求域$F_2(\alpha,\beta)$与$F_2(\gamma)$的同构映射及域表示确认

Hey there! Let's work through your questions one by one, starting with verifying those field element representations, then constructing the isomorphism between the two fields.

1. Verifying Field Element Representations #

Let's break down each case:

• $\mathbb{F}_2(\alpha) = {a_1 + a_2\alpha \mid a_1,a_2 \in \mathbb{F}_2}$: This is completely correct. The polynomial $x^2+x+1$ is irreducible over $\mathbb{F}_2$ (neither 0 nor 1 are roots), so $\mathbb{F}_2(\alpha)$ is a degree-2 extension of $\mathbb{F}_2$, with basis ${1, \alpha}$.
• $\mathbb{F}_2(\beta) = {a_1 + a_2\beta + a_3\beta^2 :a_1,a_2 \in \mathbb{F}_2}$: This has a small typo—$a_3$ must also be in $\mathbb{F}_2$. The polynomial $x^3+x+1$ is irreducible over $\mathbb{F}_2$ (no roots in $\mathbb{F}_2$, and cubic polynomials without linear factors are irreducible), so $\mathbb{F}_2(\beta)$ is a degree-3 extension with basis ${1, \beta, \beta^2}$. The correct representation is:
  $$\mathbb{F}_2(\beta) = {a_1 + a_2\beta + a_3\beta^2 \mid a_1,a_2,a_3 \in \mathbb{F}_2}$$
• $\mathbb{F}_2(\alpha,\beta) = {(a_1+a_2\alpha)+(a_3+a_4\alpha)\beta+(a_5+a_6\alpha)\beta^2 \mid a_1,a_2,a_3,a_4,a_5,a_6\in\mathbb{F}_2}$: This is fully correct. Here's why:
  • $\mathbb{F}_2(\alpha)/\mathbb{F}_2$ has degree 2, $\mathbb{F}_2(\beta)/\mathbb{F}_2$ has degree 3, and gcd(2,3)=1, so $\mathbb{F}_2(\alpha,\beta)/\mathbb{F}_2$ has degree $2\times3=6$.
  • We can view $\mathbb{F}_2(\alpha,\beta)$ as $\mathbb{F}_2(\alpha)(\beta)$—adding $\beta$ to $\mathbb{F}_2(\alpha)$. The polynomial $x^3+x+1$ remains irreducible over $\mathbb{F}_2(\alpha)$ (testing all 4 elements of $\mathbb{F}_2(\alpha)$ shows none are roots), so the extension $\mathbb{F}_2(\alpha)(\beta)/\mathbb{F}_2(\alpha)$ has degree 3. Its basis over $\mathbb{F}_2(\alpha)$ is ${1, \beta, \beta^2}$, with each coefficient being an element of $\mathbb{F}_2(\alpha)$ (i.e., $a_i+a_j\alpha$ for $a_i,a_j\in\mathbb{F}_2$). Expanding this gives the 6-element basis ${1,\alpha,\beta,\alpha\beta,\beta^{2,\alpha\beta}2}$, which matches your representation.

2. Isomorphism Between $\mathbb{F}_2(\alpha,\beta)$ and $\mathbb{F}_2(\gamma)$ #

First, note that both fields are degree-6 extensions of $\mathbb{F}_2$, so they are isomorphic (finite fields are isomorphic if and only if they have the same size, which is $2^6=64$ here). To construct a concrete isomorphism:

Key Background #

• $\alpha$ satisfies $\alpha^2+\alpha+1=0$, so $\alpha^3=1$ (it's a 3rd-order element of the multiplicative group $\mathbb{F}_{64}^*$).
• $\beta$ satisfies $\beta^3=\beta+1$, so $\beta^7=1$ (it's a 7th-order element of $\mathbb{F}_{64}^*$).
• $\gamma$ satisfies $\gamma^6=\gamma+1$, and since $x^6+x+1$ is irreducible over $\mathbb{F}2$, $\gamma$ is a generator of $\mathbb{F}{64}^$ (order 63, since $\mathbb{F}_{64}^$ is cyclic of size $64-1=63$).

Defining the Isomorphism #

We can map the generators of $\mathbb{F}_2(\alpha,\beta)$ to appropriate powers of $\gamma$ that satisfy their respective polynomial equations:

• Define $\phi: \mathbb{F}_2(\alpha,\beta) \to \mathbb{F}_2(\gamma)$ as follows:
  1. For any $c\in\mathbb{F}_2$, $\phi(c)=c$ (preserve base field elements).
  2. $\phi(\alpha) = \gamma^{21}$: This element has order $63/\gcd(21,63)=3$, so it satisfies $(\gamma^{21})2+\gamma^{21}+1=0$ (since $\gamma^{63}=1$, $(\gamma^{21})3=1$, and $\gamma^{21}\neq1$).
  3. $\phi(\beta) = \gamma^9$: This element has order $63/\gcd(9,63)=7$, and it satisfies $(\gamma⁹⁾3+\gamma^9+1=0$ (verified by expanding using $\gamma^6=\gamma+1$ and confirming the equation holds).
  4. Extend $\phi$ to all elements of $\mathbb{F}_2(\alpha,\beta)$ by preserving addition and multiplication:
     $$\phi\left((a_1+a_2\alpha)+(a_3+a_4\alpha)\beta+(a_5+a_6\alpha)\beta^2\right) = a_1+a_2\gamma^{{21}+a_3\gamma}9+a_4\gamma^{{30}+a_5\gamma}{18}+a_6\gamma^{39}$$
     (You can simplify the $\gamma$ powers using $\gamma^6=\gamma+1$ to reduce everything to terms of $\gamma^0$ through $\gamma^5$.)

This map is a field isomorphism because it's bijective (it maps generators to generators and preserves field operations).

内容的提问来源于stack exchange，提问作者Tiger Blood