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

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Infinite or finite field, i.e. characteristic $0$ or not? –  Bill Dubuque Nov 17 '12 at 18:23
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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

3 Answers 3

up vote 8 down vote accepted

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 $).

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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.

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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.

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