LEC ALG3 21

Finite fields

Quick preliminary facts:

1. Let $K$ be a finite field. The characteristic of a finite field must be prime, so $K$ will contain one of the prime fields $F={\mathbb{F}}_p$.
2. Since $K$ is finite, it will be finite dimensional when considered as a vector space over ${\mathbb{F}}_p$. Using Prp 322.4, this tells us that the extension $K/{\mathbb{F}}_p$ is algebraic.
3. Let $r$ denote the degree $[K:F]$. Recall that $[K:F]=r$ implies $K$ has an $F$-basis of size $r$. Thus, as an $F$-vector space, $K$ is isomorphic to the space $F^r$ of column vectors, which contains $p^r$ elements, customarily denoted by $q$. Fields of order $q$ are denoted by ${\mathbb{F}}_q$.
4. Conversely, if $K$ is a finite field of order $p^r$, we immediately have ${\mathbb{F}}_p\subseteq K$, since we cannot have $p^r=p_1^{r_1}$.
5. Do not confuse ${\mathbb{F}}_q$ with the ring $\mathbb{Z}/q\mathbb{Z}$, which isn’t a field.

Example 345.1(Example).

Consider ${\mathbb{F}}_4$. There is just one irreducible polynomial of degree $2$ in ${\mathbb{F}}_2[x]$, namely $f(x)=x^2+x+1$, and ${\mathbb{F}}_4$ is obtained by adjoining a root of $\alpha$ of this polynomial to $F$ ¹

$${\mathbb{F}}_4\cong {\mathbb{F}}_2[x]/(x^2+x+1).$$

By Prp 119.13, $\{1,\alpha \}$ is a ${\mathbb{F}}_2$-basis for ${\mathbb{F}}_4$. It follows that ${\mathbb{F}}_4=\{0,1,\alpha ,1+\alpha \}$.

Lemma 345.2(Lemma).

Let $q=p^r$ be a positive power of $p$.

1. The polynomial $x^q-x\in {\mathbb{F}}_p[x]$ has no multiple root in any field extension of ${\mathbb{F}}_p$.

2. In the polynomial ring ${\mathbb{F}}_p[x,y]$, $(x+y{)}^q=x^q+y^q$.

Proof.

$(1)$ The derivative of $x^q-x$ is $q{x}^{q-1}-1=-1$. Thus, $x^q-x$ and its derivative are relatively prime. Use Cor 342.7.

---

$(2)$ Expand $(x+y{)}^p$ in $\mathbb{Z}[x,y]$. The binomial coefficients $\left(\genfrac{}{}{0ex}{}{p}{r}\right)$ are divisible by $p$ for $r$ in the range $1<r<p$. Thus, the map $\mathbb{Z}[x,y]\to {\mathbb{F}}_p[x,y]$ sends these coefficients to zero, and $(x+y{)}^p=x^p+y^p$ in ${\mathbb{F}}_p[x,y]$. The result is obtained for $q=p^r$ by induction.□

0890ce

Lemma 345.3(Lemma).

Let $p$ be a prime and let $q=p^r$ be a positive power of $p$. Let $L$ be a field of characteristic $p$, and let $K$ be the set of roots of $x^q-x$ in $L$. Then $K$ is a subfield of $L$.

Proof.

Let $\alpha$ and $\beta$ be roots of the polynomial $x^q-x$ in $L$. We have to show that $\alpha +\beta$, $-\alpha$, $\alpha \beta$, ${\alpha }^{-1}$, and $1$ are roots of the same polynomial.

1. $(\alpha +\beta {)}^q={\alpha }^q+{\beta }^q=\alpha +\beta$, by Lem 2.2.
2. $(-\alpha {)}^q=(-1{)}^q{\alpha }^q=-\alpha$.
3. $(\alpha \beta {)}^q={\alpha }^q{\beta }^q=\alpha \beta$
4. $({\alpha }^{-1}{)}^q={\alpha }^{-1}$
5. $1^q=1$.

□

f27db6

Theorem 345.4(Theorem).

Let $p$ be a prime integer, and let $q=p^r$ be a positive power of $p$.

1. Let $K/{\mathbb{F}}_p$ be a field of order $q$. The elements of $K$ are roots of the polynomial $x^q-x\in {\mathbb{F}}_p[x]$.

2. The irreducible factors of the polynomial $x^q-x$ over the prime field ${\mathbb{F}}_p$ are the irreducible polynomials in ${\mathbb{F}}_p[x]$ whose degrees divide $r$.

3. Let $K$ be a field of order $q$. The multiplicative group $K^{\times }$ of nonzero elements of $K$ is a cyclic group of order $q-1$.

4. There exists a field of order $q$, and all fields of order $q$ are isomorphic.

5. A field of order $p^r$ contains a subfield of order $p^k$ iff $k$ divides $r$.

Proof.

$(1)$ The multiplicative group $K^{\times }$ has order $q-1$. Therefore the order of any element $\alpha$ of $K^{\times }$ divides $q-1$, so ${\alpha }^{q-1}-1=0$, which means $\alpha$ is a root of the polynomial $x^{q-1}-1$. The remaining root of $K$, zero, is the root of the polynomial $x$. So every element of $K$ is a root of $x^q-x$.

---

$(3)$ By Cor 103.12, $K^{\times }$ is a direct sum of cyclic subgroups, say $H_1\oplus \cdots \oplus H_k$, where $H_i$ has order $d_i$, and $d_1|d_2|\dots |d_k$. let $d:=d_k$. For every $\alpha \in K^{\times }$, we have ${\alpha }^d=1$. Therefore every element of $K^{\times }$ is a root of the polynomial $x^d-1$. This polynomial has at most $d$ roots in $K$, and therefore $|K^{\times }|=q-1⩽d$. On the other hand, $|K^{\times }|=|H_1\oplus \cdots \oplus H_k|=d_1\dots d_k$. So $d_1\dots d_k=|K^{\times }|=q-1⩽d$. Since $d=d_k$, the only possibility is that $k=1$ and $q-1=d$. Therefore $K^{\times }=H_1$, and $K^{\times }$ is cyclic.

---

$(4)$ There exists a field extension $L/{\mathbb{F}}_p$ in which $x^q-x$ splits completely by Thm 342.2. The $q$ roots of $x^q-x$ in $L$ form a field by Lem 3.

We will now show that two fields $K$ and $K^{\prime }$ of the same order $q=p^r$ are isomorphic. As we know, both can be viewed as extensions $K/{\mathbb{F}}_p$ and $K^{\prime }/{\mathbb{F}}_p$. Let $\alpha$ be a generator for the cyclic group $K^{\times }$. Then $K={\mathbb{F}}_p(\alpha )$, so the irreducible polynomial $f$ for $\alpha$ over ${\mathbb{F}}_p$ has degree $[K:{\mathbb{F}}_p]=r$. $f$ generates the ideal of polynomials in ${\mathbb{F}}_p[x]$ with root $\alpha$, since if $g\in {\mathbb{F}}_p[x]$ is such that $g(\alpha )=0$, then $f$ and $g$ are not relatively prime in ${\mathbb{F}}_p[x]$ by Prp 342.6, and $f$ being irreducible forces $f|g$. Since $\alpha$ is a root of $x^q-x$, $f$ divides $x^q-x$. Since $x^q-x$ splits completely in $K^{\prime }$, $f$ has a root ${\alpha }^{\prime }$ in $K^{\prime }$ too. Then ${\mathbb{F}}_p(\alpha )$ and ${\mathbb{F}}_p({\alpha }^{\prime })$ are both isomorphic to ${\mathbb{F}}_p[x]/(f)$, hence to each other. Since the degree of the extension ${\mathbb{F}}_p({\alpha }^{\prime })/{\mathbb{F}}_p$ is equal to $r$, $|{\mathbb{F}}_p({\alpha }^{\prime })|=p^r$, so we must have ${\mathbb{F}}_p({\alpha }^{\prime })=K^{\prime }$. Therefore, $K$ and $K^{\prime }$ are isomorphic.

---

$(5)$ Let $q=p^r$ and $q^{\prime }=p^k$. Then $[{\mathbb{F}}_q:{\mathbb{F}}_p]=r$ and $[{\mathbb{F}}_{q^{\prime }}:{\mathbb{F}}_p]=k$; we can’t have ${\mathbb{F}}_p\subseteq {\mathbb{F}}_{q^{\prime }}\subseteq {\mathbb{F}}_q$ unless $k|r$.

Suppose $k|r$, say $r=ks$. Substitution of $y=p^k$ into the equation $y^s-1=(y-1)(y^{s-1}+\cdots +y+1)$ shows that $q^{\prime }-1$ divides $q-1$. Let $K$ be a field of order $q$. Since the multiplicative group $K^{\times }$ is cyclic of order $q-1$, and since $q^{\prime }-1$ divides $q-1$, $K^{\times }$ contains an element $\beta$ of order $q^{\prime }-1$. Then $q^{\prime }-1$ powers of this element are roots of $x^{q^{\prime }-1}-1$ in $K$. Therefore $x^{q^{\prime }}-x$ splits completely in $K$. Lem 3 shows that the roots form a field of order $q^{\prime }$.

---

$(2)$ Let $g$ be an irreducible polynomial over ${\mathbb{F}}_p$ of degree $k$. The polynomial $x^q-x$ splits into linear factors in ${\mathbb{F}}_q$.

If $g$ divides $x^q-x$, it will also factor into linear factors in ${\mathbb{F}}_q$, so it will have a root $\beta$ in ${\mathbb{F}}_q$. Now,

$$\begin{array}{r}r=\underset{r}{\underbrace{[{\mathbb{F}}_q:{\mathbb{F}}_p]}}=[{\mathbb{F}}_q:{\mathbb{F}}_p(\beta )]\underset{k}{\underbrace{[{\mathbb{F}}_p(\beta ):{\mathbb{F}}_p]}},\end{array}$$

so $k|r$.

Conversely, suppose that $k$ divides $r$. Let $\beta$ be a root of $g$ in an extension field of ${\mathbb{F}}_p$. Then $[{\mathbb{F}}_p(\beta ):{\mathbb{F}}_p]=k$, so $|{\mathbb{F}}_p(\beta )|=p^k$. By $(5)$, $K$ contains a subfield $S$ of order $p^k$, and by $(4)$, there exists an isomorphism $\tau :{\mathbb{F}}_p(\beta )\to S$. Note that the isomorphism must be the identity on ${\mathbb{F}}_p$. By Rmk 338.2, ${\beta }^{\tau }\in S$ is a root of $g$. Therefore $g$ has a root in $K$, and so $g$ divides $x^q-x$ by Prp 342.6.□

977ed1

Corollary 345.5(Corollary).

For every positive integer $r$, there exists an irreducible polynomial of degree $r$ over ${\mathbb{F}}_p$.

Proof.

By Thm 4.4, there is a field $K$ of order $q=p^r$. Its degree over ${\mathbb{F}}_p$ is $r$. By Thm 4.3, $K^{\times }$ is cyclic. Clearly, a generator $\alpha$ for this cyclic group will generate $K$ as an extension field, i.e, $K={\mathbb{F}}_p(\alpha )$. Since $[K:{\mathbb{F}}_p]=r$, the degree of the irreducible polynomial of $\alpha$ over ${\mathbb{F}}_p$ is $r$ by Prp 119.13.□

4e0304

Remark 345.6(Remark).

The proof of Cor 5 tells us that a field of order $p^r$ can be obtained by going modulo an irreducible polynomial of degree $r$ in ${\mathbb{F}}_p[x]$. Conversely, if $f\in {\mathbb{F}}_p[x]$ is irreducible of degree $r$, then the quotient ${\mathbb{F}}_p[x]/(f)$ is a field (because $(f)$ is maximal), and is an ${\mathbb{F}}_p$-vector space of dimension $r$, hence has cardinality $p^r$. Therefore it is a field with $p^r$ elements, and by Thm 4.4, isomorphic to ${\mathbb{F}}_{p^r}$.

91a225

Proposition 345.7(Proposition).

No finite field is algebraically closed.

Proof.

Let’s suppose that $K$ is finite and write $K=\{{\alpha }_1,\dots ,{\alpha }_n\}$. Now take the polynomial $p(x)=(x-{\alpha }_1)\dots (x-{\alpha }_n)+1\in K[x]$. It’s easy to see that $p(x)$ doesn’t have any roots in $K$. Hence, $K$ is not algebraically closed.□

Footnotes

1. See the proof of Prp 338.5. ↩