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Question :

Let $\alpha$ be a root of $x^3+x+1$ and $\beta$ be a root of $x^3+x+3$. Show that it is not possible that $\alpha\in \mathbb Q(\beta)$.

My proof :

Given $\beta$ is a root of $x^3+x+3$.

Thus $\beta^3+\beta+3 = 0$. Then $(\beta^3+\beta+2 ) + 1 = 0$.

If $\alpha\in\mathbb Q(\beta)$, then $\alpha = r_1 + r_2\beta$ for some rationals $r_1$ and $r_2$.

Also $(r_1+r_2\beta)^3+r_1+r_2\beta+1 = 0$.

Therefore for some rationals $r_1$ and $r_2$ we have $(r_1+r_2\beta)^3+r_1 + r_2\beta = \beta^3+\beta+2$.

But equating and solving such $r_1$ and $r_2$ doesn't exist.

Thus, $\alpha$ doesn't belong to $\mathbb Q(\beta)$.

Do you think my proof is right?

If not correct me or provide a better easier proof

user26857
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Nick Diaz
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    Why is $\alpha=r_1+r_2\beta$ for some rationals $r_1$, $r_2$? – Christoph May 04 '21 at 06:49
  • If we assume α belongs to Q($\beta$). Elements of Q($\beta$) are of the form $r_1+r_2\beta$ – Nick Diaz May 04 '21 at 06:50
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    I don't think they are. Is $\beta^2$ of the form $r_1+r_2\beta$? – Christoph May 04 '21 at 06:52
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    I see Its a third degree polynomial so Elements of Q($\beta$) are of the form $r_1+r_2\beta+r_3\beta^2$ – Nick Diaz May 04 '21 at 06:59
  • $\mathbb Q(\beta)$ is the smallest subfield of $\mathbb R$ that contains $\mathbb Q$ and $\beta$, not necessarily is ${r_1+r_2\beta : r_1,r_2 \in \mathbb Q}$, the latter could not even be a field! – azif00 May 04 '21 at 07:01
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    You say "By equating and solving ...". Even if you repair your argument by looking at polynomials expressions in $\beta$ of degree $2$ instead of sometimes $1$ and sometimes $3$, this is a daunting task. – Magdiragdag May 04 '21 at 07:27
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    See related: https://math.stackexchange.com/q/2431221/72031 – Paramanand Singh May 04 '21 at 07:31
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    You can deduce this from the discriminants (or splitting of selected prime ideals), if you know some algebraic number theory. – Jyrki Lahtonen May 04 '21 at 11:30

1 Answers1

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Let $\alpha$ be a root of $x_3+x+1$ and $\beta$ be a root of $x_3+x+3$, then $$[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\beta):\mathbb{Q}]=3.$$ If $\alpha\in \mathbb{Q}(\beta)$, we know $\mathbb{Q}(\alpha)\subseteq \mathbb{Q}(\beta)$, so $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$.

the fact: Let $L$ be a number field, and we assume that $\mathcal{O}_L$ is the ring of algebraic integers in $L$. Suppose that $\alpha_1,\cdots,\alpha_n$ is the elements of $\mathcal{O}_L$, then $\alpha_1,\cdots,\alpha_n$ is a integral basis of $L$ iff $d_L(\alpha_1,\cdots,\alpha_n)$ has no square factors.

We have $d_{\mathbb{Q}(\alpha)}(1,\alpha,\alpha^2)=-31$, so $\{1,\alpha,\alpha^2\}$ is a integral basis of $\mathbb{Q}(\alpha)$ by using the fact. Thus we can get the discriminant of $\mathbb{Q}(\alpha)$: $$d(\mathbb{Q}(\alpha))=d_{\mathbb{Q}(\alpha)}(1,\alpha,\alpha^2)=-31.$$

In the similar way, we know that $d_{\mathbb{Q}(\beta)}(1,\beta,\beta^2)=-247=-13\times 19$ has no square factors, thus $$d(\mathbb{Q}(\beta))=d_{\mathbb{Q}(\beta)}(1,\beta,\beta^2)=-247\ne d(\mathbb{Q}(\alpha)).$$

That is, $\mathbb{Q}(\alpha)\ne \mathbb{Q}(\beta)$. So $\alpha \notin \mathbb{Q}(\beta)$.

DXT
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  • The solution is very good! The implication about integral bases is only valid in one way, the discriminant of a number field may have square factors. This does not affect your proof. – orangeskid Aug 03 '21 at 20:55
  • This is essentially saying that if the discriminants of two irreducible cubics are different the fields containing one of their roots are different. Correct me if I am wrong. – Paramanand Singh Aug 04 '21 at 08:31
  • I checked wiki. Here we are lucky that both the number fields under consideration have a integral power basis in which case the discriminant of the number field equals the discriminant of the minimal polynomial of the generating element. – Paramanand Singh Aug 04 '21 at 08:54