Why can't Eisenstein Criterion be used for certain polynomials (to show that it's irreducible over...
$begingroup$
Why can't Eisenstein's Criterion be used to show that
$$4x^{10} - 9x^{3} + 21x - 18$$
is irreducible over $mathbb{Q}$?
I mean even if we were to apply Eisenstein here, there doesn't exist a prime $p$ that would apply all the E.C. rules anyways.
A detailed explanation would be great! Thanks.
abstract-algebra field-theory irreducible-polynomials
edited May 13 '15 at 21:35
Ken
3,63151728
asked May 13 '15 at 21:07
dean3794dean3794
788
$endgroup$
add a comment |
$begingroup$
Why can't Eisenstein's Criterion be used to show that
$$4x^{10} - 9x^{3} + 21x - 18$$
is irreducible over $mathbb{Q}$?
I mean even if we were to apply Eisenstein here, there doesn't exist a prime $p$ that would apply all the E.C. rules anyways.
A detailed explanation would be great! Thanks.
abstract-algebra field-theory irreducible-polynomials
edited May 13 '15 at 21:35
Ken
3,63151728
asked May 13 '15 at 21:07
dean3794dean3794
788
$endgroup$
$begingroup$
@WillJagy but why would you want to do that when you have a degree as big as 10?
$endgroup$
– dean3794
May 13 '15 at 21:14
1
$begingroup$
Your cuestión is a bit weird: Eisenstein's criterion cannot be applied to that polynomial simply because there is no prime for which the required condition is satisfied.
$endgroup$
– Mariano Suárez-Álvarez
May 13 '15 at 21:39
$begingroup$
@MarianoSuárez-Alvarez hmm okay. Because working this out as a true and false question o thought it meant that using the Eisenstein Method is not sufficient to show that this polynomial is irreducible
$endgroup$
– dean3794
May 13 '15 at 21:41
$begingroup$
I'm not sure what your question is. Is it "here is a polynomial that I know to be irreducible, why can't I deduce that it's irreducible from Eisenstein?" because you seemed to have answered that already in the statement of your question - i.e. Eisenstein doesn't apply for any prime $p$, since $3^2 | 18$.
$endgroup$
– James
May 13 '15 at 21:44
add a comment |
$begingroup$
Why can't Eisenstein's Criterion be used to show that
$$4x^{10} - 9x^{3} + 21x - 18$$
is irreducible over $mathbb{Q}$?
I mean even if we were to apply Eisenstein here, there doesn't exist a prime $p$ that would apply all the E.C. rules anyways.
A detailed explanation would be great! Thanks.
abstract-algebra field-theory irreducible-polynomials
edited May 13 '15 at 21:35
Ken
3,63151728
asked May 13 '15 at 21:07
dean3794dean3794
788
$endgroup$
Why can't Eisenstein's Criterion be used to show that
$$4x^{10} - 9x^{3} + 21x - 18$$
is irreducible over $mathbb{Q}$?
I mean even if we were to apply Eisenstein here, there doesn't exist a prime $p$ that would apply all the E.C. rules anyways.
A detailed explanation would be great! Thanks.
abstract-algebra field-theory irreducible-polynomials
abstract-algebra field-theory irreducible-polynomials
edited May 13 '15 at 21:35
Ken
3,63151728
asked May 13 '15 at 21:07
dean3794dean3794
788
edited May 13 '15 at 21:35
Ken
3,63151728
asked May 13 '15 at 21:07
dean3794dean3794
788
edited May 13 '15 at 21:35
Ken
3,63151728
edited May 13 '15 at 21:35
Ken
3,63151728
edited May 13 '15 at 21:35
Ken
3,63151728
3,63151728
asked May 13 '15 at 21:07
dean3794dean3794
788
asked May 13 '15 at 21:07
dean3794dean3794
788
asked May 13 '15 at 21:07
dean3794dean3794
788
788
$begingroup$
@WillJagy but why would you want to do that when you have a degree as big as 10?
$endgroup$
– dean3794
May 13 '15 at 21:14
1
$begingroup$
Your cuestión is a bit weird: Eisenstein's criterion cannot be applied to that polynomial simply because there is no prime for which the required condition is satisfied.
$endgroup$
– Mariano Suárez-Álvarez
May 13 '15 at 21:39
$begingroup$
@MarianoSuárez-Alvarez hmm okay. Because working this out as a true and false question o thought it meant that using the Eisenstein Method is not sufficient to show that this polynomial is irreducible
$endgroup$
– dean3794
May 13 '15 at 21:41
$begingroup$
I'm not sure what your question is. Is it "here is a polynomial that I know to be irreducible, why can't I deduce that it's irreducible from Eisenstein?" because you seemed to have answered that already in the statement of your question - i.e. Eisenstein doesn't apply for any prime $p$, since $3^2 | 18$.
$endgroup$
– James
May 13 '15 at 21:44
add a comment |
$begingroup$
@WillJagy but why would you want to do that when you have a degree as big as 10?
$endgroup$
– dean3794
May 13 '15 at 21:14
1
$begingroup$
Your cuestión is a bit weird: Eisenstein's criterion cannot be applied to that polynomial simply because there is no prime for which the required condition is satisfied.
$endgroup$
– Mariano Suárez-Álvarez
May 13 '15 at 21:39
$begingroup$
@MarianoSuárez-Alvarez hmm okay. Because working this out as a true and false question o thought it meant that using the Eisenstein Method is not sufficient to show that this polynomial is irreducible
$endgroup$
– dean3794
May 13 '15 at 21:41
$begingroup$
I'm not sure what your question is. Is it "here is a polynomial that I know to be irreducible, why can't I deduce that it's irreducible from Eisenstein?" because you seemed to have answered that already in the statement of your question - i.e. Eisenstein doesn't apply for any prime $p$, since $3^2 | 18$.
$endgroup$
– James
May 13 '15 at 21:44
$begingroup$
@WillJagy but why would you want to do that when you have a degree as big as 10?
$endgroup$
– dean3794
May 13 '15 at 21:14
$begingroup$
@WillJagy but why would you want to do that when you have a degree as big as 10?
$endgroup$
– dean3794
May 13 '15 at 21:14
1
1
$begingroup$
Your cuestión is a bit weird: Eisenstein's criterion cannot be applied to that polynomial simply because there is no prime for which the required condition is satisfied.
$endgroup$
– Mariano Suárez-Álvarez
May 13 '15 at 21:39
$begingroup$
Your cuestión is a bit weird: Eisenstein's criterion cannot be applied to that polynomial simply because there is no prime for which the required condition is satisfied.
$endgroup$
– Mariano Suárez-Álvarez
May 13 '15 at 21:39
$begingroup$
@MarianoSuárez-Alvarez hmm okay. Because working this out as a true and false question o thought it meant that using the Eisenstein Method is not sufficient to show that this polynomial is irreducible
$endgroup$
– dean3794
May 13 '15 at 21:41
$begingroup$
@MarianoSuárez-Alvarez hmm okay. Because working this out as a true and false question o thought it meant that using the Eisenstein Method is not sufficient to show that this polynomial is irreducible
$endgroup$
– dean3794
May 13 '15 at 21:41
$begingroup$
I'm not sure what your question is. Is it "here is a polynomial that I know to be irreducible, why can't I deduce that it's irreducible from Eisenstein?" because you seemed to have answered that already in the statement of your question - i.e. Eisenstein doesn't apply for any prime $p$, since $3^2 | 18$.
$endgroup$
– James
May 13 '15 at 21:44
$begingroup$
I'm not sure what your question is. Is it "here is a polynomial that I know to be irreducible, why can't I deduce that it's irreducible from Eisenstein?" because you seemed to have answered that already in the statement of your question - i.e. Eisenstein doesn't apply for any prime $p$, since $3^2 | 18$.
$endgroup$
– James
May 13 '15 at 21:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $f(x)$ is a polynomial in $mathbf Z[x]$ such that $f(ax+b)$ is Eisenstein with respect to a prime $p$ for some integers (or rational numbers) $a$ and $b$, then (i) $f(x)$ is irreducible over the $p$-adic numbers $mathbf Q_p$ and (ii) $p$ divides the discriminant of $f(x)$. I find with PARI that your polynomial has discriminant with prime factors $2$, $3$, $61$, $293$, and $50997533$, and for each of these primes $p$ I find with PARI that $f(x)$ is reducible over $mathbf Q_p$ (it has a factor of degree $1$ or $2$ in each case). This proves you can't convert $f(x)$ into an Eisenstein polynomial by a linear change of variables.
answered Dec 1 '18 at 21:06
KCdKCd
16.7k4075
$endgroup$
$begingroup$
Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
$endgroup$
– Henry
Dec 3 '18 at 16:44
$begingroup$
@Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
$endgroup$
– KCd
Dec 3 '18 at 19:58
add a comment |
$begingroup$
The only prime that divides 9, 21, and 18 is 3. But $3^2 | 18$, so the Eisenstein criterion does not apply here.
answered May 13 '15 at 21:11
William StagnerWilliam Stagner
3,5831027
$endgroup$
2
$begingroup$
The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
$endgroup$
– k1.M
May 13 '15 at 21:23
$begingroup$
Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
$endgroup$
– William Stagner
May 13 '15 at 21:28
1
$begingroup$
So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
$endgroup$
– k1.M
May 13 '15 at 21:30
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
If $f(x)$ is a polynomial in $mathbf Z[x]$ such that $f(ax+b)$ is Eisenstein with respect to a prime $p$ for some integers (or rational numbers) $a$ and $b$, then (i) $f(x)$ is irreducible over the $p$-adic numbers $mathbf Q_p$ and (ii) $p$ divides the discriminant of $f(x)$. I find with PARI that your polynomial has discriminant with prime factors $2$, $3$, $61$, $293$, and $50997533$, and for each of these primes $p$ I find with PARI that $f(x)$ is reducible over $mathbf Q_p$ (it has a factor of degree $1$ or $2$ in each case). This proves you can't convert $f(x)$ into an Eisenstein polynomial by a linear change of variables.
answered Dec 1 '18 at 21:06
KCdKCd
16.7k4075
$endgroup$
$begingroup$
Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
$endgroup$
– Henry
Dec 3 '18 at 16:44
$begingroup$
@Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
$endgroup$
– KCd
Dec 3 '18 at 19:58
add a comment |
$begingroup$
If $f(x)$ is a polynomial in $mathbf Z[x]$ such that $f(ax+b)$ is Eisenstein with respect to a prime $p$ for some integers (or rational numbers) $a$ and $b$, then (i) $f(x)$ is irreducible over the $p$-adic numbers $mathbf Q_p$ and (ii) $p$ divides the discriminant of $f(x)$. I find with PARI that your polynomial has discriminant with prime factors $2$, $3$, $61$, $293$, and $50997533$, and for each of these primes $p$ I find with PARI that $f(x)$ is reducible over $mathbf Q_p$ (it has a factor of degree $1$ or $2$ in each case). This proves you can't convert $f(x)$ into an Eisenstein polynomial by a linear change of variables.
answered Dec 1 '18 at 21:06
KCdKCd
16.7k4075
$endgroup$
$begingroup$
Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
$endgroup$
– Henry
Dec 3 '18 at 16:44
$begingroup$
@Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
$endgroup$
– KCd
Dec 3 '18 at 19:58
add a comment |
$begingroup$
If $f(x)$ is a polynomial in $mathbf Z[x]$ such that $f(ax+b)$ is Eisenstein with respect to a prime $p$ for some integers (or rational numbers) $a$ and $b$, then (i) $f(x)$ is irreducible over the $p$-adic numbers $mathbf Q_p$ and (ii) $p$ divides the discriminant of $f(x)$. I find with PARI that your polynomial has discriminant with prime factors $2$, $3$, $61$, $293$, and $50997533$, and for each of these primes $p$ I find with PARI that $f(x)$ is reducible over $mathbf Q_p$ (it has a factor of degree $1$ or $2$ in each case). This proves you can't convert $f(x)$ into an Eisenstein polynomial by a linear change of variables.
answered Dec 1 '18 at 21:06
KCdKCd
16.7k4075
$endgroup$
If $f(x)$ is a polynomial in $mathbf Z[x]$ such that $f(ax+b)$ is Eisenstein with respect to a prime $p$ for some integers (or rational numbers) $a$ and $b$, then (i) $f(x)$ is irreducible over the $p$-adic numbers $mathbf Q_p$ and (ii) $p$ divides the discriminant of $f(x)$. I find with PARI that your polynomial has discriminant with prime factors $2$, $3$, $61$, $293$, and $50997533$, and for each of these primes $p$ I find with PARI that $f(x)$ is reducible over $mathbf Q_p$ (it has a factor of degree $1$ or $2$ in each case). This proves you can't convert $f(x)$ into an Eisenstein polynomial by a linear change of variables.
answered Dec 1 '18 at 21:06
KCdKCd
16.7k4075
answered Dec 1 '18 at 21:06
KCdKCd
16.7k4075
answered Dec 1 '18 at 21:06
KCdKCd
16.7k4075
answered Dec 1 '18 at 21:06
KCdKCd
16.7k4075
16.7k4075
$begingroup$
Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
$endgroup$
– Henry
Dec 3 '18 at 16:44
$begingroup$
@Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
$endgroup$
– KCd
Dec 3 '18 at 19:58
add a comment |
$begingroup$
Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
$endgroup$
– Henry
Dec 3 '18 at 16:44
$begingroup$
@Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
$endgroup$
– KCd
Dec 3 '18 at 19:58
$begingroup$
Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
$endgroup$
– Henry
Dec 3 '18 at 16:44
$begingroup$
Hello. Off topic, but I wanted to know if you had a reference for Fermat's own proof of the two square theorem, as you stated here, as a remark on page 28? I couldn't find the entire proof anywhere.
$endgroup$
– Henry
Dec 3 '18 at 16:44
$begingroup$
@Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
$endgroup$
– KCd
Dec 3 '18 at 19:58
$begingroup$
@Henry there is no record of a detailed proof by Fermat on anything. He usually just claimed to be able to do things, and a few times he very briefly sketched the idea. For the two square theorem, see Remark 5.2 of math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf.
$endgroup$
– KCd
Dec 3 '18 at 19:58
add a comment |
$begingroup$
The only prime that divides 9, 21, and 18 is 3. But $3^2 | 18$, so the Eisenstein criterion does not apply here.
answered May 13 '15 at 21:11
William StagnerWilliam Stagner
3,5831027
$endgroup$
2
$begingroup$
The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
$endgroup$
– k1.M
May 13 '15 at 21:23
$begingroup$
Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
$endgroup$
– William Stagner
May 13 '15 at 21:28
1
$begingroup$
So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
$endgroup$
– k1.M
May 13 '15 at 21:30
add a comment |
$begingroup$
The only prime that divides 9, 21, and 18 is 3. But $3^2 | 18$, so the Eisenstein criterion does not apply here.
answered May 13 '15 at 21:11
William StagnerWilliam Stagner
3,5831027
$endgroup$
2
$begingroup$
The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
$endgroup$
– k1.M
May 13 '15 at 21:23
$begingroup$
Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
$endgroup$
– William Stagner
May 13 '15 at 21:28
1
$begingroup$
So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
$endgroup$
– k1.M
May 13 '15 at 21:30
add a comment |
$begingroup$
The only prime that divides 9, 21, and 18 is 3. But $3^2 | 18$, so the Eisenstein criterion does not apply here.
answered May 13 '15 at 21:11
William StagnerWilliam Stagner
3,5831027
$endgroup$
The only prime that divides 9, 21, and 18 is 3. But $3^2 | 18$, so the Eisenstein criterion does not apply here.
answered May 13 '15 at 21:11
William StagnerWilliam Stagner
3,5831027
answered May 13 '15 at 21:11
William StagnerWilliam Stagner
3,5831027
answered May 13 '15 at 21:11
William StagnerWilliam Stagner
3,5831027
answered May 13 '15 at 21:11
William StagnerWilliam Stagner
3,5831027
3,5831027
2
$begingroup$
The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
$endgroup$
– k1.M
May 13 '15 at 21:23
$begingroup$
Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
$endgroup$
– William Stagner
May 13 '15 at 21:28
1
$begingroup$
So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
$endgroup$
– k1.M
May 13 '15 at 21:30
add a comment |
2
$begingroup$
The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
$endgroup$
– k1.M
May 13 '15 at 21:23
$begingroup$
Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
$endgroup$
– William Stagner
May 13 '15 at 21:28
1
$begingroup$
So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
$endgroup$
– k1.M
May 13 '15 at 21:30
2
2
$begingroup$
The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
$endgroup$
– k1.M
May 13 '15 at 21:23
$begingroup$
The polynomial $1+x+x^2+dots+x^{p-1}$ is irreducible by Eisenestein's theorem. But your method doesn't works.
$endgroup$
– k1.M
May 13 '15 at 21:23
$begingroup$
Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
$endgroup$
– William Stagner
May 13 '15 at 21:28
$begingroup$
Eisenstein applies there, but not directly. You must substitute $xmapsto x+1$. In general, it is not easy to see whether or not this is possible, so I answered the question for the specific polynomial specified by the OP.
$endgroup$
– William Stagner
May 13 '15 at 21:28
1
1
$begingroup$
So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
$endgroup$
– k1.M
May 13 '15 at 21:30
$begingroup$
So you can't say that a method doesn't applies, because always there exists tricky methods, and this is mathematics...
$endgroup$
– k1.M
May 13 '15 at 21:30
add a comment |
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@WillJagy but why would you want to do that when you have a degree as big as 10?
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– dean3794
May 13 '15 at 21:14
1
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Your cuestión is a bit weird: Eisenstein's criterion cannot be applied to that polynomial simply because there is no prime for which the required condition is satisfied.
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– Mariano Suárez-Álvarez
May 13 '15 at 21:39
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@MarianoSuárez-Alvarez hmm okay. Because working this out as a true and false question o thought it meant that using the Eisenstein Method is not sufficient to show that this polynomial is irreducible
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– dean3794
May 13 '15 at 21:41
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I'm not sure what your question is. Is it "here is a polynomial that I know to be irreducible, why can't I deduce that it's irreducible from Eisenstein?" because you seemed to have answered that already in the statement of your question - i.e. Eisenstein doesn't apply for any prime $p$, since $3^2 | 18$.
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– James
May 13 '15 at 21:44