Let \(F\) be a field of order 1024.
Find the prime field of \(F\).
Let \(K\) be a subfield of \(F\). Find all possible orders for \(K\).
Solution
The prime field is \(\mathbb F_2\). The possible subfield orders are \[ \boxed{2,\quad4,\quad32,\quad1024}, \] with exactly one subfield of each order.
1. The characteristic is \(2\) and \([F:\mathbb F_2]=10\).
The additive order of \(1\) is a positive divisor of \(1024\). It is greater than \(1\), since a field has \(1\ne0\), and is prime: a composite characteristic \(ab\) with \(1<a,b<ab\) would give nonzero elements \(a\cdot1,b\cdot1\) with product zero. The only prime divisor of \(1024=2^{10}\) is \(2\). The subfield generated by \(1\) is therefore \(\{0,1\}\cong\mathbb F_2\). An \(n\)-dimensional vector space over \(\mathbb F_2\) has \(2^n\) elements, so \([F:\mathbb F_2]=10\).
2. Any subfield has one of the four displayed orders.
Every subfield \(K\) contains \(\mathbb F_2\). Put \(d=[K:\mathbb F_2]\) and \(e=[F:K]\). These are finite positive integers, and the tower law gives \(de=10\). Thus \(d\in\{1,2,5,10\}\), and \(|K|=2^d\) belongs to the displayed list.
3. Each possible order occurs.
Fix a positive divisor \(d\) of \(10\) and put \(q=2^d\). The polynomial \(X^q-X\) divides \(X^{1024}-X\) in \(\mathbb F_2[X]\). Indeed, in the quotient by \((X^q-X)\) the residue class \(x\) satisfies \(x^q=x\). Induction gives \(x^{q^j}=x\) for every \(j\geq1\); taking \(j=10/d\) proves the divisibility.
Every nonzero \(a\in F\) satisfies \(a^{1023}=1\), by Lagrange’s theorem in \(F^\times\) [@DF04]. Including zero, all \(1024\) elements of \(F\) are roots of \(X^{1024}-X\), so \[ X^{1024}-X=\prod_{a\in F}(X-a). \] A monic divisor of this product is itself a product of distinct linear factors in \(F[X]\), by unique factorization for polynomials over a field [@DF04]. Consequently \(X^q-X\) has exactly \(q\) distinct roots in \(F\).
Let \(K_d=\{a\in F:a^q=a\}\). In characteristic \(2\), iterating \((a+b)^2=a^2+b^2\) gives \((a+b)^q=a^q+b^q\). Thus \(K_d\) is closed under addition; additive inverses coincide with the original elements. It is closed under multiplication because \((ab)^q=a^qb^q\), and under nonzero inversion because \((a^{-1})^q=(a^q)^{-1}\). It contains \(0,1\), so it is a subfield, and the root count gives \(|K_d|=q=2^d\).
4. There is no second subfield of any listed order.
If \(K\subseteq F\) is a subfield of order \(2^d\), then Lagrange’s theorem in \(K^\times\) gives \(a^{2^d}=a\) for each nonzero \(a\in K\), and the equation holds at zero as well. Hence \(K\subseteq K_d\). Both fields have \(2^d\) elements, so they are equal.