You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert
{{ message }}
Repository navigation
Commit a2b2642
Browse filesBrowse the repository at this point in the historyBrowse files
Copy file name to clipboardExpand all lines: content/抽象代数/Galois.md
+7-2Lines changed: 7 additions & 2 deletions
Display the source diff
Display the rich diff
Original file line number
Diff line number
Diff line change
@@ -11,7 +11,9 @@ title: Galois Theory
11
11
- Let $E$ be a field. An **automorphism** of $E$ is a isomorphism of $E$ onto itself.
12
12
- If $E$ and $K$ are both field extensions of a field $F$ and $\sigma : E \rightarrow K$ is a field isomorphism, then an element $\alpha \in E$ is **fixed** by $\sigma$ if $\sigma (\alpha) = \alpha$. An element $\alpha \in E$ is **fixed** by a collection of isomorphisms if $\alpha$ is fixed by every isomorphism in the collection. A subset $L$ of $E$ is **fixed** by a collection of isomorphisms if every $\alpha \in L$ is fixed by the collection. Often write **remains fixed** instead of simply **fixed**.
13
13
- Let $F \leq K$ be a field extension. The set $G(K/F)$ is the set of all automorphisms of the field $K$ that fix every element of the field $F$.
14
-
- Let $E$ be an algebraic extension of the field $F$. Two elements $\alpha$ and $\beta$ in $E$ are conjugates over $F$, if both have the same minimal polynomial over $F$. That is,$\text{irr}(\alpha, F) = \text{irr}(\beta, F)$.
14
+
- Let $E$ be an algebraic extension of the field $F$. Two elements $\alpha$ and $\beta$ in $E$ are **conjugates over $F$**, if both have the same minimal polynomial over $F$. That is,$\text{irr}(\alpha, F) = \text{irr}(\beta, F)$.
15
+
-**分裂域**:Let $F$ be a field and $P = \{f_1 (x), f_2 (x), \cdots, f_s (x)\}$ be a finite set of polynomials in $F[x]$. An extension field $K$ of $F$ is a splitting field of $P$ over $F$ if every polynomial $f_k (x) \in P$ factors into linear factors in $K[x]$ and for any intermediate field $E$, $F \leq E < K$, at least one polynomial $f_j (x) \in P$ does not factor into linear factors in $E[x]$. A field $K$ is a splitting field for $F$ if $E$ is a **splitting field** for some finite set of polynomials.
16
+
- Let $\sigma : F \rightarrow F$ be a field isomorphism; then $\sigma_x : F[x] \rightarrow F [x]$, defined by $\sigma_x(a_0 + a_1x + \cdots + a_n x^n ) = \sigma (a_0 ) + \sigma (a_1 )x + \cdots + \sigma (a_n )x^n$, is the **polynomial extension** of $\sigma$.
15
17
16
18
## 3. 性质
17
19
@@ -22,6 +24,9 @@ title: Galois Theory
22
24
5.**(The Conjugation Isomorphism)** Let $F$ be a field, $K$ an extension field of $F$, and $\alpha, \beta ∈ K$ algebraic over $F$ with $deg(\alpha, F) = n$. The map $\psi_{\alpha,\beta} : F(\alpha) → F(\beta)$ defined by $\psi_{\alpha,\beta} (c_0 + c_1 \alpha + c_2 \alpha^2 + \cdots + c_{n-1} \alpha^{n-1} ) = c_0 + c_1 \beta + c_2 \beta^2 + \cdots + c_{n-1} \beta^{n-1} $, for $c_i ∈ F$, is an isomorphism of $F(\alpha)$ onto $F(\beta)$ if and only if $\alpha$ and $\beta$ are conjugate over $F$.
23
25
6. Let $K$ be a field extension of $F$ with $\alpha \in K$ algebraic over $F$. Suppose that $\psi$ is an isomorphism of $F(\alpha)$ onto a subfield of $K$, with the property that every element of $F$ is fixed by $\psi$. Then $\psi$ maps $\alpha$ to a conjugate over $F$ of $\alpha$. Conversely, if $\beta \in K$ is conjugate over $F$ with $\alpha$, then there is a unique isomorphism $\psi_{\alpha,\beta}$ mapping $F(\alpha)$ onto a subfield of $K$ with the properties that each $a \in F$ is fixed by $\sigma$ and $\sigma (\alpha) = \beta$.
24
26
7. Let $f(x) \in F[x]$. If $a, b \in F$ and $f(a + bi) = 0$, then $f(a - bi) = 0$.
27
+
8. Let $F$ be a field and $P = \{f_1 (x), f_2 (x), \cdots, f_s (x)\}$ a finite set of polynomials in $F[x]$. Then there is a splitting field $K$ of $P$ over $F$. Furthermore $K$ is a finite extension of $F$.
28
+
9. if $\sigma : F \mapsto F$ is an isomorphism, then $\sigma_x : F[x] \mapsto F [x]$ is also an isomorphism.
29
+
10. Let $K = F(\alpha)$, where $\alpha$ is algebraic over $F$, and let $\sigma : F \mapsto F'$ be a field isomorphism. If $K$ is an extension field of $F$ and $\beta \in K$ is a zero of $\sigma_x (\text{irr}(\alpha, F))$, then there is a unique isomorphism $\varphi : F(\alpha) \mapsto F'(\beta)$ with $\sigma (a) = \varphi(a)$ for all $a \in F$ and $\varphi(\alpha) = \beta$.
11. Let $R$ and $R'$ be rings and let $N$ and $N'$ be ideals of $R$ and $R'$ respectively. Let $\phi$ be a homomorphism of $R$ into $R'$. Then $\phi$ induces a natural homomorphism $\phi^* : R/N \mapsto R'/N':$ $\phi^*(a+N)= \phi(a)+N'$ if $\phi[N] \subseteq N'$. $\phi[N] \subseteq N'$确保了同态映射$\phi^*$的良定义性。
9. Let $G$ and $G'$ be groups, and let $H$ and $H'$ be normal subgroups of $G$ and $G'$ respectively. Let $\phi$ be a homomorphism of $G$ into $G'$. Then $\phi$ induces a natural homomorphism $\phi^* : (G/H) \mapsto (G'/H'):$ $\phi^*(aH)= \phi(a)H'$ if $\phi[H] \subseteq H'$. $\phi[H] \subseteq H'$确保了同态映射$\phi^*$的良定义性。
0 commit comments