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。

Problem 1. Express $X_{1}^{2} X_{2} X_{3}+X_{1} X_{2}^{2} X_{3}+X_{1} X_{2} X_{3}^{2}$ in terms of elementary symmetric functions.
Proof . We have $r_{1}=2, r_{2}=1, r_{3}=1$ so $t_{1}=1, t_{2}=0, t_{3}=1 .$ The algorithm terminates in one step after after subtraction of $\left(X_{1}+X_{2}+X_{3}\right)\left(X_{1} X_{2} X_{3}\right)$. The given polynomial can be expressed as $e_{1} e_{3}$.
Problem 2. To begin the proof of Dedekind’s lemma, suppose that the $\sigma_{i}$ are linearly dependent. By renumbering the $\sigma_{i}$ if necessary, we have
$$a_{1} \sigma_{1}+\cdots a_{r} \sigma_{r}=0$$
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
$$\sum_{i=1}^{r} a_{i} \sigma_{i}(h) \sigma_{i}(g)=0$$
Proof . Equation (1) follows upon taking $\sigma_{1}(h)$ outside the summation and using the linear dependence. Equation (2) is also a consequence of the linear dependence, because $\sigma_{i}(h) \sigma_{i}(g)=\sigma_{i}(h g)$
Problem 3. Continuing Problem 2 , subtract (2) from (1) to get
$$\sum_{i=1}^{r} a_{i}\left(\sigma_{1}(h)-\sigma_{i}(h)\right) \sigma_{i}(g)=0$$
With $g$ arbitrary, reach a contradiction by an appropriate choice of $h$.
Proof . By hypothesis, the characters are distinct, so for some $h \in G$ we have $\sigma_{1}(h) \neq \sigma_{2}(h) .$ Thus in (3), each $a_{i}$ is nonzero and
$$\sigma_{1}(h)-\sigma_{i}(h)\left\{\begin{array}{ll} =0 & \text { if } i=1 \\ \neq 0 & \text { if } i=2 \end{array}\right.$$
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 $i$.)
Problem 4. If $G$ is the Galois group of $\mathbb{Q}(\sqrt[3]{2})$ over $\mathbb{Q}$, what is the fixed field of $G ?$
Proof .

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 $\mathbb{Q}(\sqrt[3]{2})$.

Problem 5.

Find the Galois group of $\mathbb{C} / \mathbb{R}$.

Proof .

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.

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