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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.

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It's usually common to define $K(\alpha)$ to be the smallest field containing both $K$ and $\alpha$; i.e. as the intersection of all fields containing both $K$ and $\alpha$. It is relatively easy to show that this is necessarily a field, saving you the hassle of checking individual cases each time. –  Matt Feb 6 '12 at 15:46
Assume $a^3 + 5b^3 + 25c^3 - 15abc=0$. Multiply by a common denominator, assume $a,b,c$ are integers, not all divisible by $5$. Get a contradiction. –  GEdgar Feb 6 '12 at 15:55
Dear mahin, The key point is that the cube root of $5$ is not a rational number. This is implicit in the arguments suggested by GEdgar in his comment above and Andre Nicolas in his answer below; note how similar the argument is to the traditional proof that $\sqrt{2}$ is irrational. A good way to appreciate this issue is to replace $5$ by $8$ (or any other number that is a cube in $\mathbb Q$), and see where the argument breaks down. Regards, –  Matt E Feb 6 '12 at 16:13
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4 Answers

up vote 10 down vote accepted

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 $(x,y,z)$.

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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 $$\mathbb{Q}(\sqrt[3]{5}) = \mathbb{Q}[\sqrt[3]{5}] = \{a + b\sqrt[3]{5}+c(\sqrt[3]{5})^2\mid a,b,c\in\mathbb{Q}\}.$$

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Thanks, I think the book is taking me toward the generalization above, but it is building the intuition behind it, which is why I wanted to work out the problem for a single case. I'll come back to your solution later when this stuff sinks in a little deeper. –  mahin Feb 6 '12 at 18:06
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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 $\alpha^{-1}=\beta\in Q [\sqrt[3]{2}]$.

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A slightly more abstract generalization : a domain algebraic over a subfield is a field. –  Georges Elencwajg Feb 6 '12 at 16:56
You're basically saying that showing that the set is a field can be reduced to showing that it is an integral domain. Which makes sense, because in linear algebra, showing that ${\bf Ax = 0}$ has no non-trivial solution makes $\bf A$ invertible, which means $\bf Ax = b$ has a unique solution. –  mahin Feb 6 '12 at 18:19
Dear @mahin, algebraicity is essential: the polynomial ring $k[X]$ over a field $k$ is a domain containing $k$ as a subfield but is not a field. –  Georges Elencwajg Feb 6 '12 at 18:41
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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).

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