** 3. Examples **

Last lecture we defined what it means that a polynomial with coefficients in a field is solvable by radicals over . Namely, there is a tower of extensions

where for each there is a positive integer such that and such that

An obvious class of examples comes from considering polynomials of the form

- If then
- If the roots of are all the numbers of the form where (Remember the convention that if then denotes the unique positive real whose -th power is ) One obvious way of adding the roots of to consists in first adding and then adding Clearly, any field containing these two numbers will contain all the numbers of the form so this will indeed give us an extension containing This shows that is solvable by radicals, since we can form the tower
where so and we can take and so so we can take Since we are done.

- If the roots of are the numbers of the form where This is because all the numbers are different, since they are -th roots of unity, and Again, is an extension by radicals, as witnessed by the tower
where and

Proposition 1If is a quadratic polynomial, then is solvable by radicals over

*Proof:* Let where and Let Remember our convention that if this is the number

Consider the tower

Note that since the roots of are the numbers Also, this is an extension by radicals, since

Proposition 2If is a cubic polynomial, then is solvable by radicals over

*Proof:* The argument is similar to the previous one. We use Cardano’s method rather than the quadratic formula, see lecture 2.1. I briefly go through the steps:

Say that where and

Let where and so Since the roots of and the roots of are the same, it suffices to show that is solvable by radicals.

Let where and Notice Also, for any complex number is a root of iff is a root of Since any field that contains will contain and vice versa. It follows that it suffices to show that is solvable by radicals.

Let so As shown in Proposition 1, is solvable by radicals and there is a tower witnessing this, with This is to say and (According to the proof of Proposition 1, but this is not important right now.)

Let be a root of so Let be a cubic root of The other two cubic roots of are and where Hence, we can extend our tower to

where and

Finally, notice that the roots of are the three numbers and These three numbers belong to and it follows that so is solvable by radicals and therefore so is

Proposition 3If is a quartic polynomial, then is solvable by radicals over .

*Proof:* This is similar to the previous case, but now we use Ferrari’s method, see lecture 2.2.

Briefly, just as in the previous case, we may assume

Let By Proposition 2, is solvable by radicals over and, in fact, there is a tower

with witnessing this. That is to say,

Fix a root of Extend the tower above as

where and This is still an extension by radicals, since Also, Using that we may assume that since otherwise we may replace with and now the equation holds.

Then factors as a product of two quadratics in (But these quadratics do not necessarily have coefficients in much less in ) Using (the proof of) Proposition 1, we can further extend the tower as

where and both quadratics factor in This shows that is solvable by radicals.

Notice that we could have made the tower above slightly shorter, since we did not need all cubic roots of any of them would have worked.

We have shown that any polynomial with coefficients in of degree at most 4 is solvable by radicals. In fact, the formulas we have for the roots of these polynomials allowed us to find the required tower of extensions in a “uniform” way.

The method above is fairly general. For example, we have the same result if instead of a polynomial over we had started with a polynomial in for any other subfield of

On the other hand, if a polynomial is not solvable by radicals, then it is certainly the case that there is no “formula” that from the coefficients of the polynomial returns the values of its roots, as long as by formula we understand any expression formed from these coefficients (and other numbers in the field ) by means of the elementary operations and the extraction of -th roots.

A final remark is in order. The method above also works for other fields; for example, for almost all finite fields. There are two problems, though. The first is that to even define we need to begin with an extension field of in which factors. (This field would play the role of .) Such a field always exists, in fact, a version of the fundamental theorem of algebra holds for any field: For any there is an extension of such that *any* polynomial with coefficients in (in particular, any polynomial with coefficients in ) factors in However, in this course we do not have the tools to show that this is the case.

The second problem is more subtle. Consider, for example, the quadratic We cannot do as usual to solve the equation (i.e., )—not even formally, without worrying about which field the solutions live in.

I mean, we cannot use the quadratic formula to solve this equation. This is because in so the expression makes no sense, since it requires that we divide by 2.

Nevertheless, it turns out that *is* solvable by radicals over , because all the nonzero elements of the field with 4 elements, satisfy the equation and, as mentioned in example 3 following Lemma 4 of the notes for last lecture, we know that the roots of are simply the two new elements of

Similarly, if we begin with a field of characteristic 2 or 3, we cannot use Cardano’s method to show that a polynomial of degree 3 is solvable by radicals over since this method requires that at some point we divide some element of the field (or of an extension) by 2 and another by 3. The same problem occurs with Ferrari’s method for equations of degree 4.

Unfortunately, it would take us far afield to see whether Propositions 1, 2, and 3 hold for all fields (finite or infinite, and of any characteristic), so I will live this question unanswered.

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