ANT.2.2

Galois Theory and Field ExtensionsMinimal Polynomial via Galois ConjugatesLemmaCyclotomic PolynomialsDefinitionPropositionTrace and NormHistorical NoteCompositional Properties of Trace and Norm in Tower ExtensionsBase Field Extension and Matrix DiagonalizationStep 1: Extension of $M$ over $L$Step 2: Extension of $K$ over $M$Step 3: Combining into one stepStep 4: Trace compositionStep 5: Norm composition

 

Galois Theory and Field Extensions

Minimal Polynomial via Galois Conjugates

Lemma

Lemma (Minimal Polynomial via Galois Conjugates): If $E/K$$E/K$ is a Galois extension and $\alpha \in E$$\alpha \in E$, then

$$m_{\alpha ,K}(X)=\prod _{\beta \in \mathrm{Orb}(\alpha )}(X-\beta )$$

where $\mathrm{Orb}(\alpha )=\{\sigma (\alpha ):\sigma \in \mathrm{Gal}(E/K)\}$$\mathrm{Orb}(\alpha) =\{\sigma(\alpha) : \sigma \in \mathrm{Gal}(E/K)\}$ is the Galois orbit of $\alpha$$\alpha$.

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Proof (Galois-theoretic approach):

Let $G=\mathrm{Gal}(E/K)$$G = \mathrm{Gal}(E/K)$ and $m=m_{\alpha ,K}(X)$$m = m_{\alpha,K}(X)$ be the minimal polynomial of $\alpha$$\alpha$ over $K$$K$.

Step 1: Let $F$$F$ be the splitting field of $m(X)$$m(X)$ over $K$$K$.

Since $m(X)$$m(X)$ is separable (as $E/K$$E/K$ being Galois implies separability), the extension $F/K$$F/K$ is Galois.

Step 2: Establish the subfield relation and group homomorphism.

Clearly $\alpha \in E$$\alpha \in E$, so $K(\alpha )\subseteq E$$K(\alpha) \subseteq E$. Since $F$$F$ is the splitting field of $m(X)$$m(X)$ and $m(X)$$m(X)$ is the minimal polynomial of $\alpha$$\alpha$, we have $F=K({\alpha }_1,\dots ,{\alpha }_d)$$F = K(\alpha_1, \ldots, \alpha_d)$, where ${\alpha }_1=\alpha ,{\alpha }_2,\dots ,{\alpha }_d$$\alpha_1 = \alpha, \alpha_2, \ldots, \alpha_d$ are all roots of $m(X)$$m(X)$.

Since these roots are all algebraic conjugates of $\alpha$$\alpha$, they all lie in $E$$E$, hence $F\subseteq E$$F \subseteq E$.

By the Fundamental Theorem of Galois Theory, there exists a surjective homomorphism:

$$\rho :G=\mathrm{Gal}(E/K)\twoheadrightarrow \mathrm{Gal}(F/K)$$

Step 3: Analyze the relationship between orbits and roots.

Let the roots of $m(X)$$m(X)$ be $\{{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_d\}$$\{\alpha_1, \alpha_2, \ldots, \alpha_d\}$, where ${\alpha }_1=\alpha$$\alpha_1 = \alpha$.

For any $\sigma \in G$$\sigma \in G$, $\sigma (\alpha )$$\sigma(\alpha)$ is an algebraic conjugate of $\alpha$$\alpha$, i.e., $m(\sigma (\alpha ))=0$$m(\sigma(\alpha)) = 0$. This follows because:

$$m(\sigma (\alpha ))=\sigma (m(\alpha ))=\sigma (0)=0$$

Therefore, $\mathrm{Orb}(\alpha )\subseteq \{{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_d\}$$\mathrm{Orb}(\alpha) \subseteq \{\alpha_1, \alpha_2, \ldots, \alpha_d\}$.

Step 4: Prove that the orbit equals the root set.

Conversely, for any root ${\alpha }_i$$\alpha_i$ of $m(X)$$m(X)$, since $F/K$$F/K$ is Galois, there exists $\tau \in \mathrm{Gal}(F/K)$$\tau \in \mathrm{Gal}(F/K)$ such that $\tau (\alpha )={\alpha }_i$$\tau(\alpha) = \alpha_i$.

Since $\rho :G\twoheadrightarrow \mathrm{Gal}(F/K)$$\rho: G \twoheadrightarrow \mathrm{Gal}(F/K)$ is surjective, there exists $\sigma \in G$$\sigma \in G$ such that $\rho (\sigma )=\tau$$\rho(\sigma) = \tau$.

This means $\sigma {|}_F=\tau$$\sigma|_F = \tau$, hence $\sigma (\alpha )=\tau (\alpha )={\alpha }_i$$\sigma(\alpha) = \tau(\alpha) = \alpha_i$.

Therefore ${\alpha }_i\in \mathrm{Orb}(\alpha )$$\alpha_i \in \mathrm{Orb}(\alpha)$, which gives us $\{{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_d\}\subseteq \mathrm{Orb}(\alpha )$$\{\alpha_1, \alpha_2, \ldots, \alpha_d\} \subseteq \mathrm{Orb}(\alpha)$.

Step 5: Conclusion.

Combining Steps 3 and 4, we obtain: $\mathrm{Orb}(\alpha )=\{{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_d\}$$\mathrm{Orb}(\alpha) = \{\alpha_1, \alpha_2, \ldots, \alpha_d\}$

Therefore:

$$m_{\alpha ,K}(X)=\prod _{i=1}^d(X-{\alpha }_i)=\prod _{\beta \in \mathrm{Orb}(\alpha )}(X-\beta )$$

This completes the proof. $◻$$\square$

Cyclotomic Polynomials

Definition

Consider the Galois extension $\mathbb{Q}\subset \mathbb{Q}({\zeta }_n)$$\mathbb{Q}\subset\mathbb{Q}(\zeta_n)$, let $G=\mathrm{Gal}(\mathbb{Q}({\zeta }_n)/\mathbb{Q})\cong U(\mathbb{Z}/n\mathbb{Z})$$G=\mathrm{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong U(\mathbb{Z}/n\mathbb{Z})$, define

$${\mathrm{\Phi }}_n=\prod _{\sigma \in G}(x-\sigma ({\zeta }_n))$$

i.e. ${\mathrm{\Phi }}_n(\alpha )=0⟺{\alpha }^n-1=0$$\Phi_n(\alpha)=0\iff \alpha^n-1=0$ with $\mathrm{ord}(\alpha )=n$$\mathrm{ord}(\alpha)=n$.

By the Lemma above, ${\mathrm{\Phi }}_n$$\Phi_n$ is the minimal polynomial of ${\zeta }_n$$\zeta_n$. Hence it is irreducible.

Proposition

$$x^n-1=\prod _{d|n}{\mathrm{\Phi }}_d$$

Proof. Let $\mathrm{Gal}(\mathbb{Q}({\zeta }_n)/\mathbb{Q})$$\mathrm{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ acts on $1,{\zeta }_n,...,{\zeta }_n^{n-1}$$1,\zeta_n,...,\zeta_n^{n-1}$, then $x^n-1=\prod _{d|n}{\mathrm{\Phi }}_d$$x^n-1=\prod_{d|n}\Phi_d$ comes from the orbit decomposition directly.

$${\zeta }_n^i\sim {\zeta }_n^j⟺\mathrm{ord}({\zeta }_n^i)=\mathrm{ord}({\zeta }_n^j)◻$$

Trace and Norm

Historical Note

Alexander Grothendieck (1928-2014) revolutionized algebraic geometry and number theory through his radical reconceptualization of mathematical foundations. In the 1950s and 1960s, working primarily at the Institut des Hautes Études Scientifiques (IHÉS) near Paris, Grothendieck developed a vast framework that transformed how mathematicians approach abstract structures.

The tensor product approach to field extensions presented here reflects Grothendieck's profound influence. While classical Galois theory had been established in the 19th century, Grothendieck's functorial perspective and scheme theory provided powerful new tools for understanding these structures. His development of étale cohomology, descent theory, and the formalism of derived categories created a language where field extensions could be viewed within a broader categorical context.

Let $L\subset K$$L\subset K$ be a finite separable extension, then $K=L[\theta ]\cong L[x]/(f(x))$$K=L[\theta]\cong L[x]/(f(x))$.

Hence

$$K{\otimes }_L\bar{L}\cong \bar{L}[x]/(x-{\theta }_1)\times ...\times \bar{L}[x]/(x-{\theta }_n)\cong {\bar{L}}^n$$

For $k\in K$$k\in K$, we have

$$k\otimes 1⟼({\sigma }_1(k),...,{\sigma }_n(k))$$

Here ${\sigma }_1,...,{\sigma }_n$$\sigma_1,...,\sigma_n$ is all the $L$$L$-embedding.

The linear map $m_k(x\otimes 1)=k\cdot (x\otimes 1)$$m_k(x\otimes 1)=k\cdot (x\otimes 1)$ correspond to the matrix

$$m_k=\mathrm{diag}({\sigma }_1(k),...,{\sigma }_n(k))$$

The trace of this matrix is $\sum _{i=1}^n{\sigma }_i(k)$$\sum_{i=1}^n\sigma_i(k)$, and $\det \left(m_k\right)=\prod _{i=1}^n{\sigma }_i(k)$$\det(m_k)=\prod_{i=1}^n\sigma_i(k)$. That is, the trace and norm of $k$$k$.

Compositional Properties of Trace and Norm in Tower Extensions

Let $L\subset M\subset K$$L\subset M\subset K$ be finite separable extensions with

$$[M:L]=m,[K:M]=n,[K:L]=mn$$

and fix an algebraic closure $\bar{L}$$\overline{L}$ to split all polynomials.

Base Field Extension and Matrix Diagonalization

Step 1: Extension of $M$$M$ over $L$$L$

Extending $M$$M$ from $L$$L$ to $\bar{L}$$\overline{L}$, we have

$$M{\otimes }_L\bar{L}\cong {\bar{L}}^m$$

corresponding to the set of $L$$L$-embeddings $\{{\sigma }_1,\dots ,{\sigma }_m\}={\text{Hom}}_L(M,\bar{L})$$\{\sigma_1,\dots,\sigma_m\}=\text{Hom}_L(M,\overline{L})$.

For any $y\in M$$y\in M$, its multiplication operator $m_y$$m_y$ on ${\bar{L}}^m$$\overline{L}^m$ is

$$m_y=\text{diag}({\sigma }_1(y),\dots ,{\sigma }_m(y))$$

Therefore

$${\text{Tr}}_{M/L}(y)=\sum _{i=1}^m{\sigma }_i(y),{\text{N}}_{M/L}(y)=\prod _{i=1}^m{\sigma }_i(y)$$

Remark.

By the infinite Galois corresponding, the fixed field of $\mathrm{Gal}(\bar{L}/L)$$\mathrm{Gal}(\overline L/L)$ is $L$$L$.

Hence we have ${\text{Tr}}_{M/L}(y)=\sum _{i=1}^m{\sigma }_i(y),{\text{N}}_{M/L}(y)=\prod _{i=1}^m{\sigma }_i(y)\in L$$\text{Tr}_{M/L}(y) = \sum_{i=1}^m\sigma_i(y), \qquad \text{N}_{M/L}(y) = \prod_{i=1}^m\sigma_i(y)\in L$.

Step 2: Extension of $K$$K$ over $M$$M$

Extending $K$$K$ from $M$$M$ to the same $\bar{L}$$\overline{L}$, we have

$$K{\otimes }_M\bar{L}\cong {\bar{L}}^n$$

corresponding to each ${\sigma }_i$$\sigma_i$ in ${\text{Hom}}_L(M,\bar{L})$$\text{Hom}_L(M,\overline{L})$, there is a set of extension embeddings

$$\{{\tau }_{i1},\dots ,{\tau }_{in}\}\subset {\text{Hom}}_M(K,\bar{L})$$

Therefore for $x\in K$$x\in K$, the multiplication operator $m_x$$m_x$ on this $n$$n$-dimensional space is

$$\text{diag}({\tau }_{i1}(x),\dots ,{\tau }_{in}(x))$$

yielding

$$\sum _{j=1}^n{\tau }_{ij}(x)={\text{Tr}}_{K/M}(x),\prod _{j=1}^n{\tau }_{ij}(x)={\text{N}}_{K/M}(x)$$

where the results belong to $M$$M$, then embedded into $\bar{L}$$\overline{L}$ by ${\sigma }_i$$\sigma_i$.

Step 3: Combining into one step

Performing the two-stage base‐change

$$M{\otimes }_L\bar{L}\cong {\bar{L}}^m⟹K{\otimes }_M\bar{L}\cong {\bar{L}}^n$$

is equivalent to a single extension

$$K{\otimes }_L\bar{L}\cong (M{\otimes }_L\bar{L}){\otimes }_M\bar{L}\cong {\bar{L}}^{mn}$$

We now spell out the two‐step unfolding of each coordinate:

1. First–level: ${\sigma }_i$$\sigma_i$–coordinates.
   Under

   $$M{\otimes }_L\bar{L}\cong {\bar{L}}^m,x\otimes 1⟼({\sigma }_1(x),\dots ,{\sigma }_m(x))$$

   each copy of $\bar{L}$$\overline{L}$ is indexed by an $L$$L$–embedding ${\sigma }_i\in {\text{Hom}}_L(M,\bar{L})$$\sigma_i\in\text{Hom}_L(M,\overline{L})$.

2. Second–level: ${\tau }_{ij}$$\tau_{ij}$–coordinates.
   Now view each factor ${\sigma }_i(x)\in \bar{L}$$\sigma_i(x)\in\overline{L}$ as coming from

   $${\sigma }_i:M\hookrightarrow \bar{L}\subset K{\otimes }_L\bar{L}$$

   and extend scalars again along $M\to \bar{L}$$M\to \overline{L}$. This splits each ${\sigma }_i$$\sigma_i$–line into $n$$n$ lines, indexed by

   $$\{{\tau }_{i1},\dots ,{\tau }_{in}\}\subset {\text{Hom}}_M(K,\bar{L})$$

   Concretely,

   $${\sigma }_i(x)⟼({\tau }_{i1}(x),\dots ,{\tau }_{in}(x))$$

3. Combined coordinate map.
   Putting these two steps together, an elementary tensor $x\otimes 1$$x\otimes1$ in $K{\otimes }_L\bar{L}$$K\otimes_L\overline{L}$ corresponds to the concatenated tuple

   $$(x\otimes 1)⟼(\underset{{\sigma }_1\text{-block}}{\underbrace{{\tau }_{11}(x),\dots ,{\tau }_{1n}(x)}}|\dots |\underset{{\sigma }_m\text{-block}}{\underbrace{{\tau }_{m1}(x),\dots ,{\tau }_{mn}(x)}})\in {\bar{L}}^{mn}$$

Thus the full block–diagonal form of the multiplication operator $m_k$$m_k$ on ${\bar{L}}^{mn}$$\overline{L}^{mn}$ has $m$$m$ blocks (one for each ${\sigma }_i$$\sigma_i$), each block being the $n\times n$$n\times n$ diagonal matrix $\text{diag}({\tau }_{ij}(k){)}_{j=1}^n$$\text{diag}\bigl(\tau_{ij}(k)\bigr)_{j=1}^n$.

Step 4: Trace composition

$$\text{Tr}(m_k)=\sum _{i=1}^m\sum _{j=1}^n{\tau }_{ij}(k)$$

$$=\sum _{i=1}^m(\sum _{j=1}^n{\tau }_{ij}(k))$$

$$=\sum _{i=1}^m{\sigma }_i({\text{Tr}}_{K/M}(k))$$

$$={\text{Tr}}_{M/L}({\text{Tr}}_{K/M}(k))$$

This gives us

$${\text{Tr}}_{K/L}(k)={\text{Tr}}_{M/L}\circ {\text{Tr}}_{K/M}(k)$$

Step 5: Norm composition

$$\det \left(m_k\right)=\prod _{i=1}^m\prod _{j=1}^n{\tau }_{ij}(k)$$

$$=\prod _{i=1}^m(\prod _{j=1}^n{\tau }_{ij}(k))$$

$$=\prod _{i=1}^m{\sigma }_i({\text{N}}_{K/M}(k))$$

$$={\text{N}}_{M/L}({\text{N}}_{K/M}(k))$$

That is,

$${\text{N}}_{K/L}(k)={\text{N}}_{M/L}\circ {\text{N}}_{K/M}(k)$$