# The smallest field containing $\sqrt[3]{2}$

Someone could explain how to build the smallest field containing to $\sqrt[3]{2}$.

-
Infinite or finite field, i.e. characteristic $0$ or not? – Bill Dubuque Nov 17 '12 at 18:23
Assuming you work over $\mathbb Q$, take $\mathbb Q(\sqrt[3] 2)$ or $\mathbb Q[X]/(X^3-2)$ or introduce a multiplication on $\mathbb Q^3$ by letting $(a,b,c)\cdot(d,e,f) = (ad+2bf+2ce, ae+be+2cf, af+eb+cd)$. If we want to have the literally smallest field $F$ such that $X^3-2$ has a solution in $F$, take $F=\mathbb F_2$. – Hagen von Eitzen Nov 17 '12 at 18:23
Should we assume "smallest field" means a field $F\supset \mathbb{Q}$ such that any other field $E\supset \mathbb{Q}$ with the property that $2^{1/3}\in E$ must have the property that $E\supset F$? Or do you mean literally "smallest" as in find a field with 8 elements in it containing that element? – Matt Nov 17 '12 at 18:24

Consider the set $F$ of expressions of the form $$\left\{ a + b \cdot \sqrt[3]{2} + c \cdot \left( \sqrt[3]{2} \right)^{2} \,\Big|\, (a,b,c) \in \mathbb{Q}^{3} \right\}.$$ This is the smallest subfield of $\mathbb{R}$ containing $\mathbb{Q}$ and $\sqrt[3]{2}$.

Clearly, $\mathbb{Q} \cup \{ \sqrt[3]{2} \} \subseteq F$. Next, observe that $F$ is closed under addition and multiplication. The only non-trivial thing that needs to be verified is that every non-zero element of $F$ has a multiplicative inverse in $F$. After having established that $F$ is a field, notice that it contains precisely all numbers that should be in any subfield of $\mathbb{R}$ containing $\mathbb{Q}$ and $\sqrt[3]{2}$. This proves that $F$ is the smallest subfield of $\mathbb{R}$ with such a property.

Notice that $F$ is a vector space of dimension $3$ over $\mathbb{Q}$, which agrees nicely with the fact that the degree of the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}$ is $3$ (the minimal polynomial is $X^{3} - 2$).

-
Haskell thanks for your answer, but not very clear to me your argument when you say: "After having established that F is a field, notice that it contains precisely all numbers that should be in any subfield of $\mathbb{R}$ containing $\mathbb{Q}$ and $\sqrt [3]{2}$. – Roiner Segura Cubero Nov 22 '12 at 5:52
Hi. What I am saying is that if $F \subseteq \mathbb{R}$ is a subfield of $\mathbb{R}$ containing $\mathbb{Q}$ and $\sqrt[3]{2}$, then it must contain any number of the form $a + b \cdot \sqrt[3]{2} + c \cdot \left( \sqrt[3]{2} \right)^{2}$, i.e., it must contain $F$ as a subset. As $F$ is a field itself, it follows that $F$ is the smallest subfield of $\mathbb{R}$ containing $\mathbb{Q}$ and $\sqrt[3]{2}$. – Haskell Curry Nov 29 '12 at 4:38

Haskell Curry gave an internal definition, but we can also define this field externally:

$$F=\bigcap\{K\mid\Bbb Q\subseteq K, K\text{ is a field}, \sqrt[3]2\in K\}$$

Namely, $F$ is the intersection of all the fields which extend $\Bbb Q$ and contain $\sqrt[3]2$. We know this collection is not empty because $\Bbb R$ is in this collection.

Now we need to show that $F$ is a field, and that it extends $\Bbb Q$ and $\sqrt[3]2\in F$. The last two things are simple. Every rational number, as well $\sqrt[3]2$ are in $K$ for every $K$ in the collection we intersected over. Therefore $\sqrt[3]2\in F$ and $\Bbb Q\subseteq F$.

To see that $F$ is a field we see that if $x,y\in F$ then $x,y\in K$ for all $K$ in the collection, and therefore $x+y,x\cdot y\in K$ (for all $K$) and therefore $F$ is closed under addition and multiplication, and the same argument shows that additive inverse and multiplicative inverse for non-zero elements also exist in $F$.

Therefore $F$ is a field and it extends the rationals and contain $\sqrt[3]2$. To see that it is minimal observe if $K$ is in the collection then $F\subseteq K$ by definition of the intersection. Therefore the minimality is assured.

Note that this construction does not specifically depend on $\Bbb Q$, and we may replace it with an arbitrary field of our liking as the "base" field. If $\sqrt[3]2$ is already in that base field then the construction will not add new elements, for obvious reasons.

-

If the base field is $F=Z_2$ then we already have the cube root of 2, which is 0. If $F=Z_3$ then $F$ already contains one cube root of 2, i.e. cube root of -1, namely -1; if we want the other cube roots of 2, then the field $F(2^{1/3})=F[2^{1/3}]=\{a+b\cdot 2^{1/3}\}$ has 9 elements.

-