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 , we can remove the linear term by setting ; our equation then becomes .

We can do something similar with any degree polynomial. Consider the polynomial . We may make the substitution to remove the -degree term.

We can also make the coefficients of the linear and constant terms equal with the substitution (where the ‘s are the coefficients of the polynomial expressed in terms of ).

Enough for motivation. Suppose is a polynomial, and let . (So, in particular, form a transcendence basis for over .) Let . A Tschirnhaus transformation is an element so that .

Applying a Tschirnhaus transformation to a polynomial of degree gives us another polynomial of degree 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 over as small as possible. (For a polynomial of degree , we will write for this number.)

Suppose . Hermite showed that it is possible to make and . Therefore . Klein showed that is in fact equal to 2.

Now suppose . Joubert showed that again we can make and . Therefore .

It is unknown whether we can make when . However, it is known that we cannot do so if is of the form for or for . It is also known (Buhler and Reichstein) that .

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Friday, April 28, 2017 at 6:52 pm

Jim CurrieHow do you compute a tschirnhausen transformation of x**3+9*x**2+6*x+3 using y=x**2+a*x+b. I have tried many ways without using a computer. Is there a way to do this on paper without using a computer.