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I'm sorry, I don't know how to write mathematical expressions on a computer, so I hope you get what I mean: What elements are in $\mathbb{Q}(\sqrt{3}+\sqrt[3]{3})$? Or can anyone give me the set in set notation?

amWhy
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Emilia
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    Welcome to the site Emilia. For future reference, writing the expressions on this website is quite easy. Use dollar signs like this: $x=2$. In your case, you wanted to write $\mathbb{Q}(\sqrt{3} + \sqrt[3]{3})$. The \mathbb just gives you the funky font. $\Bbb Q(\sqrt{3}+\sqrt[3]{3})$ is the small subfield of $\Bbb R$ which contains $\alpha:=\sqrt{3}+\sqrt[3]{3}$. Its elements look like rational linear combinations of $1,\alpha,\alpha^2,\alpha^3,\cdots$. – FShrike Dec 13 '23 at 13:06
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    Can you prove that $$\Bbb{Q}(\sqrt3+\root3\of3)=\Bbb{Q}(\sqrt3,\root3\of3)=\Bbb{Q}(\root6\of3)?$$ That will make answering the question a lot easier! – Jyrki Lahtonen Dec 13 '23 at 13:33
  • You can construct $\mathbb Z(\sqrt 3+\sqrt[3]{3})$ easily which is an integral domain... Now construct its quotient field... – Nothing special Dec 13 '23 at 13:37
  • Sorry, I dont know how to prove this... I am really bad at algebra :( – Emilia Dec 13 '23 at 13:50
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    A similar question. – Jyrki Lahtonen Dec 13 '23 at 14:32

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For $\alpha \in \mathbb{R}$, the set $\mathbb{Q}(\alpha)$ is the intersection of all fields $K \subset \mathbb{R}$ that contain $\mathbb{Q}$ and $\alpha$. This intersection is the smallest field in $\mathbb{R}$ that contains $\mathbb{Q}$ and $\alpha$. It's not hard to see this is the set of expressions $$\frac{a_0 + a_1 \alpha + ... + a_n \alpha^n}{b_0 + b_1 \alpha + ... + b_n \alpha^m}$$ where $a_i,b_i \in \mathbb{Q}$ such that the denominator is not zero.

There is a well known theorem (in my course on abstract algebra it was called the structure theorem for singular field extensions) that says that if $g \in \mathbb{Q}$ is the minimal polynomial of $\alpha$ (if it exists, and in this case it does since the sum of algebraic numbers is again algebraic), and has degree $n$, then $1, \alpha, ... , \alpha^{n-1}$ is a base of $\mathbb{Q}(\alpha)$ as vector space over $\mathbb{Q}$. This means that all $x \in \mathbb{Q}(\alpha)$ can be written (uniquely) as $x = a_0 + a_1 \alpha + a_{n-1}\alpha^{n-1}$ with $a_i \in \mathbb{Q}$.

PS: To write mathematical notation, you can use latex code, similarly as you would write in the document section of the Overleaf text editor.

Steve
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  • You should show how to construct polynomial $g$ in an explicit way. – Jean Marie Dec 13 '23 at 13:27
  • The degree of $g$ is smaller than or equal to 6 since $\mathbb{Q}(\alpha) \subset \mathbb{Q}(\sqrt{3} , \sqrt[3]{3})$ and the dimension $[\mathbb{Q}(\sqrt{3} , \sqrt[3]{3}) : \mathbb{Q}] = [\mathbb{Q}(\sqrt{3} , \sqrt[3]{3}) : \mathbb{Q} : \mathbb{Q}(\sqrt{3} ) : \mathbb{Q} ][\mathbb{Q}(\sqrt{3}) : \mathbb{Q}] \leq 3 \cdot 2$. However, I don't thinks there is a nice way to construct $g$ since the square and cube of $\alpha$ is not nice. The only option I see is the naive option to fill in $\alpha$ in $a_0 + a_1 x + ... + a_6 x^6$ and calculate the $a_i$. – Steve Dec 13 '23 at 13:42
  • Thanks for your answer. – Jean Marie Dec 13 '23 at 14:03
  • The two instructions of Sagemath : K = QQbar(sqrt(3)+3^(1/3)) p = K.minpoly();p give the minimal polynomial $x^6 - 9x^4 - 6x^3 + 27x^2 - 54x - 18$ – Jean Marie Dec 13 '23 at 14:24