$, then the mapping $\varphi(n/m) = n \cdot (m \cdot 1_F)^{-1}$ is a field homomorphism from $\mathbb{Q}$ into $F$. Since $1_F \in \operatorname{im} \varphi$, we see that $\operatorname{im} \varphi$ is a field extension of $P$. Now, $\operatorname{im} \varphi \cong \mathbb{Q}$ by Proposition 20. Since $\mathbb{Q}$ contains no proper subfield (Example 10), it follows that $\operatorname{im} \varphi = P$.

If $\operatorname{char}(F) = p$, then the mapping $\phi(n) = n \cdot 1_F$ is a field homomorphism from $\mathbb{Z}/p\mathbb{Z}$ into $F$. Analogously as above, we can conclude that $P = \operatorname{im} \phi \cong \mathbb{Z}/p\mathbb{Z}$. $\square$

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