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.
$$
a_{1} \sigma_{1}+\cdots a_{r} \sigma_{r}=0
$$
where all $a_{i}$ are nonzero and $r$ is as small as possible. Show that for every $h$ and $g \in G$ we have $e$
$$
\sum_{i=1}^{r} a_{i} \sigma_{1}(h) \sigma_{i}(g)=0
$$
and
$$
\sum_{i=1}^{r} a_{i} \sigma_{i}(h) \sigma_{i}(g)=0
$$
$$
\sum_{i=1}^{r} a_{i}\left(\sigma_{1}(h)-\sigma_{i}(h)\right) \sigma_{i}(g)=0
$$
With $g$ arbitrary, reach a contradiction by an appropriate choice of $h$.
$$
\sigma_{1}(h)-\sigma_{i}(h)\left\{\begin{array}{ll}
=0 & \text { if } i=1 \\
\neq 0 & \text { if } i=2
\end{array}\right.
$$
This contradicts the minimality of $r .$ (Note that the $i=2$ case is important, since there is no contradiction if $\sigma_{1}(h)-\sigma_{i}(h)=0$ for all $i$.)
Let $E=\mathbb{Q}(\sqrt[3]{2}),$ as in $(3.5 .3) .$ The Galois group of the extension consists of the identity automorphism alone. For any $\mathbb{Q}$ -monomorphism $\sigma$ of $E$ must take $\sqrt[3]{2}$ into a root of $X^{3}-2 .$ Since the other two roots are complex and do not belong to $E, \sqrt[3]{2}$ must map to itself. But $\sigma$ is completely determined by its action on $\sqrt[3]{2},$ and the result follows.
If $E / F$ is not normal, we can always enlarge $E$ to produce a normal extension of $F$. If $C$ is an algebraic closure of $E,$ then $C$ contains all the roots of every polynomial in $F[X],$ so $C / F$ is normal. Let us try to look for a smaller normal extension.
the Galois group consists of the identity alone. Since the identity fixes all elements, the fixed field of $G$ is $\mathbb{Q}(\sqrt[3]{2})$.
Find the Galois group of $\mathbb{C} / \mathbb{R}$.
Since $\mathbb{C}=\mathbb{R}[i],$ an $\mathbb{R}$ -automorphism $\sigma$ of $\mathbb{C}$ is determined by its action on $i$. Since $\sigma$ must permute the roots of $X^{2}+1$ , we have $\sigma(i)=i$ or $-i .$ Thus the Galois group has two elements, the identity automorphism and complex conjugation.
abstract algebra代写请认准UpriviateTA. UpriviateTA为您的留学生涯保驾护航。
更多内容请参阅另外一份Galois代写.
