This is based on a talk by Zinovy Reichstein from the PIMS Algebra Summer School in Edmonton.

The motivation comes from looking at ways to simplify polynomials. For example, if we start with a quadratic equation x^2+ax+b, we can remove the linear term by setting y=x+\frac{a}{2}; our equation then becomes y^2+b'.

We can do something similar with any degree polynomial. Consider the polynomial x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_n. We may make the substitution y=x+\frac{a_1}{n} to remove the (n-1)-degree term.

We can also make the coefficients of the linear and constant terms equal with the substitution z=\frac{b_n}{b_{n-1}}y (where the b‘s are the coefficients of the polynomial expressed in terms of y).

Enough for motivation. Suppose f(x)=x^n+a_1x^{n-1}+\cdots+a_n is a polynomial, and let K=\mathbb{C}(a_1,\ldots,a_n). (So, in particular, a_1,\ldots,a_n form a transcendence basis for K over \mathbb{C}.) Let L=K[x]/(f(x)). A Tschirnhaus transformation is an element y\in L so that L=K(y).

Applying a Tschirnhaus transformation to a polynomial of degree n gives us another polynomial of degree n with different coefficients. They also allow us to simplify polynomial expressions in various senses. We will use the following two criteria of simplification:

1) A simplification involves making as many coefficients as possible 0.

2) A simplification involves making the transcendence degree of \mathbb{C}(b_1,\ldots,b_n) over \mathbb{C} as small as possible. (For a polynomial of degree n, we will write d(n) for this number.)

Suppose n=5. Hermite showed that it is possible to make b_1=b_3=0 and b_4=b_5. Therefore d(5)\le 2. Klein showed that d(5) is in fact equal to 2.

Now suppose n=6. Joubert showed that again we can make b_1=b_3=0 and b_5=b_6. Therefore d(6)\le 3.

It is unknown whether we can make b_1=b_3=0 when n=7. However, it is known that we cannot do so if n is of the form n=3^r+3^s for r>s\ge 0 or n=3^r for r\ge 0. It is also known (Buhler and Reichstein) that d(n)\ge\left\lfloor\frac{n}{2}\right\rfloor.