"Abstract algebra" 09-Galois theory

1. Normal domain expansion

In the study domain \ (f\) of the algebraic expansion \ (e\), the first premise is that the expansion of \ (e\) is present, followed by all the expansion of the same space, that is, they are operational. The space that satisfies such conditions is the algebraic closure of \ (f\), using the language of set theory, the algebraic closure can be described as the sum of the divisions of all the polynomial. The legitimacy of this definition still needs to be elaborated, you can combine the nature of the algebraic expansion of the self-discussion, here first assume its existence. Second, the different closures are not necessarily interoperable, and the following discussion will avoid this "parallel world" discussion, limiting the scope to a selected algebraic closure \ (\omega\).

Even if there is only one closure, there are a number of options for expanding the domain that satisfies a particular condition, a mapping that corresponds to a field to a closure is generally called a domain embedding , and a different embedding is called a conjugate domain . It not only finds a unified closure for the domain, but also is an important method to study the extended domain structure (the conjugate domain is of course the same (f\) remains intact). In the previous construction of the single-spread domain, you may have found that the built-up domain is not related to the selection of the root, they are the conjugate domain. If you embed a single domain into a closed domain, each embedding method corresponds to a root of \ (f (x) \), which may have different elements between them, or it may be the same element, but the embedding method differs.

This is because, perhaps not all of the roots are in the same single domain, we naturally ask: so will there be different elements of split domain embedding? More generally, consider the split domain \ (e\) of the polynomial set \ (S\subseteq f[x]\), assuming that \ (e\) is isomorphic to another split domain \ (E ' \) and the isomorphism is mapped to \ (\sigma\). Because the coefficients of any \ (f (x) = (x-a_1) \cdots (x-a_n) \in s\) are in \ (f\), there is always \ (\sigma (f (x)) =f (x) \), so \ ((\sigma (a_1), \cdots,\sigma (A_n)) \) is just a permutation of \ ((A_1,\cdots,a_n) \). Thus, if all the roots of \ (s\) are \ (r\), then there is the following derivation process, i.e. \ (E ' \) is the automorphism of \ (e\).

\[e ' =\sigma (E) =\sigma (F (R)) =f (\sigma (R)) =f (R) =e\tag{1}\]

Only the self-isomorphic conjugate domain is called the self-conjugate domain , and the invariant domain such as the split domain is called \ (f\)-Self-conjugate domain. The above conclusion proves that the splitting domain of polynomial sets is self-conjugate domain. It is easy to prove that automorphism and \ (f\)-automorphism all form a group, in which the automorphism group is recorded as \ (\text{aut} (E) \), \ (f\)-automorphism group is also called Gamma Galois , which is generally remembered as \ (\text{gal} (f/s), this group will be the focus of our research. If \ (e\) is \ (f (x) \) the Split field on \ (f\), \ (\text{gal} (E/f) \) is also called the polynomial \ (f (x) \) of the gamma Galois, recorded as \ (\text{gal} (f) \) or \ (\text{gal} (f,f) \).

• Prove that \ (\bbb{z},\bbb{q},\bbb{r}\) has only an identity automorphism, and \ (\bbb{c}\) has infinitely many self-isomorphism.

\ (f\)-The self-conjugate domain embodies the uniqueness of the extended domain, and we know that algebraic expansion can begin with the single spread of any algebraic element. Investigation \ (f\)-Self-conjugate spread domain \ (e\) any irreducible polynomial \ (f (x) \), if it has a root \ (a\) on \ (e\), then \ (e\) can be generated starting with \ (f (a) \). It is known in the previous discussion that it is conjugated to a widening domain generated from \ (f (a ') \) (\ (a ' \) is another root of \ (f (x) \)), by \ (f\)-The uniqueness of the self-conjugate domain is known \ (a ' \in e\), so \ (F (x) \) is divided in \ (e\). For any irreducible polynomial \ (f (x) \in f[x]\), if it has root in the expanding domain \ (e\), it will be possible to conclude that the other root is also in \ (e\), this expansion is called the Normal expansion of the domain (note that if \ (f (x) \) in \ (e\) does not have a root, does not mean \ (f (x) \) in Not decomposed in \ (e\). Just now the conclusion is that \ (f\)-The self-conjugate domain is normal expansion, it is also easy to prove that the normal expansion can be regarded as the generation domain of all of its divisive polynomial, combined with the preceding conclusion, the following three propositions are equivalent (\ (e\) is the algebraic extension of \ (f\)).

(1) \ (e\) is the formal expansion of \ (f\);

(2) \ (e\) is a split-domain of a polynomial set in \ (f[x]\);

(3) \ (e\) is \ (f\)-Self-conjugate domain.

In particular, if the expansion is limited expansion, then the second proposition can be changed into a split domain of a polynomial. Through these equivalence definitions it is easy to prove that formal expansion of the intersection is also the formal expansion. The intersection of all formal expansions containing \ (e\) is known as regular closures , and it is easy to prove that finite expansion is a split domain of the smallest polynomial set of the generating element (which can be used as an exercise).

2. Galois theory 2.1 gamma Galois and fixed subdomain

As mentioned earlier, \ (f\)-The automorphism group is a subgroup of automorphism group \ (\text{aut} (E) \), and different subdomains \ (f\) correspond to different subgroups. This reminds us to study the correlation, but note that there are two ways of associating it, one is determined by \ (f\) Galois \ (\text{gal} (E/f) \), and the other is a subdomain of \ (g\) to determine a sub-group \ (\text{ INV} (G) \), which is known as a fixed subdomain of \ (g\). These two mappings are not necessarily the same, and at least some conditions are required, which will be the focus of this section.

\[\TEXT{INV} (G) =\{a\in e\mid \sigma\in G\rightarrow\sigma (a) =a\}\tag{2}\]

Let's look at the basic properties of these mappings, first of all, it is obvious that the containing relationship of the image is opposite to the containment of the original image (equation (3), the following is \ (\text{gal} (E/E) \) abbreviated to \ (\text{gal} (F) \)). It is also easy to prove that the combination of the two mappings amplifies the range of the original image (equation (4)). For complex operations like this, the "Elimination Law" (Formula (5)) is easily obtained by combining the previous two containing relationships with two different angles. These basic properties are very important in the discussion below, and you need to memorize the heart without confusion.

\[g_1\subseteq G_2\LEFTRIGHTARROW\TEXT{INV} (g_1) \SUPSETEQ\TEXT{INV} (g_2), \quad F_1\subseteq F_2\Leftrightarrow\ Text{gal} (F_1) \supseteq\text{gal} (f_2) \tag{3}\]

\[f\subseteq\text{inv}\circ\text{gal} (F), \quad G\subseteq \TEXT{GAL}\CIRC\TEXT{INV} (G) \tag{4}\]

\[\text{gal}\circ\text{inv}\circ\text{gal} (f) =\text{gal} (f), \quad \text{inv}\circ\text{gal}\circ\text{inv} (G) =\ TEXT{INV} (G) \tag{5}\]

2.2 Galois dilation and Artin theorem

In order to study the relationship between automorphism subgroups and sub-domains, we need to do some further research on their characteristics. First to examine the gamma Galois \ (\text{gal} (f/s) \), each of its elements is a \ (f\)-automorphism, the order of the group is the number of automorphism. For finite domain (e=f (a_1,a_2,\cdots,a_n) \), all embedding can be split into a series of single-Spread (F (a_1,\cdots,a_{k-1}) (A_k) \) embedding. The previous conclusion tells us that the number of embedded in each single-spread domain \ (c_k\) is not less than \ (a_k\) the minimum polynomial \ (f (x) \) (D_k=[f (A_1,\cdots,a_k): F (a_1,\cdots,a_{k-1})]\), the equal condition is \ (f (x) \) does not have a heavy root. If a self-isomorphic embedding is also required, then the root of \ (f (x) \) is also required to be in \ (e\).

The total number of embedding is naturally \ (\prod C_k\leqslant\prod d_k=[e:f]\), the number of gamma Galois is not greater than the total number of embedded numbers, the same condition is \ (e\) is the normal expansion of the domain. The conclusion of the above discussion is that the formula (6) is established, and a sufficient condition for the establishment of the equal sign is: \ (e\) is both the normal and the extended domain. This kind of normal expansion is called Galois expansion , and of course we only focus on finite Galois expansion.

\[\left|\text{gal} (/f) \right|\leqslant[e:f]\tag{6}\]

Now in turn, the finite subgroups of the \ (e\) automorphism group (g\), examine the relationship between \ (F=\text{inv} (G) \) and \ (e\). If \ (e\) to \ (f\) is limited expansion, by formula and easy to get \ (| G|\leqslant|\text{gal} (F) |\leqslant[e:f]\). Artin to the opposite conclusion, he proved that \ ([e:f]\leqslant| g|\) (At this time \ (e\) natural is \ (f\) Limited expansion), combined with these two points are constant formula (7) is established. The proof process takes full advantage of the nature of the extended domain and the automorphism, and can be used as a good example to illustrate the general idea.

\[| G|=[E:\TEXT{INV} (G)]\tag{7}\]

Set \ (n=| g|\), first to examine the dimensions of the linear space on the e\ \ (f\), if the number of dimensions is limited, and the (m\) is greater than the dimension, then any \ (e\) element \ (m\) in \ (a_i\) is linearly related. A precise description is that the linear equation \ (\sum\limits_{i=1}^m{a_ix_i}=0, (a_i\in E) \) has a total of 0 solutions on \ (f\), and now we are going to prove that \ (m>n\) has a solution. In order to contact \ (g\), set its \ (n\) element is \ (\{\sigma_j\}\), the original equation is equivalent to the equation set \ (\sum{\sigma_j (a_ix_i)}=\sum{\sigma_j (a_i) x_i}=0\) on \ (f\) has a solution. Because \ (m>n\), the equation set in \ (e\) must have a non-0 solution, we need to construct the solution on \ (f\).

Any \ (\sigma_k\) action on the equation set (\sum{\sigma_k\sigma_j (a_i) \sigma_k (x_i)}=0\), because \ ((\sigma_k\sigma_1,\cdots,\sigma_k\ Sigma_n) \) Just a permutation of \ ((\sigma_1,\cdots,\sigma_n) \), the equations in addition to the order has not changed, so \ ((\sigma_k (x_1), \cdots, (\sigma_k (x_m)) \) is also the solution of the original Equation group. Because \ ((x_1,\cdots,x_m) \) Nonzero, can be set \ (X_1\ne 0\), then \ (\bar{x}= (1,x ' _2=\dfrac{x_2}{x_1},\cdots,x ' _m=\dfrac{x_m}{x_1}) \) is also the solution of the equation Group. If \ (x ' _i\in f\) is established, our conclusion is proved. otherwise set \ (x ' _2\not\in f\), which means existence \ (\sigma_k\) makes \ (\sigma_k (x ' _2) \ne x ' _2\). Since \ ((1,\sigma_k (x ' _2), \cdots,\sigma_k (x ' _m) \) is also the root of the equation set, subtracting from \ (\bar{x}\) is another non-0 solution ((0,x ' _2-\sigma_k (x ' _2), \cdots) \) , where the number of non-zero elements is less than \ (\bar{x}\). This process can only be limited step, in the end must be able to obtain \ (f\) on the non-0 solution, theArtin theorem is proof.

　　• \ (k\) is the domain of \ (f\), \ (F (x) \in f[x]\), verify: \ (\text{gal} (f,f) \leqslant \text{gal} (f,k) \).

2.3 Galois theory

With the formula (6) and (7), now come back to discuss the relationship between the automorphism subgroup and the subdomain, since the formula (6) equals a sufficient condition for the formation of Galois expansion, and the Galois expansion cannot be established everywhere, so we limit the study to some Galois expansion. The subdomain \ (f\) corresponds to one of its Galois domains \ (G=\text{gal} (E/F) \), and vice versa \ (g\) corresponds to its fixed subdomain \ (F ' =\text{inv} (G) \). Now compare \ ([e:f]\) and \ ([e:f ']\), according to the formula and respectively ([e:f]=| g|\) and \ ([E:f ']=| g|\), and the formula description \ (F\subseteq F ' \), so there are \ (F=f ' \), subdomains and automorphism subgroups are established on the finite Galois expansion.

If you set \ (e,f\) all intermediate domains \ (f\leqslant F ' \leqslant e\) make up the set \ (\sigma\), it is easy to prove that all elements in \ (e\) pair \ (\sigma\) are finite Galois expansions. If all subgroups of \ (g\) constitute a set \ (\gamma\), then the above conclusion establishes a single shot \ (\varphi\) from \ (\sigma\) to \ (\gamma\), which satisfies the formula (8). Conversely to any \ (G ' \in\gamma\), first there \ (| G ' |=[e:\text{inv} (g ')]\), and by formula (6) (|\text{gal}\circ\text{inv} (g ') |=[e:\text{inv} (g ')]\), so there \ (g ' =\text{gal}\ CIRC\TEXT{INV} (g ') =\varphi (\text{inv} (g ')) \). This shows that \ (\varphi\) is full-shot, and thus is the one by one mapping, all \ (\sigma\) and \ (\gamma\) exist between one by one mappings, satisfies the formula (8).

\[\varphi (F ') =\text{gal} (E/E '), \quad\varphi^{-1} (g ') =\text{inv} (g ') \tag{8}\]

According to the definition of \ (\varphi\), it is easy to have the formula (9) established, where \ (\cup\) represents the generating group (domain). In addition, because \ ([e:f]=| G|,[e:f ']=| G ' |\), then \ ([F ': F]=[g:g ']\) (the latter represents the exponent of the subgroup). When you see this, you may ask a question: \ (F ' \) is the correlation between Galois expansion and \ (G ' \) is the normal subgroup? Easy to verify, for any \ (\sigma\in g\), \ (\sigma G ' \sigma^{-1}\) in the map \ (\varphi\) The original image is \ (\sigma (F ') \). So the equivalent condition of \ (G ' \) for normal subgroup is \ (\sigma (F ') =f ' \), that is, \ (f ' \) for the normal expansion of the domain, and then by \ (f ' \) is obviously the separation and expansion of the domain, so \ (G ' \) for the normal subgroup of the equivalent condition is \ (f ' \) for the Galois expansion domain.

\[f_1\cap F_2=\TEXT{INV} (G_1\cup g_2), \quad F_1\cup F_2=\text{inv} (G_1\cap g_2) \tag{9}\]

Further, set \ (H=\text{gal} (f '/F) \), construct homomorphic mappings \ (\eta:h\to g\), so that \ (\sigma=\eta (H) \) satisfies \ (\sigma (f ') \), obviously homomorphic kernel is \ (G ' \), thereby \ (=F) Isomorphism with \ (g/g ' \) (formula (10)).

\[\text{gal} (f '/f) \cong g/g ' \tag{10}\]

3. Classic Application 3.1 Positive polygon drawing

The construction of the polygon is as old and famous as the "three drawing puzzles", and sometimes they are called "the four big drawing puzzles". First, it is easy to prove that if \ (p,q\) coprime and positive \ (p,q\) edge can be made, then the positive \ (pq\) edge can also be made. According to the arithmetic basic theorem, \ (N=2^{e}p_1^{e_1}\cdots p_m^{e_m}\), and the positive \ (2^e\) edge shape is easy to make, so we only need to study the positive \ (p_k^{e_k}\) side shape of the drawing.

Gauss at the age of 20 made a positive \ (17\) edge shape, and gave a positive \ (m\) edge shape can be plotted the necessary and sufficient conditions, here we use the language of the domain to re-describe the argument of the idea. To make a positive \ (p_s\) edge, it is actually making \ (f (x) \) root \ (\omega\) (Type (11)). Obviously \ (\omega\) is the generator of the \ (f (x) \) split domain, which is \ (E=\bbb{q} (\omega) \). In the previous section of the drawing theory we know that \ (\omega\) can be plotted the necessary and sufficient conditions are: \ ([e:\bbb{q}]=2^t\).

\[f (x) =x^{p^s}-1,\quad\omega=e^{\frac{2\pi}{p^s}i}\tag{11}\]

Because \ (e\) is a split domain, it is Galois expansion, so there is \ ([E:\bbb{q}]=\text{gal} (E/\bbb{q}). \ (e\) \ (\bbb{q}\)-automorphism \ (\sigma\) is determined by \ (\sigma (\omega) \) only, \ (\sigma (\omega) \) can only take \ (\omega^{k}\), where \ ((k,p^s) =1\). By the knowledge of elementary number theory, \ (k\) desirable \ (\varphi (P^s) =p^{s-1} (p-1) \), so \ (2^t=p^{s-1} (p-1) \). First there \ (s=1\), again by the knowledge of the elementary number theory, must have \ (t=2^n\), and \ (2^{2^n}+1\) is a prime number.

satisfies the form (12) The number called the horse number , above conclusion namely (P^s\) the necessary and sufficient condition that the shape can be plotted is: \ (s=1\) and \ (p\) is the cost horse prime. Then \ (n\) side shape can be plotted by the equation (13), where \ (P_k\) is the Fermat of the different primes. Before \ (5\) The number of horses is exactly the prime, Fermat at that time asserted that all the fees are prime, but so far has not found the first 6\ of the prime number of horses.

\[f_n=2^{2^n}+1\quad (f_0=3,\,f_1=5,\,f_2=17,\,f_3=257,\,f_4=65537,\,\cdots) \tag{12}\]

\[m=2^sp_1p_2\cdots p_n,\:(n\geqslant 0) \tag{13}\]

3.2 Seeking root of polynomial

The polynomial seeking root is an important content of ancient algebra, as early as the ancient Babylon in BC, people have mastered the root of the equation two times. The Italians in the Renaissance gave the general method and formula to solve the location equation, and the main ideas were the descending method. For the three-time equation, a simple substitution \ (y=x+\dfrac{a}{3}\) is used to eliminate two entries ((14)), and then (y\) parameterized \ (y=\sqrt[3]{m}+\sqrt[3]{n}\) using the formal characteristics of the cubic and formula. Since \ (m,n\) can change continuously and add the restriction condition \ (3\sqrt[3]{mn}=p\), the original equation is equivalent to the simpler equation set (15).

\[x^3+ax^2+bx+c=0\:\rightarrow\: y^3=py+q\tag{14}\]

\[mn= (\dfrac{p}{3}) ^3,\quad m+n=q\tag{15}\]

For the four-time equation also use \ (y=x+\dfrac{a}{4}\) to eliminate three entries, and then introduce the parameter \ (t\) and recipe (formula (16)). Find the appropriate \ (t\) to make the equation to the right of the formula, so that the four-time equation is reduced to two-time equation. When the formula is established \ (t\) satisfies a three-time equation, the above has given its solution method, so that the four-time equation has been successfully solved. The complete formula for the location equation is complex and is not given here (and not necessary).

\[x^4+ax^3+bx^2+cx+d=0\:\rightarrow\: (y^2+t) ^2= (2t+p) y^2+qy+ (t^2+r) \tag{16}\]

When people can't wait to reach the general five-time equation, they find that the formula is not found in any way. The so-called "formula" is the expression of arithmetic and radical, in order to make use of the theory of expanding domain, we need to define a kind of expanding domain for the radical. Set \ (A\in f\), an algebraic closure of \ (x^n=a\) of any one of the \ (\sqrt[n]{a}\), the single-domain \ (F (\sqrt[n]{a}) \) called the Radical expansion . If the root of a polynomial is represented by a formula, it means that there is a radical expansion chain (17), which can contain a split domain \ (e\). Such a polynomial is called a radical solvable , and our question is: what kind of polynomial radicals to solve?

\[f=f_0\leqslant f_1\leqslant\cdots\leqslant f_n=k,\quad E\subseteq k\tag{17}\]

Let's start with some general discussion of the radical expansion, and provide useful tools for the following arguments, the following discussion is the default extension of the domain, so the division domain is the Galois expansion domain. First consider the equation \ (x^n=1\), whose root is called \ (n\) The secondary unit root . In a complex field, all the unit roots form a cyclic group, where the generators are called \ (n\) secondary primitive roots (\ (\omega\)). In fact, this conclusion is also established in the general domain, because \ (n=\prod{p_k^{e_k}}\), so we just need to find \ (p^e\) secondary primitive root can be. It is easy to prove that the root of \ ((x^{p^e}-1)/(x^{p^{e-1}}-1) =0\) is the primitive root, so that the division domain of \ (x^n-1\) is actually \ (e=f (\omega) \).

Each element of \ (F (\omega) \) Gamma Galois is uniquely determined by \ (\sigma (\omega) =\omega^l, (l,n) =1\) and has a single homomorphism mapping to \ (z_n^{*}\), so it is a commutative group, an expansion called an abelian expansion . For \ (x^n=a\) root \ (d=\sqrt[n]{a}\), easy to know \ (D\omega^k\) is also the root of the equation. In order to also use a single spread domain to represent the separation domain, assume \ (\omega\in f\) in advance, so \ (X^n-a\) has a split domain of \ (F (d) \). Each element of the \ (F (d) \) Gamma Galois is uniquely determined by \ (\sigma (d) =d\omega^l, (l,n) =1\) and has a single homomorphism mapping to \ (z_n^{+}\), so it is a cyclic group, which is called a cyclic expansion .

Focus on the radical expansion \ (f (d=\sqrt[p]{a}) \), the above conclusion shows that when \ (\omega\in f,d\not\in f\) \ (\text{gal} (f (d)//) \) is the \ (p\) Order Loop group. Conversely, if \ (\text{gal} (E) \) is \ (p\) Order Loop Group \ (\left\langle\sigma\right\rangle\), take any one \ (c\in e-f\), Kee \ (c_k=\sigma^k (c) \), construct the following \ (D_ K\) (Type (18)). Consider them to be \ (c_0,c_1,\cdots,c_{p-1}\) of the equations, because Vandermonde determinant (reference linear algebra) nonzero, there must be some \ (D=d_k\not\in f\). It is also possible to verify \ (\sigma (d^p) =\sigma (d) ^p= (\omega^{-1}d) ^p=d^p\), so it is known by Galois theory (d^p\in f\), so \ (e\) is a radical expansion. Summary above is, if \ (\omega\in f\), then the radical expansion is equivalent to the \ (p\) Order cycle expansion.

\[d_k=c_0+c_1\omega^k+c_2\omega^{2k}+\cdots+c_{p-1}\omega^{(p-1) K},\quad k=0,1,\cdots,p-1\tag{18}\]

Now it is time to discuss what polynomial is a radical solution, and the radical can be solved by means of a radical expansion chain \ (F=f_0\leqslant\cdots\leqslant f_n=k\). In order to use the Galois theory, other roots can be added to the expansion chain, assuming that \ (k\) is already Galois expansion. In order to use the above conclusion, the least common multiple of all the Roots \ (m_k\) is \ (m\) and \ (m\) the primitive root is \ (\omega\), each expansion of the list is a single expansion \ (F ' _k=f_k (\omega) \), obviously \ (m_k\) The primitive root is also in \ (f\ In Each step of the new expansion chain (19) is Galois dilation, which, according to Galois theory, forms a regular group of all Gamma Galois. And because each gamma Galois is a commutative group, so \ (\text{gal} (K (\omega), F) \) is a solvable group, so the subgroup \ (\text{gal} (e,f) \) is also a solvable group.

\[f\leqslant F ' _0\leqslant F ' _1\leqslant\cdots\leqslant F ' _n=k (\omega) \tag{19}\]

Conversely if \ (\text{gal} (e,f) \) is solvable group, take \ ([e:f]\) the primitive root \ (\omega\), by the previous exercise know \ (\text{gal} (E (\omega)/F (\omega)) \) is \ (\text{gal} (e/ F) \) Subgroup, it is also solvable group. According to Galois theory, the existence of \ (F (\omega) \) to \ (E (\omega) \) Galois expansion chain, each expanded gamma Galois is a prime order cycle group. Again by the above exercises that each Galois expansion of the order \ (m_k\) is the factor of \ ([e:f]\), so \ (m_k\) Order primitive root in \ (F (\omega) \), so each expansion to a radical expansion. Because \ (F (\omega) \) is also a radical expansion, so \ (E (\omega) \) can be expanded by \ (f\) radicals, so the equation radical solvable.

This gives the conclusion that Galois's genius: the necessary and sufficient condition for a polynomial to have a radical solution is that its gamma Galois is a solvable group. This conclusion can be applied to any specific polynomial, but the "formula" solution of the equation is actually the general polynomial of the argument (f (x) \) (Formula (20)), where \ (T_k\) is an indeterminate element. The invariant domain of the equation is \ (f=\bbb{q} (T_1,t_2,\cdots,t_n) \), and we need to determine if \ (F (x) \) in \ (f\) The gamma Galois are solvable. Because \ (t_k\) can be represented by \ (y_k\) with a basic inequality, split domain \ (F (Y_1,y_2,\cdots,y_n) =\bbb{q} (Y_1,y_2,\cdots,y_n) \).

\[f (x) =x^n-t_1x^{n-1}+t_2x^{n-2}+\cdots+ ( -1) ^nt_n,\quad t_k=\sigma_k (y_1,y_2,\cdots,y_n) \tag{20}\]

\[g (x) =x^n-p_1x^{n-1}+p_2x^{n-2}+\cdots+ ( -1) ^np_n,\quad p_k=\sigma_k (x_1,x_2,\cdots,x_n) \tag{21}\]

However, because \ (y_k\) values and interrelationships are derived from \ (t_k\), the gamma Galois of \ (f (x) \) is not a good analysis. We would prefer \ (y_k\) to be an independent invariant, for which we use the indefinite element \ (X_k\) to establish a polynomial (g (x) \) (Formula (21)), whose coefficient \ (p_k\) is the basic inequality of \ (x_k\) (\ (p_k\) is not an indeterminate element). It is also possible to have the invariant domain of this equation as \ (\bbb{q} (P_1,p_2,\cdots,p_n) \), expanding to \ (\bbb{q} (X_1,x_2,\cdots,x_n) \). It can be argued (omitted) that the gamma Galois of these two polynomial are isomorphic (formula (22)), while the latter is isomorphic to \ (s_n\) (\ (x_k\) is an indeterminate element), so \ (f (x) \) has \ (n\) a different root. Again because \ (n\geqslant 5\), \ (s_n\) is not solvable group, so \ (f (x) \) can not be solved by the formula.

\[\BBB{Q} (Y_1,y_2,\cdots,y_n)/\bbb{q} (T_1,t_2,\cdots,t_n) \cong \bbb{q} (X_1,x_2,\cdots,x_n)/\Bbb{Q} (p_1,p_2,\ Cdots,p_n) \tag{22}\]

So far, we've covered the basic concepts of abstract algebra, but these are just warm-up exercises for abstract algebra. As the cornerstone of modern mathematics, it has a very broad content and unlimited wisdom, the ultimate goal of learning it is to exercise our abstract thinking and scientific mathematical view. With this kind of edification to learn other subjects, you will have a different height, the understanding of things no longer floating on the surface. Abstract algebra is a basic method, it has more vivid and deep content, I will continue to dive into the content of each branch.

"Complete"

"Abstract algebra" 09-Galois theory