Am I the only one constantly forgetting the Eisenstein criterion? [on hold]
Clash Royale CLAN TAG#URR8PPP
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Do you guys have tricks to remember the Eisenstein criterion? I constantly forget it and I am looking for some logic in it to never forget it again.
abstract-algebra intuition mnemonic
put on hold as off-topic by rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom, amWhy yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." – rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom
add a comment |
up vote
5
down vote
favorite
Do you guys have tricks to remember the Eisenstein criterion? I constantly forget it and I am looking for some logic in it to never forget it again.
abstract-algebra intuition mnemonic
put on hold as off-topic by rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom, amWhy yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." – rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom
1
Maybe memorizing a few specific examples (and non-examples) can help.
– Berci
2 days ago
1
Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this.
– André 3000
2 days ago
2
I'm pretty sure if you review every day for a week, and periodically thereafter, you will be able to remember any list of three things as simple as the three criteria for Eisenstein. I don't think it's worthwhile looking for a weird mnemonic. If it turns out you have some weird version of Eisenstein in mind that I don't know about, apologies. And of course I applaud your attempt to cement it with some sort of heuristic.
– rschwieb
2 days ago
1
How about: $x^2-2$ is irreducible, $x^2-4=(x-2)(x+2)$ obviously isn't, neither is $2x^2-2=2(x-1)(x+1)$. That's pretty much it re the first and the last coefficient. Those in between are the easy part, right?
– Jyrki Lahtonen
2 days ago
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Do you guys have tricks to remember the Eisenstein criterion? I constantly forget it and I am looking for some logic in it to never forget it again.
abstract-algebra intuition mnemonic
Do you guys have tricks to remember the Eisenstein criterion? I constantly forget it and I am looking for some logic in it to never forget it again.
abstract-algebra intuition mnemonic
abstract-algebra intuition mnemonic
asked 2 days ago
roi_saumon
1547
1547
put on hold as off-topic by rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom, amWhy yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." – rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom
put on hold as off-topic by rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom, amWhy yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is not about mathematics, within the scope defined in the help center." – rschwieb, Dietrich Burde, Morgan Rodgers, Tom-Tom
1
Maybe memorizing a few specific examples (and non-examples) can help.
– Berci
2 days ago
1
Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this.
– André 3000
2 days ago
2
I'm pretty sure if you review every day for a week, and periodically thereafter, you will be able to remember any list of three things as simple as the three criteria for Eisenstein. I don't think it's worthwhile looking for a weird mnemonic. If it turns out you have some weird version of Eisenstein in mind that I don't know about, apologies. And of course I applaud your attempt to cement it with some sort of heuristic.
– rschwieb
2 days ago
1
How about: $x^2-2$ is irreducible, $x^2-4=(x-2)(x+2)$ obviously isn't, neither is $2x^2-2=2(x-1)(x+1)$. That's pretty much it re the first and the last coefficient. Those in between are the easy part, right?
– Jyrki Lahtonen
2 days ago
add a comment |
1
Maybe memorizing a few specific examples (and non-examples) can help.
– Berci
2 days ago
1
Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this.
– André 3000
2 days ago
2
I'm pretty sure if you review every day for a week, and periodically thereafter, you will be able to remember any list of three things as simple as the three criteria for Eisenstein. I don't think it's worthwhile looking for a weird mnemonic. If it turns out you have some weird version of Eisenstein in mind that I don't know about, apologies. And of course I applaud your attempt to cement it with some sort of heuristic.
– rschwieb
2 days ago
1
How about: $x^2-2$ is irreducible, $x^2-4=(x-2)(x+2)$ obviously isn't, neither is $2x^2-2=2(x-1)(x+1)$. That's pretty much it re the first and the last coefficient. Those in between are the easy part, right?
– Jyrki Lahtonen
2 days ago
1
1
Maybe memorizing a few specific examples (and non-examples) can help.
– Berci
2 days ago
Maybe memorizing a few specific examples (and non-examples) can help.
– Berci
2 days ago
1
1
Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this.
– André 3000
2 days ago
Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this.
– André 3000
2 days ago
2
2
I'm pretty sure if you review every day for a week, and periodically thereafter, you will be able to remember any list of three things as simple as the three criteria for Eisenstein. I don't think it's worthwhile looking for a weird mnemonic. If it turns out you have some weird version of Eisenstein in mind that I don't know about, apologies. And of course I applaud your attempt to cement it with some sort of heuristic.
– rschwieb
2 days ago
I'm pretty sure if you review every day for a week, and periodically thereafter, you will be able to remember any list of three things as simple as the three criteria for Eisenstein. I don't think it's worthwhile looking for a weird mnemonic. If it turns out you have some weird version of Eisenstein in mind that I don't know about, apologies. And of course I applaud your attempt to cement it with some sort of heuristic.
– rschwieb
2 days ago
1
1
How about: $x^2-2$ is irreducible, $x^2-4=(x-2)(x+2)$ obviously isn't, neither is $2x^2-2=2(x-1)(x+1)$. That's pretty much it re the first and the last coefficient. Those in between are the easy part, right?
– Jyrki Lahtonen
2 days ago
How about: $x^2-2$ is irreducible, $x^2-4=(x-2)(x+2)$ obviously isn't, neither is $2x^2-2=2(x-1)(x+1)$. That's pretty much it re the first and the last coefficient. Those in between are the easy part, right?
– Jyrki Lahtonen
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
8
down vote
accepted
It's memorable if you view the proof as a consequence of the unique factorization of prime products.
The first part of the hypothesis says: $bmod p!: f equiv a x^n,$ is (an associate of) a prime power $x^n$
By uniqueness a $rmcolor#c00proper$ factorization has form $, ghequiv (b x^i) (c x^j),$ for $,color#c00i,j ge 1$
But $,color#c00i,j ge 1,Rightarrow, pmid g(0),h(0),Rightarrow, p^2mid g(0)h(0)!=!f(0),,$ contra hypothesis.
Said in ideal language it is $,(p,x)^2equiv (p^2),pmod! x$
Remark $ $ This view immediately leads to the discriminant-based test for finding shifts $xmapsto x+c$ that are Eisenstein, e.g. see this answer.
Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
– AOrtiz
2 days ago
@AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
– Bill Dubuque
2 days ago
Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
– Bill Dubuque
2 days ago
@BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
– roi_saumon
12 hours ago
@roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
– Bill Dubuque
11 hours ago
add a comment |
up vote
6
down vote
The way I always remember it is via the Newton polygon.
The basic idea is this: You plot a graph of dots with coordinates $(n,v_p(a_n))$ for $a_n$ the coefficients of your polynomial and $v_p$ the valuation with respect to $p$. Then you look at the lower convex hull, if this is a single line segment with slope $-1/n$ then the polynomial is irreducible, http://www-users.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf for more details.
Now this probably sounds more complicated! But its very geometric and thus easier to visualise whats going on (it also generalises quite well!) and just remembering what the picture has to look like is more than enough to recover the condition.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
It's memorable if you view the proof as a consequence of the unique factorization of prime products.
The first part of the hypothesis says: $bmod p!: f equiv a x^n,$ is (an associate of) a prime power $x^n$
By uniqueness a $rmcolor#c00proper$ factorization has form $, ghequiv (b x^i) (c x^j),$ for $,color#c00i,j ge 1$
But $,color#c00i,j ge 1,Rightarrow, pmid g(0),h(0),Rightarrow, p^2mid g(0)h(0)!=!f(0),,$ contra hypothesis.
Said in ideal language it is $,(p,x)^2equiv (p^2),pmod! x$
Remark $ $ This view immediately leads to the discriminant-based test for finding shifts $xmapsto x+c$ that are Eisenstein, e.g. see this answer.
Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
– AOrtiz
2 days ago
@AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
– Bill Dubuque
2 days ago
Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
– Bill Dubuque
2 days ago
@BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
– roi_saumon
12 hours ago
@roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
– Bill Dubuque
11 hours ago
add a comment |
up vote
8
down vote
accepted
It's memorable if you view the proof as a consequence of the unique factorization of prime products.
The first part of the hypothesis says: $bmod p!: f equiv a x^n,$ is (an associate of) a prime power $x^n$
By uniqueness a $rmcolor#c00proper$ factorization has form $, ghequiv (b x^i) (c x^j),$ for $,color#c00i,j ge 1$
But $,color#c00i,j ge 1,Rightarrow, pmid g(0),h(0),Rightarrow, p^2mid g(0)h(0)!=!f(0),,$ contra hypothesis.
Said in ideal language it is $,(p,x)^2equiv (p^2),pmod! x$
Remark $ $ This view immediately leads to the discriminant-based test for finding shifts $xmapsto x+c$ that are Eisenstein, e.g. see this answer.
Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
– AOrtiz
2 days ago
@AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
– Bill Dubuque
2 days ago
Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
– Bill Dubuque
2 days ago
@BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
– roi_saumon
12 hours ago
@roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
– Bill Dubuque
11 hours ago
add a comment |
up vote
8
down vote
accepted
up vote
8
down vote
accepted
It's memorable if you view the proof as a consequence of the unique factorization of prime products.
The first part of the hypothesis says: $bmod p!: f equiv a x^n,$ is (an associate of) a prime power $x^n$
By uniqueness a $rmcolor#c00proper$ factorization has form $, ghequiv (b x^i) (c x^j),$ for $,color#c00i,j ge 1$
But $,color#c00i,j ge 1,Rightarrow, pmid g(0),h(0),Rightarrow, p^2mid g(0)h(0)!=!f(0),,$ contra hypothesis.
Said in ideal language it is $,(p,x)^2equiv (p^2),pmod! x$
Remark $ $ This view immediately leads to the discriminant-based test for finding shifts $xmapsto x+c$ that are Eisenstein, e.g. see this answer.
It's memorable if you view the proof as a consequence of the unique factorization of prime products.
The first part of the hypothesis says: $bmod p!: f equiv a x^n,$ is (an associate of) a prime power $x^n$
By uniqueness a $rmcolor#c00proper$ factorization has form $, ghequiv (b x^i) (c x^j),$ for $,color#c00i,j ge 1$
But $,color#c00i,j ge 1,Rightarrow, pmid g(0),h(0),Rightarrow, p^2mid g(0)h(0)!=!f(0),,$ contra hypothesis.
Said in ideal language it is $,(p,x)^2equiv (p^2),pmod! x$
Remark $ $ This view immediately leads to the discriminant-based test for finding shifts $xmapsto x+c$ that are Eisenstein, e.g. see this answer.
edited 2 days ago
answered 2 days ago
Bill Dubuque
206k29189619
206k29189619
Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
– AOrtiz
2 days ago
@AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
– Bill Dubuque
2 days ago
Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
– Bill Dubuque
2 days ago
@BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
– roi_saumon
12 hours ago
@roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
– Bill Dubuque
11 hours ago
add a comment |
Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
– AOrtiz
2 days ago
@AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
– Bill Dubuque
2 days ago
Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
– Bill Dubuque
2 days ago
@BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
– roi_saumon
12 hours ago
@roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
– Bill Dubuque
11 hours ago
Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
– AOrtiz
2 days ago
Why does $i,jge 1$ imply that $p$ divides both $g(0)$ and $h(0)$?
– AOrtiz
2 days ago
@AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
– Bill Dubuque
2 days ago
@AOrtiz $bmod p!: gequiv b x^large i, ige 1,Rightarrow g(0)equiv bcdot 0^large iequiv 0 $ so $ pmid g(0), $ i.e. a nonconstant monomial $,bx^large i$ has zero constant term (in $Bbb F_p = Bbb Z/p)$
– Bill Dubuque
2 days ago
Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
– Bill Dubuque
2 days ago
Note: In the answer I (intentionally) skipped an initial application of Gauss's Lemma.
– Bill Dubuque
2 days ago
@BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
– roi_saumon
12 hours ago
@BillDubuque This explanation seems nice, but there is some things I don't understand. When you talk about the unique factorization of prime products you mean this, or could you point me to the statement? Why if $f$ is irreducible it is associated to $x^n$?
– roi_saumon
12 hours ago
@roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
– Bill Dubuque
11 hours ago
@roi Yes, $D = Bbb F_p[x],$ is a UFD. The image of $f$ in $D$ is $ ax^n$ since $p=0$ in $D$ and all non-lead coef's are divisible by $p$ so $0.,$ Since $x$ is irreducible (so prime) in the UFD $D$, every factorization of $x^n$ must have form $,x^i x^n-i,$ (up to unit factors).
– Bill Dubuque
11 hours ago
add a comment |
up vote
6
down vote
The way I always remember it is via the Newton polygon.
The basic idea is this: You plot a graph of dots with coordinates $(n,v_p(a_n))$ for $a_n$ the coefficients of your polynomial and $v_p$ the valuation with respect to $p$. Then you look at the lower convex hull, if this is a single line segment with slope $-1/n$ then the polynomial is irreducible, http://www-users.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf for more details.
Now this probably sounds more complicated! But its very geometric and thus easier to visualise whats going on (it also generalises quite well!) and just remembering what the picture has to look like is more than enough to recover the condition.
add a comment |
up vote
6
down vote
The way I always remember it is via the Newton polygon.
The basic idea is this: You plot a graph of dots with coordinates $(n,v_p(a_n))$ for $a_n$ the coefficients of your polynomial and $v_p$ the valuation with respect to $p$. Then you look at the lower convex hull, if this is a single line segment with slope $-1/n$ then the polynomial is irreducible, http://www-users.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf for more details.
Now this probably sounds more complicated! But its very geometric and thus easier to visualise whats going on (it also generalises quite well!) and just remembering what the picture has to look like is more than enough to recover the condition.
add a comment |
up vote
6
down vote
up vote
6
down vote
The way I always remember it is via the Newton polygon.
The basic idea is this: You plot a graph of dots with coordinates $(n,v_p(a_n))$ for $a_n$ the coefficients of your polynomial and $v_p$ the valuation with respect to $p$. Then you look at the lower convex hull, if this is a single line segment with slope $-1/n$ then the polynomial is irreducible, http://www-users.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf for more details.
Now this probably sounds more complicated! But its very geometric and thus easier to visualise whats going on (it also generalises quite well!) and just remembering what the picture has to look like is more than enough to recover the condition.
The way I always remember it is via the Newton polygon.
The basic idea is this: You plot a graph of dots with coordinates $(n,v_p(a_n))$ for $a_n$ the coefficients of your polynomial and $v_p$ the valuation with respect to $p$. Then you look at the lower convex hull, if this is a single line segment with slope $-1/n$ then the polynomial is irreducible, http://www-users.math.umn.edu/~garrett/m/number_theory/newton_polygon.pdf for more details.
Now this probably sounds more complicated! But its very geometric and thus easier to visualise whats going on (it also generalises quite well!) and just remembering what the picture has to look like is more than enough to recover the condition.
answered 2 days ago
Alex J Best
1,65711122
1,65711122
add a comment |
add a comment |
1
Maybe memorizing a few specific examples (and non-examples) can help.
– Berci
2 days ago
1
Maybe learning about the connection between Eisenstein polynomials and totally ramified primes would be helpful. Keith Conrad has notes on this.
– André 3000
2 days ago
2
I'm pretty sure if you review every day for a week, and periodically thereafter, you will be able to remember any list of three things as simple as the three criteria for Eisenstein. I don't think it's worthwhile looking for a weird mnemonic. If it turns out you have some weird version of Eisenstein in mind that I don't know about, apologies. And of course I applaud your attempt to cement it with some sort of heuristic.
– rschwieb
2 days ago
1
How about: $x^2-2$ is irreducible, $x^2-4=(x-2)(x+2)$ obviously isn't, neither is $2x^2-2=2(x-1)(x+1)$. That's pretty much it re the first and the last coefficient. Those in between are the easy part, right?
– Jyrki Lahtonen
2 days ago