Subfields

1. What is a subfield?

A large field can contain a smaller field. For instance, $\mathbb{F}_{2^k}$ always contains as a sub-field the 2-element subset $\mathcal{S} = \left\{0,1\right\}$ consisting of constant polynomials — it is easy to check that this subset satisfies the definition of field (it is closed under addition and multiplication; it contains a zero element and an identity element; every element has an opposite element in $\mathcal{S}$ since $-0 = 0$ and $-1 = 1$ in $\mathbb{F}_{2^k}$; addition and multiplication are commutative etc.). Another way to see that this two-element subset is a field is to notice that it can be identified with $\mathbb{F}_{2} = \mathbb{Z}_{2}$, as the addition and multiplication of elements $0$ and $1$ are defined in $\mathbb{F}_{2^k}$ the same way as in $\mathbb{F}_{2}$ (so one can say that $\mathbb{F}_{2^k}$ contains $\mathbb{F}_{2}$ as a subfield). More generally:

Let $\mathcal F$ be a field. A subset $\mathcal{S} \subseteq \mathcal F$ is called a subfield of $\mathcal F$ if $\mathcal{S}$ is a field itself with respect to the operations of $\mathcal F$.

By this definition, every field is a subfield of itself. But it may also contain strictly smaller subfields. Those are called the proper subfields. For example, as we saw, $\mathbb{F}_{2}$ is a proper subfield of $\mathbb{F}_{2^k}$ for $k > 1$.

A practical criterion for checking whether a subset of a field is a subfield is the following:

Proposition (Subfield criterion). Let $\mathcal F$ be a field. A subset $\mathcal{S} \subseteq \mathcal F$ is a subfield of $\mathcal F$ if and only if it contains the zero and identity element of $\mathcal F$, is closed under the multiplication, addition and taking opposite elements of $\mathcal F$, and $\mathcal{S}\setminus \{0\}$ (the set of non-zero elements belonging to $\mathcal{S}$) is closed under taking inverses in $\mathcal F$.

Proof

Let's show necessity, i. e. that every subfield $\mathcal{S} \subseteq \mathcal F$ satisfies the conditions given in the proposition.

Suppose $\mathcal{S} \subseteq \mathcal F$ is a subfield of $\mathcal F$. Any field contains exactly two elements $x$ satisfying $x \cdot x = x$, namely its zero and identity element (indeed, if $x \neq 0$ and $x \cdot x = x$ then $x = 1 \cdot x = (x^{-1} \cdot x) \cdot x = x^{-1} \cdot (x \cdot x) = x^{-1} \cdot x = 1$), and therefore $\mathcal G$ must contain the zero and the identity element of $\mathcal F$. Moreover, actually the zero element of $\mathcal G$ is the zero of $\mathcal F$ and the identity element of $\mathcal G$ is the identity of $\mathcal F$, since in any field only the zero element satisfies $x + x = x$.

Obviously $\mathcal{S}$ must be closed under multiplication and addition by the definition of field. To prove that $\mathcal{S}$ is closed under taking opposite elements in $\mathcal F$, we will show that the opposite element of every element of $\mathcal{S}$ is its opposite element in $\mathcal F$. Let $a \in \mathcal{S}$ be arbitrary. By the definition of opposite element, the opposite element of $a$ in $\mathcal S$ is an element $b \in \mathcal S$ sucht that $a + b$ is the zero element of $\mathcal S$. Since the zero element of $\mathcal S$ coincides with the zero element of $\mathcal F$, such an element is also the zero element of $a$ in $\mathcal F$. By the uniqueness of opposite element, the opposite element of $a$ in $\mathcal S$ must therefore coincide with the opposite element in $\mathcal F$.

Analogously one shows that taking inverse elements in $\mathcal S$ is the same as taking inverse elements in $\mathcal F$ and therefore $\mathcal S\setminus \{0\}$ is closed under taking inverses in $\mathcal F$. The necessity part of the proposition has thus been proven.

Sufficiency is obvious: if a subset $\mathcal{S} \subseteq \mathcal F$ satisfies the conditions given in the proposition then it is straightforward to check that it satisfies all field axioms.

Which of the following sets of real numbers are subfields of the field $\mathbb R$?

$\mathbb Q$, the set of rational numbers Answer Yes, $\mathbb Q$ is a subfield of $\mathbb R$.
$\mathbb Z$, the set of integers Answer $\mathbb Z$ is not a subfield of $\mathbb R$ since it is not closed under taking inverses: e. g. $2^{-1} \not\in \mathbb Z$.
$\{ \ldots, \frac14, \frac12, 1, 2, 4, 8, \ldots \}$, the set of integer powers of $2$ Answer This is not a subfield since it is not closed under addition: e. g. $2 + 4 = 6$ which is not an integer power of $2$.
$\mathbb R$, the set of all real numbers Answer Yes, $\mathbb R$ itself is a subfield of $\mathbb R$.
$\{0\}$, the one-element set consisting of number zero Answer This is not a subfield since it does not contain the identity element $1 \in \mathbb R$.
$\{0, 1\}$ Answer Yes, this is a subfield of $\mathbb R$.

2. Subfields in finite fields

The definition and criterion above hold for both finite and infinite fields. But the simplest way to find a sub-field of a finite field is to find an non-invertible element that generates the entire multiplicative group of a sub-field. Such an element must exist, since the multiplicative group of a finite field is always generated by a single element, as we know from the previous lesson Multiplicative group in finite fields.

For example, $\mathbb{F}_{16}$ specified by the polynomial $\alpha^4+\alpha^3+ \alpha^2+\alpha+ 1$ might contain subfields of size $2$ and $4$, as elements in the multiplicative group have orders $1$, $3$, $5$ and field sizes must be powers of primes. For instance, let us consider the element $\alpha^3+ \alpha^2$ and see whether it generates the multiplicative group of some subfield. The element $\alpha^3+ \alpha^2$ has order $3$ in the multiplicative group of $\mathbb{F}_{16}$ since \[\begin{align*} (\alpha^3+ \alpha^2)(\alpha^3+ \alpha^2)&=\alpha^3+\alpha+ 1,\\ (\alpha^3+ \alpha^2)(\alpha^3+\alpha+ 1)&=1\enspace. \end{align*}\]

So it is plausible that these three powers of $\alpha^3+ \alpha^2$ together with zero could form a subfield of $\mathbb{F}_{16}$, which is a representation of the four-element field $\mathbb{F}_4$. To check that this four-element subset $\{0,\ 1,\ \alpha^3+ \alpha^2,\ \alpha^3+\alpha+ 1\}$ is indeed a subfield by using Proposition (Subfield criterion), we need to check that it is closed under addition and taking opposite element, since closedness of the three-element subset $\{1,\ \alpha^3+ \alpha^2,\ \alpha^3+\alpha+ 1\}$ under multiplication and inverses follows from the fact that it is the fact that by construction it is a subgroup of the multiplicative group of $\mathbb{F}_{16}$ (namely the one generated by the element $\alpha^3+ \alpha^2$) and a subset of a field that is closed under multiplication obviously stays so if we add zero to it. Closedness under taking opposite element is also obvious, since in the field $\mathbb{F}_{16}$, the opposite of any element is the element itself. Closedness under addition is easy to verify by direct computation (basically by computing the 4-by-4 addition table of this four-element set and check that all the entries lie in the set — but for most entries this is trivially true, so not much actual computation is needed). Therefore the four-element subset $\{0,\ 1,\ \alpha^3+ \alpha^2,\ \alpha^3+\alpha+ 1\} \subset \mathbb{F}_{16}$ turns indeed out to be a subfield. So $\mathbb{F}_{16}$ contains $\mathbb{F}_{4}$ as a subfield.

Actually it turns out that this manual verification of closedness under addition and opposite element was unnecessary, thanks to the following result, which we will not prove here.

Theorem. Let $p$ be a prime, $k$ a positive integer and $a$ a non-zero element of $\mathbb F_{p^k}$. The set of all integer powers of $a$ together with the zero element is a subfield of $\mathcal F$ if and only if the order of $a$ in the multiplicative group of $\mathbb F_{p^k}$ is of the form $p^\ell - 1$ where $j$ is a positive integer.

Indeed, the theorem says that since the order of $\alpha^3+ \alpha^2$ in the multiplicative group is of the form $2^\ell-1$ (because $3 = 2^2-1$), the powers of $\alpha^3+ \alpha^2$ together with $0$ must form a subfield of $\mathbb F_{16}$.

Prove that $\mathbb{F}_{p^\ell}$ is a sub-field of $\mathbb{F}_{p^k}$ if and only if $p^\ell-1$ divides $p^k-1$.

Hint: Consider the orders of generators.

Which of the fields $\mathbb{F}_4,\mathbb{F}_8,\mathbb{F}_{16},\ldots,\mathbb{F}_{256}$ contain proper subfields of size more than two? Solution. Answer: $\mathbb{F}_{16}$, $\mathbb{F}_{64}$ and $\mathbb{F}_{256}$. Indeed, from the theorem above we know that all subfields of $\mathbb{F}_{2^k}$ are of the form $\mathbb{F}_{2^\ell}$ (there are no subfields $\mathbb{F}_{p^\ell}$ where $p > 2$). So we can use the statement proven in the previous exercise to determine whether a field $\mathbb{F}_{2^k}$ contains a proper subfield of size more than two: this happens if and only if $2^k-1$ is divisible by $2^\ell-1$ for some $\ell$ with $0 < \ell < k$.

Since we know that our example subfield $\{0,\ 1,\ \alpha^3+ \alpha^2,\ \alpha^3+\alpha+ 1\} \subset \mathcal F_{16}$ is the four-element field $\mathbb F_4$, we could obviously use raw encoding of these polynomials $0000$, $0001$, $1100$ and $1011$ to represent $\mathbb{F}_4$. However, this representation is wasteful: since $\mathbb{F}_4$ contains four elements, these elements can be represented by two bits instead of four. Moreover, we know that $\mathbb{F}_4$ can be constructed as a set of polynomials of first degree with coefficients in $\mathbb Z_2$; these coefficients can be stored in two bits in a natural way, so it is also convenient to make computations in the two-bit representation (in other words, packing the elements into two bits does not come at the expense of CPU usage). So we haven't found a better representation of $\mathbb{F}_4$ than we knew before, but knowing that the field $\mathbb{F}_{16}$ contains $\mathbb{F}_4$ is still a nice mathematical insight, which may be useful. For example, if we have variables with values in the field $\mathbb{F}_{16}$ and at some point we realize that the values of the variables actually only belong to the set $\{0000, 0001, 1100, 1011\}$ then we can re-encode the variables into two bits to enchance memory and CPU efficiency of the field operations.

Let $\mathbb{F}_4$ be specified by the polynomial $\beta^2+\beta+ 1$. Define the corresponding isomorphism from $\mathbb{F}_4$ to the subset $\{0,\ 1,\ \alpha^3+ \alpha^2,\ \alpha^3+\alpha+ 1\}$ of $\mathbb{F}_{16}$.

Solution.

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Subfields

1. What is a subfield?

2. Subfields in finite fields