I want to show $5^{1/3}$ is not in the field $\mathbb{Q}(2^{1/3})$. My reasoning was $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=3$ and $[\mathbb{Q}(5^{1/3}):\mathbb{Q}]=3$. Similarly we know that $[\mathbb{Q}(2^{1/3}, 5^{1/3}):\mathbb{Q}]=9$ so by multiplicativity of dimensions, $[\mathbb{Q}(5^{1/3}):\mathbb{Q(2^{1/3})}]=3$ which implies that $5^{1/3} \not\in \mathbb{Q}(2^{1/3})$. Is this the right approach?
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3You meant you want to show that $5^{1/3}$ is not in $\Bbb Q(2^{1/3})$, right? – coldnumber Aug 08 '15 at 00:49
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7How do you know that the dimension of $\mathbb{Q}(2^{1/3},5^{1/3})$ over the rationals is $9$? – André Nicolas Aug 08 '15 at 00:52
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It is a right approach if you succeed in proving the dimension assertion. But I would choose crude and direct. – André Nicolas Aug 08 '15 at 01:00
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What do you mean crude and direct,? @AndréNicolas – Aug 08 '15 at 01:38
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2Suppose to the contrary that $5^{1/3}=a+b\cdot2^{1/3}+c\cdot2^{2/3}$ where $a,b,c$ are rational. – André Nicolas Aug 08 '15 at 01:39
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Oh, I see, yeah I thought about it but I would have had to raise the whole thing to power of 3 which would end up being long, I was seeing if I could avoid it. – Aug 08 '15 at 01:41
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What you wrote works as an outline, but I think your claims about the degree of each field extension would need explanation.
For $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]$ and $[\mathbb{Q}(5^{1/3}):\mathbb{Q}]$ you need to show that the irreducible polynomials of $2^{1/3}$ and $5^{1/3}$ over $\Bbb Q$ have degree 3.
(Since you know that $2^{1/3}$ is a root of $x^3-2 \in \Bbb Q[x]$, you just need to explain why $x^3-2$ is irreducible over $\Bbb Q$. You can use Eisenstein's criterion. Same for $5^{1/3}$.)
The more difficult part is explaining why $[\mathbb{Q}(2^{1/3}, 5^{1/3}):\mathbb{Q}]=9$.
I think I'd write the extensions in a different way and then use the multiplicative property of the degree:
$\Bbb Q \subseteq \Bbb Q(2^{1/3}) \subseteq [\Bbb Q(2^{1/3})](5^{1/3})=\mathbb{Q}(2^{1/3}, 5^{1/3})$, so the problem becomes showing that $5^{1/3}$ has degree $3$ over $\Bbb Q(2^{1/3})$.
$[\Bbb Q(2^{1/3}):\Bbb Q]=3$, so $\Bbb Q(2^{1/3})$ has a basis of three elements: $1, 2^{1/3}$ and $2^{2/3}$.
You can use this to show that $5^{1/3}$ is not an element of $\Bbb Q(2^{1/3})$: you could try writing it as a linear combination of the basis elements with coefficients in $\Bbb Q$ and deriving a contradiction. (This might be long and tedious.)
coldnumber
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1While I think degrees of field extensions could be helpful for the OP question, I note that your last two paragraphs stand completely on their own as a proof strategy, without needing any of the stuff above it. – Greg Martin Aug 08 '15 at 01:22
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@GregMartin interesting... I guess I was more trying to show how I thought each of the claims OP made could be justified than to write a nice cohesive proof :P. I think from the first part it's still necessary to explain the degree of $2^{1/3}$ over $\Bbb Q$ using the irreducible polynomial, no? Explaining that part shows an awareness of Eisenstein. – coldnumber Aug 08 '15 at 01:31
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It's true that one needs to justify why ${1,2^{1/3},2^{2/3}}$ is a $\Bbb Q$-basis for $\Bbb Q(2^{1/3})$. Although for extensions generated by a root of an integer, that's pretty easy to see by hand. – Greg Martin Aug 08 '15 at 05:57
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In general, if $p$ and $q$ are two different primes and $k>1$, $p^{1/k}\not\in\mathbb{Q}(q^{1/k})$.
Assuming the opposite, for any sufficiently big prime number $r\equiv 1\pmod{k}$, with $q$ being a $k$-th residue $\pmod{r}$, $p$ would be a $k$-th residue, too, but we may provide counter-examples to that situation through the Chinese theorem.
Jack D'Aurizio
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@GregMartin: because we assumed $p^{1/k}\in\mathbb{Q}(q^(1/k))$, hence if $q^{1/k}$ lives in $\mathbb{F}_r^$ and not in some extension, $p^{1/k}$ lives in $\mathbb{F}_r^$, too. – Jack D'Aurizio Aug 08 '15 at 07:15
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This should also work for $p^{1/k_1}\notin\mathbb Q(q^{1/k_2})$. – Hagen von Eitzen Aug 08 '15 at 11:19
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@JackD'Aurizio Can you please tell me what is the definition of a $k$-th residue, that is, given integers $r$ and $k$, when do we say that an integer $q$ is a $k$-th residue of mod $r$? Thank you. – caffeinemachine Sep 14 '16 at 21:59
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@caffeinemachine: we say that $a$ is a $k$-th residue $!!\pmod{p}$ if $a\equiv b^k\pmod{p}$ for some $b$. – Jack D'Aurizio Sep 14 '16 at 22:00
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@JackD'Aurizio Can you please elaborate on how we are applying the Chinese remainder theorem to get a contradiction? Thank you. – caffeinemachine Sep 17 '16 at 07:24