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How to prove that $x^5-5$ is irreducible over $mathbb Q(sqrt 2, sqrt[3] 3) $
How to prove that $x^5-5$ is irreducible over $K = mathbb Q(sqrt 2, sqrt[3] 3)? $
I came across this problem while solving another one and I dunno exactly how to proceed. I know by Eisenstein's criterion that $p_c(x)= x^5-5 $ is irreducible over $mathbb Q$. This gives me for instance that any field extension containing $sqrt[5]5$ must be at least of degree $5$, but since the degree of $K$ over $mathbb Q$ is $6$, this won't help me to guarantee that $sqrt[5] 5 not in K$.
Any further directions? thanks.
abstract-algebra polynomials field-theory factoring irreducible-polynomials
edited Nov 21 '18 at 0:31
Batominovski
33.9k33292
asked Nov 20 '18 at 23:47
math.h
1,122517
add a comment |
How to prove that $x^5-5$ is irreducible over $K = mathbb Q(sqrt 2, sqrt[3] 3)? $
I came across this problem while solving another one and I dunno exactly how to proceed. I know by Eisenstein's criterion that $p_c(x)= x^5-5 $ is irreducible over $mathbb Q$. This gives me for instance that any field extension containing $sqrt[5]5$ must be at least of degree $5$, but since the degree of $K$ over $mathbb Q$ is $6$, this won't help me to guarantee that $sqrt[5] 5 not in K$.
Any further directions? thanks.
abstract-algebra polynomials field-theory factoring irreducible-polynomials
edited Nov 21 '18 at 0:31
Batominovski
33.9k33292
asked Nov 20 '18 at 23:47
math.h
1,122517
3
Well $Q(sqrt[5]{5})$ would them be a subfield. Is that possible?
– Sorfosh
Nov 20 '18 at 23:51
1
@Sorfosh I know that the answer is no beacuse $5$ does not divide $6$. But I still can't figure out why $p_c$ being reducible would imply that $Q(sqrt[5]5)$ is a subfield. Would you mind completing your hint?
– math.h
Nov 21 '18 at 0:15
1
I am not sure where the confusion is. If $sqrt[5]{5} in Q(sqrt{2},sqrt[3]{3})$ then $Q(sqrt[5]{5})$ must be a subfield of $Q(sqrt{2},sqrt[3]{3})$ by definition. It is the smallest field (with $Q$) that has $sqrt[5]{5}$
– Sorfosh
Nov 21 '18 at 1:33
add a comment |
How to prove that $x^5-5$ is irreducible over $K = mathbb Q(sqrt 2, sqrt[3] 3)? $
I came across this problem while solving another one and I dunno exactly how to proceed. I know by Eisenstein's criterion that $p_c(x)= x^5-5 $ is irreducible over $mathbb Q$. This gives me for instance that any field extension containing $sqrt[5]5$ must be at least of degree $5$, but since the degree of $K$ over $mathbb Q$ is $6$, this won't help me to guarantee that $sqrt[5] 5 not in K$.
Any further directions? thanks.
abstract-algebra polynomials field-theory factoring irreducible-polynomials
edited Nov 21 '18 at 0:31
Batominovski
33.9k33292
asked Nov 20 '18 at 23:47
math.h
1,122517
How to prove that $x^5-5$ is irreducible over $K = mathbb Q(sqrt 2, sqrt[3] 3)? $
I came across this problem while solving another one and I dunno exactly how to proceed. I know by Eisenstein's criterion that $p_c(x)= x^5-5 $ is irreducible over $mathbb Q$. This gives me for instance that any field extension containing $sqrt[5]5$ must be at least of degree $5$, but since the degree of $K$ over $mathbb Q$ is $6$, this won't help me to guarantee that $sqrt[5] 5 not in K$.
Any further directions? thanks.
abstract-algebra polynomials field-theory factoring irreducible-polynomials
abstract-algebra polynomials field-theory factoring irreducible-polynomials
edited Nov 21 '18 at 0:31
Batominovski
33.9k33292
asked Nov 20 '18 at 23:47
math.h
1,122517
edited Nov 21 '18 at 0:31
Batominovski
33.9k33292
asked Nov 20 '18 at 23:47
math.h
1,122517
edited Nov 21 '18 at 0:31
Batominovski
33.9k33292
edited Nov 21 '18 at 0:31
Batominovski
33.9k33292
edited Nov 21 '18 at 0:31
Batominovski
33.9k33292
33.9k33292
asked Nov 20 '18 at 23:47
math.h
1,122517
asked Nov 20 '18 at 23:47
math.h
1,122517
asked Nov 20 '18 at 23:47
math.h
1,122517
1,122517
3
Well $Q(sqrt[5]{5})$ would them be a subfield. Is that possible?
– Sorfosh
Nov 20 '18 at 23:51
1
@Sorfosh I know that the answer is no beacuse $5$ does not divide $6$. But I still can't figure out why $p_c$ being reducible would imply that $Q(sqrt[5]5)$ is a subfield. Would you mind completing your hint?
– math.h
Nov 21 '18 at 0:15
1
I am not sure where the confusion is. If $sqrt[5]{5} in Q(sqrt{2},sqrt[3]{3})$ then $Q(sqrt[5]{5})$ must be a subfield of $Q(sqrt{2},sqrt[3]{3})$ by definition. It is the smallest field (with $Q$) that has $sqrt[5]{5}$
– Sorfosh
Nov 21 '18 at 1:33
add a comment |
3
Well $Q(sqrt[5]{5})$ would them be a subfield. Is that possible?
– Sorfosh
Nov 20 '18 at 23:51
1
@Sorfosh I know that the answer is no beacuse $5$ does not divide $6$. But I still can't figure out why $p_c$ being reducible would imply that $Q(sqrt[5]5)$ is a subfield. Would you mind completing your hint?
– math.h
Nov 21 '18 at 0:15
1
I am not sure where the confusion is. If $sqrt[5]{5} in Q(sqrt{2},sqrt[3]{3})$ then $Q(sqrt[5]{5})$ must be a subfield of $Q(sqrt{2},sqrt[3]{3})$ by definition. It is the smallest field (with $Q$) that has $sqrt[5]{5}$
– Sorfosh
Nov 21 '18 at 1:33
3
3
Well $Q(sqrt[5]{5})$ would them be a subfield. Is that possible?
– Sorfosh
Nov 20 '18 at 23:51
Well $Q(sqrt[5]{5})$ would them be a subfield. Is that possible?
– Sorfosh
Nov 20 '18 at 23:51
1
1
@Sorfosh I know that the answer is no beacuse $5$ does not divide $6$. But I still can't figure out why $p_c$ being reducible would imply that $Q(sqrt[5]5)$ is a subfield. Would you mind completing your hint?
– math.h
Nov 21 '18 at 0:15
@Sorfosh I know that the answer is no beacuse $5$ does not divide $6$. But I still can't figure out why $p_c$ being reducible would imply that $Q(sqrt[5]5)$ is a subfield. Would you mind completing your hint?
– math.h
Nov 21 '18 at 0:15
1
1
I am not sure where the confusion is. If $sqrt[5]{5} in Q(sqrt{2},sqrt[3]{3})$ then $Q(sqrt[5]{5})$ must be a subfield of $Q(sqrt{2},sqrt[3]{3})$ by definition. It is the smallest field (with $Q$) that has $sqrt[5]{5}$
– Sorfosh
Nov 21 '18 at 1:33
I am not sure where the confusion is. If $sqrt[5]{5} in Q(sqrt{2},sqrt[3]{3})$ then $Q(sqrt[5]{5})$ must be a subfield of $Q(sqrt{2},sqrt[3]{3})$ by definition. It is the smallest field (with $Q$) that has $sqrt[5]{5}$
– Sorfosh
Nov 21 '18 at 1:33
add a comment |
1 Answer
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Let $p(x):=x^5-5$ (why do you need the unnecessary subscript $c$ anyway?). If $p(x)$ is reducible over $K$, then $p(x)$ has a linear factor over $K$ or an irreducible quadratic factor over $K$.
In the case where $p(x)$ has a linear factor over $K$, it follows that $p(x)$ has a root in $K$. Since $Ksubseteqmathbb{R}$ and $sqrt[5]{5}$ is the unique real root of $p(x)$, it follows that $sqrt[5]{5}in K$, which you know why this is false.
In the case where $p(x)$ has an irreducible quadratic factor $q(x)in K[x]$. Since $Ksubseteq mathbb{R}$, we must have
$$q(x)=(x-omegasqrt[5]{5})(x-bar{omega}sqrt[5]{5}),,$$
where $omega$ is a primitive $5$-th root of unity. This shows that the constant term $sqrt[5]{25}$ of $q(x)$ is in $K$. You probably know why $sqrt[5]{25}$ is not in $K$.
Here is a proof. If $sqrt[5]{25}$ is in $K$, then $sqrt[5]{5}=dfrac{(sqrt[5]{25})^3}{5} in K$.
answered Nov 21 '18 at 0:25
Batominovski
33.9k33292
add a comment |
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Let $p(x):=x^5-5$ (why do you need the unnecessary subscript $c$ anyway?). If $p(x)$ is reducible over $K$, then $p(x)$ has a linear factor over $K$ or an irreducible quadratic factor over $K$.
In the case where $p(x)$ has a linear factor over $K$, it follows that $p(x)$ has a root in $K$. Since $Ksubseteqmathbb{R}$ and $sqrt[5]{5}$ is the unique real root of $p(x)$, it follows that $sqrt[5]{5}in K$, which you know why this is false.
In the case where $p(x)$ has an irreducible quadratic factor $q(x)in K[x]$. Since $Ksubseteq mathbb{R}$, we must have
$$q(x)=(x-omegasqrt[5]{5})(x-bar{omega}sqrt[5]{5}),,$$
where $omega$ is a primitive $5$-th root of unity. This shows that the constant term $sqrt[5]{25}$ of $q(x)$ is in $K$. You probably know why $sqrt[5]{25}$ is not in $K$.
Here is a proof. If $sqrt[5]{25}$ is in $K$, then $sqrt[5]{5}=dfrac{(sqrt[5]{25})^3}{5} in K$.
answered Nov 21 '18 at 0:25
Batominovski
33.9k33292
add a comment |
Let $p(x):=x^5-5$ (why do you need the unnecessary subscript $c$ anyway?). If $p(x)$ is reducible over $K$, then $p(x)$ has a linear factor over $K$ or an irreducible quadratic factor over $K$.
In the case where $p(x)$ has a linear factor over $K$, it follows that $p(x)$ has a root in $K$. Since $Ksubseteqmathbb{R}$ and $sqrt[5]{5}$ is the unique real root of $p(x)$, it follows that $sqrt[5]{5}in K$, which you know why this is false.
In the case where $p(x)$ has an irreducible quadratic factor $q(x)in K[x]$. Since $Ksubseteq mathbb{R}$, we must have
$$q(x)=(x-omegasqrt[5]{5})(x-bar{omega}sqrt[5]{5}),,$$
where $omega$ is a primitive $5$-th root of unity. This shows that the constant term $sqrt[5]{25}$ of $q(x)$ is in $K$. You probably know why $sqrt[5]{25}$ is not in $K$.
Here is a proof. If $sqrt[5]{25}$ is in $K$, then $sqrt[5]{5}=dfrac{(sqrt[5]{25})^3}{5} in K$.
answered Nov 21 '18 at 0:25
Batominovski
33.9k33292
add a comment |
Let $p(x):=x^5-5$ (why do you need the unnecessary subscript $c$ anyway?). If $p(x)$ is reducible over $K$, then $p(x)$ has a linear factor over $K$ or an irreducible quadratic factor over $K$.
In the case where $p(x)$ has a linear factor over $K$, it follows that $p(x)$ has a root in $K$. Since $Ksubseteqmathbb{R}$ and $sqrt[5]{5}$ is the unique real root of $p(x)$, it follows that $sqrt[5]{5}in K$, which you know why this is false.
In the case where $p(x)$ has an irreducible quadratic factor $q(x)in K[x]$. Since $Ksubseteq mathbb{R}$, we must have
$$q(x)=(x-omegasqrt[5]{5})(x-bar{omega}sqrt[5]{5}),,$$
where $omega$ is a primitive $5$-th root of unity. This shows that the constant term $sqrt[5]{25}$ of $q(x)$ is in $K$. You probably know why $sqrt[5]{25}$ is not in $K$.
Here is a proof. If $sqrt[5]{25}$ is in $K$, then $sqrt[5]{5}=dfrac{(sqrt[5]{25})^3}{5} in K$.
answered Nov 21 '18 at 0:25
Batominovski
33.9k33292
Let $p(x):=x^5-5$ (why do you need the unnecessary subscript $c$ anyway?). If $p(x)$ is reducible over $K$, then $p(x)$ has a linear factor over $K$ or an irreducible quadratic factor over $K$.
In the case where $p(x)$ has a linear factor over $K$, it follows that $p(x)$ has a root in $K$. Since $Ksubseteqmathbb{R}$ and $sqrt[5]{5}$ is the unique real root of $p(x)$, it follows that $sqrt[5]{5}in K$, which you know why this is false.
In the case where $p(x)$ has an irreducible quadratic factor $q(x)in K[x]$. Since $Ksubseteq mathbb{R}$, we must have
$$q(x)=(x-omegasqrt[5]{5})(x-bar{omega}sqrt[5]{5}),,$$
where $omega$ is a primitive $5$-th root of unity. This shows that the constant term $sqrt[5]{25}$ of $q(x)$ is in $K$. You probably know why $sqrt[5]{25}$ is not in $K$.
Here is a proof. If $sqrt[5]{25}$ is in $K$, then $sqrt[5]{5}=dfrac{(sqrt[5]{25})^3}{5} in K$.
answered Nov 21 '18 at 0:25
Batominovski
33.9k33292
answered Nov 21 '18 at 0:25
Batominovski
33.9k33292
answered Nov 21 '18 at 0:25
Batominovski
33.9k33292
answered Nov 21 '18 at 0:25
Batominovski
33.9k33292
33.9k33292
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add a comment |
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3
Well $Q(sqrt[5]{5})$ would them be a subfield. Is that possible?
– Sorfosh
Nov 20 '18 at 23:51
1
@Sorfosh I know that the answer is no beacuse $5$ does not divide $6$. But I still can't figure out why $p_c$ being reducible would imply that $Q(sqrt[5]5)$ is a subfield. Would you mind completing your hint?
– math.h
Nov 21 '18 at 0:15
1
I am not sure where the confusion is. If $sqrt[5]{5} in Q(sqrt{2},sqrt[3]{3})$ then $Q(sqrt[5]{5})$ must be a subfield of $Q(sqrt{2},sqrt[3]{3})$ by definition. It is the smallest field (with $Q$) that has $sqrt[5]{5}$
– Sorfosh
Nov 21 '18 at 1:33