Exploration 6: Finite fields

• Recall that quotienting a Euclidean domain by a prime ideal results in a Euclidean domain. But $\ZZ_p\iso\ZZ/(p)$ where $\ZZ$ is a Euclidean domain (since integer division exists) and $(p)$ is a prime ideal, therefore $\ZZ_p$ is a Euclidean domain as well.
• To show that $\ZZ_p$ is also a field, recall that in a Euclidean domain, Bezout’s identity for coprime integers $n,m$ gives $1=an+bm$ for some $a,b\in R$. Then since $[p]=[0]$ in $\ZZ_p$, and $p$ is coprime to every positive integer $k$ less than $p$, we can use Bezout’s identity with $[p]$ and $[k]$ to get $[1]=[a][0]+[b][k]=[b][k]$ which implies every $[k]$ for $1\le k<p$ is a unit in $\ZZ_p$. Therefore every nonzero element in $\ZZ_p$ is a unit, making $\ZZ_p$ a field.

• Every ring contains the elements $\{1,(1+1),(1+1+1),\ldots\}$, but for prime characteristic rings where $p\cdot 1=0$, this only goes up to adding $p-1$ copies of $1$.
• These are exactly the elements that form $\ZZ_p$, the integers mod $p$, so every prime characteristic ring contains $\ZZ_p$ in the sense that you can define an isomorphism between those elements and $\ZZ_p$.

• By the binomial theorem, $(a+b)^p=\sum_{k=0}^p{p\choose k}a^kb^{p-k}$. Let’s focus on the coefficients $p\choose k$.
• ${p\choose k}=\frac{p!}{k!(p-k)!}$ always includes a factor $p$ in the numerator, but $p$ (being prime) is not a factor of the denominator unless $k=0$ or $k=p$.
• Thus for other values of $k$, the coefficient $p\choose k$ must be zero since it includes a factor $p$ in a field of characteristic $p$. For the special cases $k=0$ and $k=p$, the coefficient $p\choose k$ is equal to $1$.
• Therefore $(a+b)^p=1a^p+0+\ldots+0+1b^p=a^p+b^p$.

• Let $\phi$ be the map $a\mapsto a^p$. By the previous proof we observe $\phi(a+b)=\phi(a)+\phi(b)$.
• To be a homomorphism, $\phi$ must also preserve additive inverses and multiplication.
  • Preserving subtraction is easy for odd primes $p$, because $\phi(-a)=(-a)^p=-a^p=-\phi(a)$. For the remaining prime $p=2$, recall that every element is its own additive inverse in characteristic $2$, so we can directly write $\phi(a)=-\phi(a)$.
  • Preserving multiplication is straightforward. $\phi(ka)=(ka)^p=k^pa^p=\phi(k)\phi(a)$.
• Thus $\phi$ is a homomorphism from the ring to itself, i.e. an endomorphism.

• Fermat’s little theorem implies that for any integer $a$ not divisible by $p$ (i.e. any nonzero, since $p$ is prime), we have $a^{p-1}-1\equiv 0\mod p$.
• In other words, the polynomial $x^{p-1}-1$ has every nonzero element of $\ZZ_p[x]$ as a root. We can add the zero element as a root if we multiply by $x$ to get $x^p-x$.
• Thus, if you have a polynomial ring defined over a prime characteristic field like $\ZZ_p[x]$, $x^p-x$ is a polynomial that has all $p$ distinct roots.

$F[x]/(f)$ consists of cosets $[f]$ whose representatives are polynomials $f\in F[x]$. Earlier we proved that the degree of every representative is less than $\deg f$, and therefore has at most $\deg f$ terms with $|F|$ possible coefficients for each term. This implies that there are $|F|^{\deg f}$ distinct cosets in $F[x]/(f)$.

$F^\times$, like all finite abelian groups, decomposes into cyclic groups of prime power order: $$F^\times\iso C_{q_1}\times C_{q_2}\times\ldots\times C_{q_n}$$ Since the order of the component cyclic fields $q_1,q_2,\ldots,q_i$ are pairwise coprime, their direct product is cyclic.

• We just showed that for an irreducible degree $n$ polynomial $g\in F[x]$, the quotient $\ZZ_p[x]/(g)$ is a finite field of order $p^n$. So finite fields of order $p^n$ exist for every prime $p$ and integer $n\ge 1$.
• If we can define a field homomorphism between two arbitrary fields of prime power order $p^n$, then they are isomorphic because all field homomorphisms are injective and the domain and codomain have the same order $p^n$.
• Consider the multiplicative group $K^\times$ of an arbitrary field $K$ of prime power order $p^n$. The order of $K^\times$ is exactly $p^n-1$. By Lagrange’s theorem, the order of every element $a$ of $K^\times$ must divide the order of $K^\times$, so $a^{p^n-1}=1$, which can be rearranged to $a^{p^n}-a=0$.
• Then $a^{p^n}-a=0$ holds for every $a\in K^\times$. But since $0^{p^n}-0=0$, this extends to all $a\in K$. Since this holds for arbitrary fields of prime power order, let $K_1,K_2$ be two such fields.
• Then every element of $a\in K_1$ or $b\in K_2$ is a distinct root of the polynomial $f=x^{p^n}-x$, in the sense that $f(a)=0$ for all $a\in K_1$ and $f(b)=0$ for all $b\in K_2$.
• Recall that ring homomorphisms (and therefore field homomorphisms) preserve rational expressions, including polynomial expressions. This means if $f(a)=0$, then applying $\phi$ to both sides gives $\phi(f(a))=f(\phi(a))=0$, implying that $\phi(a)$ is a root of $f$ as well.
• Since $f$, by construction, has $p^n$ distinct roots, $\phi$ maps all $p^n$ elements of $K_1$ to some element in $K_2$. Therefore $\phi$ is a well-defined field homomorphism and we are done.

• If a ring $R$ is not of a prime power order, then the prime factorization of its order $p_1^{k_1}p_2^{k_2}\ldots p_n^{k_n}$ must contain at least two distinct prime powers.
• Consider the principal ideal $p_i^{k_i}R$ for some $i$. Since prime powers are pairwise coprime, that means the Chinese Remainder Theorem can be used to show $R\iso R/p_1^{k_1}R\times R/p_2^{k_2}R\times\ldots\times R/p_n^{k_n}R$.
• But since a direct product of rings contains zero divisors, and zero divisors cannot be units, $R$ contains nonzero nonunits and is therefore not a field.

The result follows pretty quickly after we prove a number theoretic lemma:

Then:

• ($\to$) By Lagrange’s theorem, the order of the subfield’s multiplicative group $|\FF_{p^n}^\times|$ divides that of the larger field $|\FF_{p^m}^\times|$. Thus $p^n-1\mid p^m-1$, so $n\mid m$ by the lemma.
• ($\from$) If $n\mid m$, then $$\begin{aligned} n&\mid m\\ p^n-1&\mid p^m-1&\text{ by the lemma}\\ x^{p^n-1}-1&\mid x^{p^m-1}-1&\text{ by the lemma}\\ f\cdot (x^{p^n-1}-1)&=x^{p^m-1}-1&\text{ for some polynomial }f\in\ZZ[x]\\ f\cdot (x^{p^n}-x)&=x^{p^m}-x&\text{ multiply both sides by }x\\ \end{aligned}$$ which shows that every root of $x^{p^n}-x$ is a root of $x^{p^m}-x$. But since the roots of $x^{p^n}-x$ are exactly $\FF_{p^n}$, this implies every element of $\FF_{p^n}$ is in $\FF_{p^m}$, i.e. $\FF_{p^n}$ is a subfield of $\FF_{p^m}$.

• If $\FF_{p^n}$ is larger, then $\FF_{p^m}$ is a subfield of $\FF_{p^n}$. Since $\alpha$ generates $\FF_{p^n}$, the resulting field is $\FF_{p^n}$, since that already includes $\FF_{p^m}$.
• If $\FF_{p^m}$ is larger, then $\FF_{p^n}$ is a subfield of $\FF_{p^m}$ and therefore $\alpha$ is already in $\FF_{p^m}$, so the resulting field is $\FF_{p^m}$.

This follows directly from the previous result. WLOG assume $\alpha$ generates the larger field. Then $\alpha$ generates $\beta$, thus you need only adjoin $\alpha$.