# Why is $L=\mathbb{Q}(\sqrt[1]{2})\cup\mathbb{Q}(\sqrt[2]{2})\cup\mathbb{Q}(\sqrt[3]{2})\cup\cdots$ a field?

Why is $L=\mathbb{Q}(\sqrt[1]{2})\cup\mathbb{Q}(\sqrt[2]{2})\cup\mathbb{Q}(\sqrt[3]{2})\cup\cdots$ a field?

The hint provided in my textbook is: $\mathbb{Q}(\sqrt[n]{2})\cup\mathbb{Q}(\sqrt[m]{2})\subset\mathbb{Q}(\sqrt[mn]{2})$, but this doesn't really get me anywhere. Actually, I have no idea what to do whatsoever. Could anyone clarify or give some hint please?

• Hint: Suppose $\alpha^2=2, \beta^3 = 2$. What is $(\alpha \beta)^6$? Can you use that to find $\gamma$ in your union such that $\gamma^6=2$?
– lulu
Mar 17, 2017 at 0:54
• Have you tried verifying each of the axioms of a field one by one? Which ones are giving you trouble? Mar 17, 2017 at 0:56
• That's because it's a direct limit of fields Mar 17, 2017 at 0:57
• @Tyron don't try to avoid what seems like the more tedious method. If you can't explain why that union is closed under addition, for example, then you really owe it to yourself to figure that out. Later you may understand more slick solutions, but it's better to have a solution than to have none at all.
– KCd
Mar 17, 2017 at 2:00
• More generally, if $\{L_i:i\in I\}$ is any collection of fields indexed by $I$ with the property that for all $i,j\in I$ there exists a $k\in I$ such that $L_i,L_j\subset L_k$, then $L=\bigcup_{i\in I}L_i$ is a field. It is actually easy to verify it is closed under addition, multiplication, inverses etc. Suppose $a,b\in L$... (go from there) Mar 17, 2017 at 3:01

Suppose you take any two elements $x, y\in L.$ This means that there are some positive integers $m$ and $n$ such that $x\in \mathbb{Q}(\sqrt[m]{2})$ and $y\in \mathbb{Q}(\sqrt[n]{2})$.

From the hint in your textbook (that you wrote in your question), this implies that $x,y\in \mathbb{Q}(\sqrt[mn]{2}).$

Now, your end goal is to show that $x+y$ and $xy$ are both elements of $L$ as well (and that $L$ is closed under inverses for nonzero elements).

Is there any information about $\mathbb{Q}(\sqrt[mn]{2})$ that you can use which might help here (since $x$ and $y$ are elements of this set)?

• Very economical. +1. Mar 17, 2017 at 16:15

The smallest field containing $K(a^{1/n})$ and $K(a^{1/m})$ is $K(a^{1/n},a^{1/m}) = K(a^{1/lcm(n,m)})$

• Hello, thank you for your reply. Unfortunately, I do not fully understand what you are getting at. It seems to me that you are proofing the hint my textbook provided, but my guess would be that your statement contains more information than just that. Could you please elaborate? Mar 17, 2017 at 0:55
• $K(a^{1/n},a^{1/m}) \subset K(a^{1/lcm(n,m)})$ is obvious. What you need is proving the other direction @Tyron Mar 17, 2017 at 0:58