Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
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 $).

share|cite|improve this answer
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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.