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# abstract algebra代写|galois theory代写|抽象代数代写|伽罗瓦基本定理代写

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。

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
compute
$$\Psi^{\prime}(\Psi(\sigma)) \tau^{-1} y=\tau^{-1} \Psi(\sigma) y=\tau^{-1} \Psi(\sigma) \tau x=\tau^{-1} \tau \sigma(x)=\sigma(x)=\sigma\left(\tau^{-1} y\right)$$
Thus $\Psi^{\prime} \Psi$ is the identity on $G$.

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 $N$.

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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