Field extension obtained by adjoining a cubic root to the rationals.

I hope it's not too long winded, but I prefer to give a short intro hoping for a last chance to go over this in my head and catch any error.

Here's a primitive example of a field extension: $\mathbb{Q}(\sqrt 2) = \{a + b\sqrt 2 \;|\; a,b \in \mathbb{Q}\}$. It's easy to show that it is a commutative additive group with identity $0$. It's a little more involved to show that once $0$ is taken out (which rules out $a = b = 0$), what's left is a multiplicative group with identity $1$ and multiplicative inverse $$\dfrac{1}{a + b\sqrt 2} = \dfrac{1}{a + b\sqrt 2}\dfrac{a - b\sqrt 2}{a - b\sqrt 2} = \dfrac{a - b\sqrt 2}{a^2 - 2b^2} = \dfrac{a}{a^2 - 2b^2} + \dfrac{- b}{a^2 - 2b^2}\sqrt 2$$ which always exists because $a,b\in\mathbb{Q}$ ensures that the denominator in the above equation can never equal zero and since $a$ and $b$ cannot both be $0$ neither can the inverse, giving us closure. So $\mathbb{Q}(\sqrt 2)$ is a field.

Now we seek to replicate this with $\sqrt[3] 5$. A little bit of algebra will quickly show that if we define $\mathbb{Q}(\sqrt[3]5)$ with elements $a + b\sqrt[3]5$ as before, we'll run into problems with closure when multiplying two such elements. Instead we define

$$\mathbb{Q}(\sqrt[3]5) = \{a + b\sqrt[3]5 + c\sqrt[3]{25} \;|\; a,b,c \in \mathbb{Q}\}$$

Once again, checking that the above set is an additive abelian group is easy. To show that the set minus $0$ is a multiplicative group we need to do some linear algebra: We want to show that given $a,b$ and $c$ (not all three $0$) there exists a unique $x,y$ and $z$ such that

$$(a + b\sqrt[3]5 + c\sqrt[3]{25})(x + y\sqrt[3]5 + z\sqrt[3]{25}) = (1 + 0\sqrt[3]5 + 0\sqrt[3]{25})$$

where the right-hand side of the above equation is just $1$, namely the multiplicative identity. Some tedious algebra allows us to rewrite the left-hand side as $$(ax + 5cy + 5bz) + (bx + ay + 5cz)\sqrt[3]{5} + (cx + by + az)\sqrt[3]{25} = 1$$ So we can rewrite the above as a system of equations ${\bf A x = b}$ given by $$\begin{pmatrix}a & 5c & 5b \\ b & a & 5c \\ c & b & a\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$$ This reduces the problem of showing that there always exists a unique multiplicative inverse to one of showing that the above square matrix is invertible (which would guarantee us a unique solution.) So let's find its determinant:

$$\det({\bf A}) = a(a^2 - 5bc) + b(5b^2 - 5ac) + c(25c^2 - 5ab) = a^3 + 5b^3 + 25c^3 - 15abc.$$

Finally, we get to where I'm stuck. How can we guarantee that the above is always non-zero as long as $a,b$ and $c$ are not all zero? The notes I'm going over skipped this part and just said that $\mathbb{Q}(\sqrt[3]{5})$ is a field.

Solution 1:

If we really want to do it more or less in the way that you describe, it can be done. What we want to show is that $$a^3 + 5b^3 + 25c^3 - 15abc=0$$ has no non-zero rational solutions. If we do not insist on rationality, of course there are many solutions.

Suppose to the contrary there is such a solution. Express $a$, $b$, and $c$ using a common denominator $d$, and multiply by $d^3$. We end up looking for integer solutions of the above equation.

Note that $5$ divides all the coefficients except the one for $a^3$. So $5$ must divide $a$, say $a=5a_1$. Substitute. Divide through by $5$.

We find that $5$ now divides all the coefficients except the one for $b^3$. So $5$ divides $b$, say $b=5b_1$. Substitute, divide by $5$.

Now we are in the same situation with $c^3$. Let $c=5c_1$, divide through by $5$.

We conclude that if $(a,b,c)$ is an integer solution, so is $(a_1,b_1,c_1)$. Now we can play the same game with $(a_1,b_1,c_1)$. And with $(a_2,b_2,c_2)$. And so on, forever.

This is impossible unless $(a,b,c)=(0,0,0)$, for the only integer divisible by arbitrarily high powers of $5$ is $0$.

Note that we have used the classic descent argument for proving irrationality.

Remark: For polynomials that are only a little more complicated than $x^3-5$, imitating your computation and the subsequent descent argument sounds very unpleasant. But the Bezout Theorem approach so nicely described by Arturo Magidin gives a uniform proof for all polynomials irreducible over the rationals. Moreover, the Bezout process uses a simple algorithm, essentially the Extended Euclidean Algorithm, which efficiently produces your desired $(x,y,z)$.

Solution 2:

Take any non-zero $\alpha \in \mathbb Q [\sqrt[3]{2}]$ and consider the $\mathbb Q$-linear endomorphism $$m_{\alpha}:\mathbb Q [\sqrt[3]{2}]\to \mathbb Q [\sqrt[3]{2}]:x\mapsto x\cdot \alpha$$ It is injective because $\mathbb Q [\sqrt[3]{2}]\subset \mathbb R $ is a domain, hence surjective because $\mathbb Q [\sqrt[3]{2}]$ is is finite-dimensional.
So there exists $\beta \in \mathbb Q [\sqrt[3]{2}]$ such that $m_{\alpha}( \beta )=\beta\cdot \alpha=1 $ and we have proved that $\alpha^{-1}=\beta\in Q [\sqrt[3]{2}]$.

Solution 3:

Here's a simpler way to deal with all of these issues.

Recall that if $F$ is a field, an element $\alpha$ of an overfield $K$ is said to be algebraic if and only if there exists a monic polynomial $f(x)$ with coefficients in $F$ such that $f(\alpha)=0$. If so, then it is easy to show (using the division algorithm, for example) that every polynomial that has $\alpha$ as a root is a multiple of a particular monic polynomial, which will be irreducible over $F$ (cannot be written as a product of nonconstant polynomials of strictly smaller degree); this polynomial is called "the monic irreducible (polynomial) of $\alpha$ over $F$."

Theorem. Let $F$ be a field, let $K$ be an algebraic closure of $F$, and let $\alpha\in K$. Let $f(x) = a_0+a_1x+\cdots a_{n-1}x^{n-1}+x^n$ be the monic irreducible polynomial of $\alpha$ with coefficients in $K$. Then $$\{b_0 + b_1\alpha+\cdots + b_{n-1}\alpha^{n-1}\mid b_i\in F\}$$ is a field, and is the smallest subfield of $K$ that contains $F$ and $\alpha$. That is, $$F(\alpha)=F[\alpha] = \{b_0 + b_1\alpha+\cdots + b_{n-1}\alpha^{n-1}\mid b_i\in F\}$$

(Note: By definition, $F(\alpha)$ is the smallest subfield of $K$ that contains $F$ and $\alpha$, and $F[\alpha]$ is the smallest subring of $K$ that contains $F$ and $\alpha$.)

Proof. It is straightforward that this collection forms a group; showing it is a ring is not hard, we can use the fact that $f(\alpha)=0$ to get $$\alpha^n = -(a_0 + a_1\alpha+\cdots + a_{n-1}\alpha^{n-1}),$$ which allows us to write any power of $\alpha$ greater than $n-1$ in terms of smaller powers.

The real issue is multiplicative inverses. Let $$b_0 + b_1\alpha+\cdots + b_{n-1}\alpha^{n-1}$$ be a nonzero element of our set. Let $p(x) = b_0+b_1x+\cdots+b_{n-1}x^{n-1}$.

I claim that there exist polynomials $m(x)$ and $n(x)$ with coefficients in $F$ such that $m(x)p(x) + n(x)f(x) = 1$. Indeed, $F[x]$ is a Euclidean domain; since $f(x)$ is irreducible, and $p(x)$ is not zero and of degree strictly smaller than $f(x)$, then $f(x)$ and $p(x)$ are relatively prime. And since $F[x]$ is a Euclidean domain, the extended Euclidean algorithm will produce $m(x)$ and $n(x)$.

Now plug in $\alpha$. We have $1 = m(\alpha)p(\alpha) + n(\alpha)f(\alpha) = m(\alpha)p(\alpha)$. Thus, $p(\alpha) = b_0+b_1\alpha+\cdots+b_{n-1}\alpha^{n-1}$ has a multiplicative inverse which is a polynomial in $\alpha$ with coefficients in $F$. We can then rewrite any power of $\alpha$ greater than $n-1$ as smaller powers to get that $p(\alpha)$ has an inverse in our set, and we are done.

Now, clearly any subring of $F$ that contains $F$ and contains $\alpha$ will contain our set, so then $$\{a_0+a_1\alpha+\cdots+a_{n-1}\alpha^{n-1}\mid a_i\in F\} \subseteq F[\alpha]\subseteq F(\alpha)\subseteq \{a_0+a_1\alpha+\cdots+a_{n-1}\alpha^{n-1}\mid a_i\in F\}$$ giving equality. $\Box$

In your case, $F=\mathbb{Q}$, $\alpha=\sqrt[3]{5}$, which satisfies $f(x) = x^3-5$, which is monic and irreducible over $\mathbb{Q}$. So $$\mathbb{Q}(\sqrt[3]{5}) = \mathbb{Q}[\sqrt[3]{5}] = \{a + b\sqrt[3]{5}+c(\sqrt[3]{5})^2\mid a,b,c\in\mathbb{Q}\}.$$

Solution 4:

It is really simple: generalizing rationalizing denominators, one can easily read off the inverse of $\alpha$ from any minimal polynomial. Let $\alpha\ne 0$ be an element of a domain $D$ that is algebraic over some subfield $F$. Being algebraic, $f(\alpha) = 0$ for some $f(x)\in F[x],\ f\ne 0$. Since $D$ is a domain we may assume $f(0) \ne 0$ since $f(\alpha)\:\alpha^n = 0$ $\Rightarrow$ $f(\alpha) = 0$. Writing $f(x) = x\ g(x) - a,\ a \ne 0$ and evaluating at $x = \alpha$ we deduce $\alpha\ g(\alpha) = a$ hence $\alpha\ g(\alpha)\ a^{-1} = 1$, i.e. $\alpha^{-1} = g(\alpha)\ a^{-1}.$

This is a generalization of rationalizing denominators in the quadratic / splitting field case, where the constant term of a minimal polynomial is a norm (product of conjugates). For much further discussion of this viewpoint see this answer.

This fails in non-domains, e.g. $\alpha\in \mathbb Q[\alpha]/(\alpha^2)$ has minimal polynomial $f(x) = x^2$ with $f(0) = 0$ so the above method fails. Indeed, zero-divisors are never invertible (except in the zero ring).