Abstract Algebra: Theory and Applications, Remixed!

One possible basis is \(\{1, \sqrt{7}\}\text{,}\) making \([\Q(\sqrt{7}):\Q] = 2\)

One possible basis is \(\{1, \sqrt[3]{5}, \sqrt[3]{5}^2\}\text{.}\) Thus \([\Q(\sqrt[3]{5}):\Q] = 3\text{.}\)

The polynomial \(p(x)\) is irreducible by Eisenstein's criterion. Therefore, it is the minimal polynomial for any of its roots, including \(\alpha\text{.}\) We then know that \(\{1, \alpha, \alpha^2, \alpha^3, \alpha^4\}\) will be a basis for \(\Q(\alpha)\text{,}\) making \([\Q(\alpha):\Q] = 5\text{.}\)

We claim that \([\Q(c):\Q]\) is equal to the degree of the minimal polynomial for \(c\) (so the degree of the extension is the degree of the polynomial). In fact, we know that \(\{1, c, c^2, \ldots, c^{n-1}\}\) will always be a basis (where \(n\) is the degree).

Why is this? We know that \(\Q(c)\) must contain all powers of \(c\text{.}\) However, \(c^m\) for \(m\ge n\) can always be expressed as a linear combination of \(c^i\) for \(i \lt n\text{,}\) using the minimal polynomial for \(c\text{.}\) This proves that \(\{1, c, c^2, \ldots, c^{n-1}\}\) spans \(\Q(c)\text{.}\) To see why this set is linearly independent, note that if it were not, then there would be a linear combination of powers of \(c\) that was zero. But that would give a polynomial equal to zero at \(c\text{,}\) making \(c\) a root of a polynomial of degree less than \(n\text{.}\)

Clearly this cannot be true, since we already said that these degrees were \(2\) and \(3\) respectively. They key thing that made our argument work above was that the polynomial was a minimal polynomial.

To further see that this is silly, what would the basis for \(\Q(\sqrt{7})\) be if it had size 7? Would it be \(\{1, \sqrt{7}, \sqrt{7}^2,\ldots, \sqrt{7}^4\}\text{?}\) That is clearly linearly dependent, since \(\sqrt{7}^2\) is a multiple of 1.

Over \(\Q(\sqrt[3]{5})\text{,}\) the minimal polynomial for \(\sqrt{7}\) is still \(x^2 - 7\text{.}\) For what else could it be? The only way that there could be a smaller degree polynomial would be for there to be a degree 1 polynomial, but in that case, we would have \(\sqrt{7} \in \Q(\sqrt[3]{5})\text{.}\) It is not obvious, but this is not the case.

Given that, we know that the \([\Q(\sqrt[3]{5}, \sqrt{7}):\Q(\sqrt[3]{5})] = 2\text{.}\)

A basis will be \(\{1, \sqrt{7}\}\text{.}\)

A basis for \(\Q(\sqrt[3]{5})\) over \(\Q\) is \(\{1, \sqrt[3]{5}, \sqrt[3]{5}^2\}\text{.}\) So every element of \(\Q(\sqrt[3]{5})\) can be written in the form \(a+b\sqrt[3]{5} + c \sqrt[3]{5}^2\text{.}\) These are the coefficients we can multiply by basis elements \(\{1, \sqrt{7}\}\) for \(\Q(\sqrt[3]{5},\sqrt{7})\) over \(\Q(\sqrt[3]{5})\text{.}\) So a general element in \(\Q(\sqrt[3]{5}, \sqrt{7})\) has the form

A basis is thus

From looking at the basis above, we see that the degree over \(\Q\) is 6, which is of course, \(2 \cdot 3\text{.}\) It certainly appears this will always work. In other words, if \(K\) is an extension of \(F\text{,}\) which is an extension of \(E\text{,}\) we have

We do indeed get the same thing. The minimal polynomial for \(\sqrt[3]{5}\) is \(x^3 - 5\text{,}\) even over the extension field \(\Q(\sqrt{7})\text{.}\) Thus \([\Q(\sqrt{7},\sqrt[3]{5}):\Q(\sqrt{7})] = 3\text{.}\) A basis is \(\{1, \sqrt[3]{5}, \sqrt[3]{5}^2\}\text{.}\)

If we combine this with the basis for \(\Q(\sqrt{7})\) over \(\Q\text{,}\) we get a basis for \(\Q(\sqrt[3]{5}, \sqrt{7})\) over \(\Q\text{:}\) \(\{1, \sqrt{7}, \sqrt[3]{5}, \sqrt{7}\sqrt[3]{5}, \sqrt[3]{5}^2, \sqrt{7}\sqrt[3]{5}^2\}\text{,}\) containing 6 elements, as expected.

At first, this might look like a counterexample to the “tower rule” we gave above. The minimum polynomial for \(\sqrt[4]{2}\) over \(\Q\) is \(x^4 - 2\text{,}\) which has degree 4. The minimum polynomial for \(\sqrt{2}\) is \(x^2 - 2\text{.}\) However, \([\Q(\sqrt{2}, \sqrt[4]{2}):\Q] = 4\text{,}\) not 8. This can be seen easily by noticing that \(\Q(\sqrt{2}, \sqrt[4]{2}) = \Q(\sqrt[4]{2})\text{,}\) as this already contains \(\sqrt{2}\text{.}\)

The point is that the tower rule is correct: you do multiply the degrees of each extension. But these degrees are not always the degree of the minimal polynomials over the base field. Each time you go up one floor in the tower, you must again ask what the minimum polynomial for the next element you adjoin is.