### 数学代写|伽罗瓦理论代写Galois Theory代考|МАТН3040

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## 数学代写|伽罗瓦理论代写Galois Theory代考|Quintic Equations

So far, we have a series of special tricks, different in each case. We can approach the general quintic equation
$$t^5+a t^4+b t^3+c t^2+d t+e=0$$
in a similar way. A Tschirnhaus transformation $y=t+a / 5$ reduces it to
$$y^5+p y^3+q y^2+r y+s=0$$
However, all variations on the tricks that we used for the quadratic, cubic, and quartic equations grind to a halt.

In 1770-1771 Lagrange analysed all of the above special tricks, showing that they can all be ‘explained’ using general principles about symmetric functions of the roots. When he applied this method to the quintic, however, he found that it ‘reduced’ the problem to a sextic – an equation of degree 6. Instead of helping, the method made the problem worse.

Lagrange observed that all methods for solving polynomial equations by radicals involve constructing rational functions of the roots that take a small number of values when the roots $\alpha_j$ are permuted. Prominent among these is the expression
$$\delta=\prod_{1 \leq j<k \leq n}\left(\alpha_j-\alpha_k\right)$$
where $n$ is the degree. This takes just two values, $\pm \delta$ : plus for even permutations and minus for odd ones. Therefore $\Delta=\delta^2$ (known as the discriminant because it is nonzero precisely when the roots are distinct, so it ‘discriminates’ among the roots) is a rational function of the coefficients. This gets us started, and it yields a complete solution for the quadratic, but for cubics upwards it does not help much unless we can find other expressions in the roots with similar properties under permutation.

Lagrange worked out what these expressions look like for the cubic and the quartic, and noticed a pattern. For example, if a cubic polynomial has roots $\alpha_1, \alpha_2, \alpha_3$, and $\omega$ is a primitive cube root of unity, then the expression
$$u=\left(\alpha_1+\omega \alpha_2+\omega^2 \alpha_3\right)^3$$ takes exactly two distinct values.

## 数学代写|伽罗瓦理论代写Galois Theory代考|The Fundamental Theorem of Algebra

At the time of Galois, the natural setting for most mathematical investigations was the complex number system. The real numbers were inadequate for many questions, because $-1$ has no real square root. The arithmetic, algebra, anddecisively – analysis of complex numbers were richer, more elegant, and more complete than the corresponding theories for real numbers.

In this chapter we establish one of the key properties of $\mathbb{C}$, known as the Fundamental Theorem of Algebra. This theorem asserts that every polynomial equation with coefficients in $\mathbb{C}$ has a solution in $\mathbb{C}$. This theorem is, of course, false over $\mathbb{R}$-consider the equation $t^2+1=0$. It was fundamental to classical algebra, but the name is somewhat archaic, and modern algebra bypasses $\mathbb{C}$ altogether, preferring greater generality. Because we find it convenient to work in the same setting as Galois, the theorem is fundamental for us.

All rigorous proofs of the Fundamental Theorem of Algebra require quite a lot of background. Here, we give a proof that uses a few simple ideas from algebra and trigonometry, estimates of the kind that are familiar from any first course in analysis, and one simple basic result from point-set topology. Later, we give an almost purely algebraic proof, but the price is the need for much more machinery: see Chapter 23. Ironically, that proof uses Galois theory to prove the Fundamental Theorem of Algebra, the exact opposite of what Galois did. The logic is not circular, because the proof in Chapter 23 rests on the abstract approach to Galois theory described in the second part of this book, which makes no use of the Fundamental Theorem of Algebra.

# 伽罗瓦理论代考

## 数学代写|伽罗瓦理论代写Galois Theory代考|Quintic Equations

$$t^5+a t^4+b t^3+c t^2+d t+e=0$$

$$y^5+p y^3+q y^2+r y+s=0$$

$$\delta=\prod_{1 \leq j<k \leq n}\left(\alpha_j-\alpha_k\right)$$

$$u=\left(\alpha_1+\omega \alpha_2+\omega^2 \alpha_3\right)^3$$

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