$X^2-q in mathbb{Q}(sqrt{p})[X]$ is irreducible and $[mathbb{Q}(sqrt{p}, sqrt{q}): mathbb{Q}]=4$












1












$begingroup$


I don't know how to prove the following:



Let $p neq q$ be prime numbers, then $sqrt{p}, sqrt{q} in mathbb{R}$ and $mathbb{Q}(sqrt{p}),mathbb{Q}( sqrt{q}),mathbb{Q}(sqrt{p}, sqrt{q}) subset mathbb{R}$.



Show that $X^2-q in mathbb{Q}(sqrt{p})[X]$ is irreducible, $[mathbb{Q}(sqrt{p}, sqrt{q}): mathbb{Q}]=4$ and that $1, sqrt{p}, sqrt{q}, sqrt{pq}$ is a basis for $mathbb{Q}(sqrt{p}, sqrt{q})$.



I keep finding posts about Galois groups but we didn't introduce that concept yet.
Thanks in advance for any help.






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Should it say $mathbb{Q} (sqrt{p}) [X]$ in the title?
    $endgroup$
    – Rellek
    Dec 13 '18 at 22:35










  • $begingroup$
    Yes, you're right. I'll fix it, thank you.
    $endgroup$
    – Anzu
    Dec 13 '18 at 22:38










  • $begingroup$
    I assume that the paragraph starting "Let" is background, not something you want help proving, but that could be clearer. Then there are three subquestions: irreducibility, extension degree, and basis. Do you have any idea for how to tackle any of the three?
    $endgroup$
    – Peter Taylor
    Dec 13 '18 at 22:44










  • $begingroup$
    Yes, exactly. The first paragraph is given. It could have been clearer, my bad. Reliek's solution is comprehensible, it answers my questions.
    $endgroup$
    – Anzu
    Dec 13 '18 at 22:57
















1












$begingroup$


I don't know how to prove the following:



Let $p neq q$ be prime numbers, then $sqrt{p}, sqrt{q} in mathbb{R}$ and $mathbb{Q}(sqrt{p}),mathbb{Q}( sqrt{q}),mathbb{Q}(sqrt{p}, sqrt{q}) subset mathbb{R}$.



Show that $X^2-q in mathbb{Q}(sqrt{p})[X]$ is irreducible, $[mathbb{Q}(sqrt{p}, sqrt{q}): mathbb{Q}]=4$ and that $1, sqrt{p}, sqrt{q}, sqrt{pq}$ is a basis for $mathbb{Q}(sqrt{p}, sqrt{q})$.



I keep finding posts about Galois groups but we didn't introduce that concept yet.
Thanks in advance for any help.






abstract-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Should it say $mathbb{Q} (sqrt{p}) [X]$ in the title?
    $endgroup$
    – Rellek
    Dec 13 '18 at 22:35










  • $begingroup$
    Yes, you're right. I'll fix it, thank you.
    $endgroup$
    – Anzu
    Dec 13 '18 at 22:38










  • $begingroup$
    I assume that the paragraph starting "Let" is background, not something you want help proving, but that could be clearer. Then there are three subquestions: irreducibility, extension degree, and basis. Do you have any idea for how to tackle any of the three?
    $endgroup$
    – Peter Taylor
    Dec 13 '18 at 22:44










  • $begingroup$
    Yes, exactly. The first paragraph is given. It could have been clearer, my bad. Reliek's solution is comprehensible, it answers my questions.
    $endgroup$
    – Anzu
    Dec 13 '18 at 22:57














1












1








1


0



$begingroup$


I don't know how to prove the following:



Let $p neq q$ be prime numbers, then $sqrt{p}, sqrt{q} in mathbb{R}$ and $mathbb{Q}(sqrt{p}),mathbb{Q}( sqrt{q}),mathbb{Q}(sqrt{p}, sqrt{q}) subset mathbb{R}$.



Show that $X^2-q in mathbb{Q}(sqrt{p})[X]$ is irreducible, $[mathbb{Q}(sqrt{p}, sqrt{q}): mathbb{Q}]=4$ and that $1, sqrt{p}, sqrt{q}, sqrt{pq}$ is a basis for $mathbb{Q}(sqrt{p}, sqrt{q})$.



I keep finding posts about Galois groups but we didn't introduce that concept yet.
Thanks in advance for any help.






abstract-algebra







share|cite|improve this question











$endgroup$




I don't know how to prove the following:



Let $p neq q$ be prime numbers, then $sqrt{p}, sqrt{q} in mathbb{R}$ and $mathbb{Q}(sqrt{p}),mathbb{Q}( sqrt{q}),mathbb{Q}(sqrt{p}, sqrt{q}) subset mathbb{R}$.



Show that $X^2-q in mathbb{Q}(sqrt{p})[X]$ is irreducible, $[mathbb{Q}(sqrt{p}, sqrt{q}): mathbb{Q}]=4$ and that $1, sqrt{p}, sqrt{q}, sqrt{pq}$ is a basis for $mathbb{Q}(sqrt{p}, sqrt{q})$.



I keep finding posts about Galois groups but we didn't introduce that concept yet.
Thanks in advance for any help.






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 22:38







Anzu

















asked Dec 13 '18 at 22:17









AnzuAnzu

246




246












  • $begingroup$
    Should it say $mathbb{Q} (sqrt{p}) [X]$ in the title?
    $endgroup$
    – Rellek
    Dec 13 '18 at 22:35










  • $begingroup$
    Yes, you're right. I'll fix it, thank you.
    $endgroup$
    – Anzu
    Dec 13 '18 at 22:38










  • $begingroup$
    I assume that the paragraph starting "Let" is background, not something you want help proving, but that could be clearer. Then there are three subquestions: irreducibility, extension degree, and basis. Do you have any idea for how to tackle any of the three?
    $endgroup$
    – Peter Taylor
    Dec 13 '18 at 22:44










  • $begingroup$
    Yes, exactly. The first paragraph is given. It could have been clearer, my bad. Reliek's solution is comprehensible, it answers my questions.
    $endgroup$
    – Anzu
    Dec 13 '18 at 22:57


















  • $begingroup$
    Should it say $mathbb{Q} (sqrt{p}) [X]$ in the title?
    $endgroup$
    – Rellek
    Dec 13 '18 at 22:35










  • $begingroup$
    Yes, you're right. I'll fix it, thank you.
    $endgroup$
    – Anzu
    Dec 13 '18 at 22:38










  • $begingroup$
    I assume that the paragraph starting "Let" is background, not something you want help proving, but that could be clearer. Then there are three subquestions: irreducibility, extension degree, and basis. Do you have any idea for how to tackle any of the three?
    $endgroup$
    – Peter Taylor
    Dec 13 '18 at 22:44










  • $begingroup$
    Yes, exactly. The first paragraph is given. It could have been clearer, my bad. Reliek's solution is comprehensible, it answers my questions.
    $endgroup$
    – Anzu
    Dec 13 '18 at 22:57
















$begingroup$
Should it say $mathbb{Q} (sqrt{p}) [X]$ in the title?
$endgroup$
– Rellek
Dec 13 '18 at 22:35




$begingroup$
Should it say $mathbb{Q} (sqrt{p}) [X]$ in the title?
$endgroup$
– Rellek
Dec 13 '18 at 22:35












$begingroup$
Yes, you're right. I'll fix it, thank you.
$endgroup$
– Anzu
Dec 13 '18 at 22:38




$begingroup$
Yes, you're right. I'll fix it, thank you.
$endgroup$
– Anzu
Dec 13 '18 at 22:38












$begingroup$
I assume that the paragraph starting "Let" is background, not something you want help proving, but that could be clearer. Then there are three subquestions: irreducibility, extension degree, and basis. Do you have any idea for how to tackle any of the three?
$endgroup$
– Peter Taylor
Dec 13 '18 at 22:44




$begingroup$
I assume that the paragraph starting "Let" is background, not something you want help proving, but that could be clearer. Then there are three subquestions: irreducibility, extension degree, and basis. Do you have any idea for how to tackle any of the three?
$endgroup$
– Peter Taylor
Dec 13 '18 at 22:44












$begingroup$
Yes, exactly. The first paragraph is given. It could have been clearer, my bad. Reliek's solution is comprehensible, it answers my questions.
$endgroup$
– Anzu
Dec 13 '18 at 22:57




$begingroup$
Yes, exactly. The first paragraph is given. It could have been clearer, my bad. Reliek's solution is comprehensible, it answers my questions.
$endgroup$
– Anzu
Dec 13 '18 at 22:57










1 Answer
1






active

oldest

votes


















2












$begingroup$

Notice that $X^2 - q$, $X^2 - p$ are irreducible over $mathbb{Q}$ because they have no roots, since any root would imply that $sqrt{p}$ and $sqrt{q}$ are rational numbers. Using the same proof as for the irrationality of $sqrt{2}$, we know this isn't true.



We then just need to show that $X^2 - q$ remains irreducible over $mathbb{Q} (sqrt{p})$. Again, it suffices to show there are no roots. A general element of $mathbb{Q} (sqrt{p})$ looks like $a + b sqrt{p}$ for $a, b in mathbb{Q}$; plugging this in:
$$(a+ b sqrt{p})^2 - q = 0 implies a^2 + b^2 p - q + 2 ab sqrt{p} = 0$$
which tells us $ab = 0$ and $a^2 + b^2 p = q$. This means $a=0$ or $b=0$. In the first case, we find that $p$ divides $q$, a contradiction. In the second case, we find that $a^2 = p$ for some rational number, again, impossible. This means that $[mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ( sqrt{p} ) ] =2$. Using the tower law,
$$[mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ] = [mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ( sqrt{p} ) ] cdot [ mathbb{Q} ( sqrt{p}) : mathbb{Q} ] = 4$$
Then to show that the set mentioned in the original question is a basis, it suffices to show $mathbb{Q}$-linear independence. I leave that to you.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I like this answer for its generality. There are shorter proofs, either very special, or depending on fancy number theory. But this does the trick admirably.
    $endgroup$
    – Lubin
    Dec 13 '18 at 22:50












  • $begingroup$
    Thanks, this helps a lot. I should be able to show the linear independence.
    $endgroup$
    – Anzu
    Dec 13 '18 at 23:01












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1 Answer
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1 Answer
1






active

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active

oldest

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active

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2












$begingroup$

Notice that $X^2 - q$, $X^2 - p$ are irreducible over $mathbb{Q}$ because they have no roots, since any root would imply that $sqrt{p}$ and $sqrt{q}$ are rational numbers. Using the same proof as for the irrationality of $sqrt{2}$, we know this isn't true.



We then just need to show that $X^2 - q$ remains irreducible over $mathbb{Q} (sqrt{p})$. Again, it suffices to show there are no roots. A general element of $mathbb{Q} (sqrt{p})$ looks like $a + b sqrt{p}$ for $a, b in mathbb{Q}$; plugging this in:
$$(a+ b sqrt{p})^2 - q = 0 implies a^2 + b^2 p - q + 2 ab sqrt{p} = 0$$
which tells us $ab = 0$ and $a^2 + b^2 p = q$. This means $a=0$ or $b=0$. In the first case, we find that $p$ divides $q$, a contradiction. In the second case, we find that $a^2 = p$ for some rational number, again, impossible. This means that $[mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ( sqrt{p} ) ] =2$. Using the tower law,
$$[mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ] = [mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ( sqrt{p} ) ] cdot [ mathbb{Q} ( sqrt{p}) : mathbb{Q} ] = 4$$
Then to show that the set mentioned in the original question is a basis, it suffices to show $mathbb{Q}$-linear independence. I leave that to you.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I like this answer for its generality. There are shorter proofs, either very special, or depending on fancy number theory. But this does the trick admirably.
    $endgroup$
    – Lubin
    Dec 13 '18 at 22:50












  • $begingroup$
    Thanks, this helps a lot. I should be able to show the linear independence.
    $endgroup$
    – Anzu
    Dec 13 '18 at 23:01
















2












$begingroup$

Notice that $X^2 - q$, $X^2 - p$ are irreducible over $mathbb{Q}$ because they have no roots, since any root would imply that $sqrt{p}$ and $sqrt{q}$ are rational numbers. Using the same proof as for the irrationality of $sqrt{2}$, we know this isn't true.



We then just need to show that $X^2 - q$ remains irreducible over $mathbb{Q} (sqrt{p})$. Again, it suffices to show there are no roots. A general element of $mathbb{Q} (sqrt{p})$ looks like $a + b sqrt{p}$ for $a, b in mathbb{Q}$; plugging this in:
$$(a+ b sqrt{p})^2 - q = 0 implies a^2 + b^2 p - q + 2 ab sqrt{p} = 0$$
which tells us $ab = 0$ and $a^2 + b^2 p = q$. This means $a=0$ or $b=0$. In the first case, we find that $p$ divides $q$, a contradiction. In the second case, we find that $a^2 = p$ for some rational number, again, impossible. This means that $[mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ( sqrt{p} ) ] =2$. Using the tower law,
$$[mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ] = [mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ( sqrt{p} ) ] cdot [ mathbb{Q} ( sqrt{p}) : mathbb{Q} ] = 4$$
Then to show that the set mentioned in the original question is a basis, it suffices to show $mathbb{Q}$-linear independence. I leave that to you.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I like this answer for its generality. There are shorter proofs, either very special, or depending on fancy number theory. But this does the trick admirably.
    $endgroup$
    – Lubin
    Dec 13 '18 at 22:50












  • $begingroup$
    Thanks, this helps a lot. I should be able to show the linear independence.
    $endgroup$
    – Anzu
    Dec 13 '18 at 23:01














2












2








2





$begingroup$

Notice that $X^2 - q$, $X^2 - p$ are irreducible over $mathbb{Q}$ because they have no roots, since any root would imply that $sqrt{p}$ and $sqrt{q}$ are rational numbers. Using the same proof as for the irrationality of $sqrt{2}$, we know this isn't true.



We then just need to show that $X^2 - q$ remains irreducible over $mathbb{Q} (sqrt{p})$. Again, it suffices to show there are no roots. A general element of $mathbb{Q} (sqrt{p})$ looks like $a + b sqrt{p}$ for $a, b in mathbb{Q}$; plugging this in:
$$(a+ b sqrt{p})^2 - q = 0 implies a^2 + b^2 p - q + 2 ab sqrt{p} = 0$$
which tells us $ab = 0$ and $a^2 + b^2 p = q$. This means $a=0$ or $b=0$. In the first case, we find that $p$ divides $q$, a contradiction. In the second case, we find that $a^2 = p$ for some rational number, again, impossible. This means that $[mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ( sqrt{p} ) ] =2$. Using the tower law,
$$[mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ] = [mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ( sqrt{p} ) ] cdot [ mathbb{Q} ( sqrt{p}) : mathbb{Q} ] = 4$$
Then to show that the set mentioned in the original question is a basis, it suffices to show $mathbb{Q}$-linear independence. I leave that to you.






share|cite|improve this answer









$endgroup$



Notice that $X^2 - q$, $X^2 - p$ are irreducible over $mathbb{Q}$ because they have no roots, since any root would imply that $sqrt{p}$ and $sqrt{q}$ are rational numbers. Using the same proof as for the irrationality of $sqrt{2}$, we know this isn't true.



We then just need to show that $X^2 - q$ remains irreducible over $mathbb{Q} (sqrt{p})$. Again, it suffices to show there are no roots. A general element of $mathbb{Q} (sqrt{p})$ looks like $a + b sqrt{p}$ for $a, b in mathbb{Q}$; plugging this in:
$$(a+ b sqrt{p})^2 - q = 0 implies a^2 + b^2 p - q + 2 ab sqrt{p} = 0$$
which tells us $ab = 0$ and $a^2 + b^2 p = q$. This means $a=0$ or $b=0$. In the first case, we find that $p$ divides $q$, a contradiction. In the second case, we find that $a^2 = p$ for some rational number, again, impossible. This means that $[mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ( sqrt{p} ) ] =2$. Using the tower law,
$$[mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ] = [mathbb{Q} ( sqrt{p} , sqrt{q} ) : mathbb{Q} ( sqrt{p} ) ] cdot [ mathbb{Q} ( sqrt{p}) : mathbb{Q} ] = 4$$
Then to show that the set mentioned in the original question is a basis, it suffices to show $mathbb{Q}$-linear independence. I leave that to you.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 22:47









RellekRellek

1,505517




1,505517












  • $begingroup$
    I like this answer for its generality. There are shorter proofs, either very special, or depending on fancy number theory. But this does the trick admirably.
    $endgroup$
    – Lubin
    Dec 13 '18 at 22:50












  • $begingroup$
    Thanks, this helps a lot. I should be able to show the linear independence.
    $endgroup$
    – Anzu
    Dec 13 '18 at 23:01


















  • $begingroup$
    I like this answer for its generality. There are shorter proofs, either very special, or depending on fancy number theory. But this does the trick admirably.
    $endgroup$
    – Lubin
    Dec 13 '18 at 22:50












  • $begingroup$
    Thanks, this helps a lot. I should be able to show the linear independence.
    $endgroup$
    – Anzu
    Dec 13 '18 at 23:01
















$begingroup$
I like this answer for its generality. There are shorter proofs, either very special, or depending on fancy number theory. But this does the trick admirably.
$endgroup$
– Lubin
Dec 13 '18 at 22:50






$begingroup$
I like this answer for its generality. There are shorter proofs, either very special, or depending on fancy number theory. But this does the trick admirably.
$endgroup$
– Lubin
Dec 13 '18 at 22:50














$begingroup$
Thanks, this helps a lot. I should be able to show the linear independence.
$endgroup$
– Anzu
Dec 13 '18 at 23:01




$begingroup$
Thanks, this helps a lot. I should be able to show the linear independence.
$endgroup$
– Anzu
Dec 13 '18 at 23:01


















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