Abstract algebra不算是一门简单的学科,这门学科在国内叫做抽象代数,经常有很多学生在学linear algebra或者analysis(advance calculus)的时候觉得并不困难,但是却觉得Abstract algebra很难,这是因为没有找到正确的方法学习Abstract algebra,UpriviateTA有一系列非常擅长Abstract algebra的老师,可以确保您在Abstract algebra取得满意的成绩。

以下是UCLA的一次assignment.更多的经典案例请参阅以往案例,关于abstract algebra的更多的以往案例可以参阅相关文章。abstract algebra代写请认准UpriviateTA.

Problem 1. Let $E=F\left(\alpha_{1}, \ldots, \alpha_{n}\right),$ where each $\alpha_{i}$ is algebraic over $F .$ If for each $i=1, \ldots, n,$ all the conjugates of $\alpha_{i}$ (the roots of the minimal polynomial of $\alpha_{i}$ over $F$ ) belong to $E$, show that $E / F$ is normal.
Proof . Let $f_{i}$ be the minimal polynomial of $\alpha_{i}$ over $F$. Then $E$ is a splitting field for $f=$ $f_{1} \cdots f_{n}$ over $F,$ and the result follows.
Problem 2. Suppose that $F=K_{0} \leq K_{1} \leq \cdots \leq K_{n}=E,$ where $E / F$ is a finite Galois extension, and that the intermediate field $K_{i}$ corresponds to the subgroup $H_{i}$ under the Galois correspondence. Show that $K_{i} / K_{i-1}$ is normal (hence Galois) if and only if $H_{i} \leq H_{i-1}$, and in this case, $\operatorname{Gal}\left(K_{i} / K_{i-1}\right)$ is isomorphic to $H_{i-1} / H_{i}$
Proof . This is a corollary of part 2 of the fundamental theorem, with $F$ replaced by $K_{i-1}$ and $G$ replaced by $\operatorname{Gal}\left(E / K_{i-1}\right)=H_{i-1}$
Problem 3. Let $E$ and $K$ be extensions of $F,$ and assume that the composite $E K$ is defined. If $A$ is any set of generators for $K$ over $F$ (for example, $A=K$ ), show that $E K=E(A)$, the field formed from $E$ by adjoining the elements of $A$.
Proof . $E(A)$ is a field containing $E \geq F$ and $A,$ hence $E(A)$ contains $E$ and $K,$ so that by definition of composite, $E K \leq E(A)$. But any field (in particular $E K$ ) that contains $E$ and $K$ contains $E$ and $A$, hence contains $E(A)$. Thus $E(A) \leq E K$.
Problem 4. Let $\mathrm{E} / \mathrm{F}$ be a finite Galois extension with Galois group $G,$ and let $E^{\prime} / F^{\prime}$ be a finite Galois extension with Galois group $G^{\prime}$. If $\tau$ is an isomorphism of $E$ and $E^{\prime}$ with $\tau(F)=F^{\prime},$ we expect intuitively that $G \cong G^{\prime} .$ Prove this formally.
Proof .

If $\sigma \in G,$ define $\Psi(\sigma)(\tau(x))=\tau \sigma(x), x \in E .$ Then $\psi(\sigma) \in G^{\prime} .$ [If $y=\tau(x) \in F^{\prime}$ with
$x \in F,$ then $\Psi(\sigma) y=\Psi(\sigma) \tau x=\tau \sigma(x)=\tau(x)=y .]$ Now $\Psi\left(\sigma_{1} \sigma_{2}\right) \tau(x)=\tau \sigma_{1} \sigma_{2}(x)$ and
$\Psi\left(\sigma_{1}\right) \Psi\left(\sigma_{2}\right) \tau(x)=\Psi\left(\sigma_{1}\right) \tau \sigma_{2}(x)=\tau \sigma_{1} \sigma_{2}(x),$ so $\Psi$ is a group homomorphism. The
inverse of $\Psi$ is given by $\Psi^{\prime}\left(\sigma^{\prime}\right) \tau^{-1} y=\tau^{-1} \sigma^{\prime}(y), \sigma^{\prime} \in G^{\prime}, y \in E^{\prime} .$ To see this, we
\Psi^{\prime}(\Psi(\sigma)) \tau^{-1} y=\tau^{-1} \Psi(\sigma) y=\tau^{-1} \Psi(\sigma) \tau x=\tau^{-1} \tau \sigma(x)=\sigma(x)=\sigma\left(\tau^{-1} y\right)
Thus $\Psi^{\prime} \Psi$ is the identity on $G$.

Problem 5.

Let $K / F$ be a finite separable extension. Although $K$ need not be a normal extension of $F,$ we can form the normal closure $N$ of $K$ over $F.$ Then $N / F$ is a Galois extension (see Problem 8 of Section 6.3 ); let $G$ be its Galois group. Let $H=\operatorname{Gal}(N / K),$ so that the fixed field of $H$ is $K .$ If $H^{\prime}$ is a normal subgroup of $G$ that is contained in $H,$ show that the fixed field of $H^{\prime}$ is $N$.

Proof .

Since $H^{\prime}$ is a normal subgroup of $G,$ its fixed field $L=\mathcal{F}\left(H^{\prime}\right)$ is normal over $F,$ so by minimality of the normal closure, we have $N \subseteq L .$ But all fixed fields are subfields of $N,$ so $L \subseteq N,$ and consequently $L=N$.

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