$X^2-q in mathbb{Q}(sqrt{p})[X]$ is irreducible and $[mathbb{Q}(sqrt{p}, sqrt{q}): mathbb{Q}]=4$
$begingroup$
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
asked Dec 13 '18 at 22:17
AnzuAnzu
246
$endgroup$
add a comment |
$begingroup$
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
asked Dec 13 '18 at 22:17
AnzuAnzu
246
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 13 '18 at 22:17
AnzuAnzu
246
$endgroup$
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
asked Dec 13 '18 at 22:17
AnzuAnzu
246
asked Dec 13 '18 at 22:17
AnzuAnzu
246
edited Dec 13 '18 at 22:38
Anzu
asked Dec 13 '18 at 22:17
AnzuAnzu
246
asked Dec 13 '18 at 22:17
AnzuAnzu
246
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 13 '18 at 22:47
RellekRellek
1,505517
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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.
answered Dec 13 '18 at 22:47
RellekRellek
1,505517
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 13 '18 at 22:47
RellekRellek
1,505517
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 13 '18 at 22:47
RellekRellek
1,505517
$endgroup$
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.
answered Dec 13 '18 at 22:47
RellekRellek
1,505517
answered Dec 13 '18 at 22:47
RellekRellek
1,505517
answered Dec 13 '18 at 22:47
RellekRellek
1,505517
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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