Eisenstein's Irreducibility Criterion

  • Thread starter roam
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In summary, the conversation discusses the application of Eisenstein's Criterion to prove that g(x)=x^4+4x^3+7x+5 is irreducible over the rational numbers. The method fails because there is no prime number that satisfies the criteria. Alternative methods, such as using g(x+1) or g(x+2), are suggested but also do not work. The conversation then turns to using a theorem that states if g(x+1) is irreducible, then g(x) is irreducible. The conversation ends with a discussion on using the Mod p irreducibility test, where p is a prime, to show that g(x) is irreducible over the rational numbers.
  • #1
roam
1,271
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I have a question about the application of Eisenstein’s Criterion. I want to show that

g(x)=x4+4x3+7x+5 is irreducible over [tex]\mathbb{Q}[/tex].

That means I need to find a prime number p such that

[tex]p \nmid 1[/tex] , [tex]p | 4[/tex] , [tex]p | 7[/tex], [tex]p | 5[/tex] and [tex]p^2 \nmid 5[/tex].

But unfortunately I can't see any prime number which would satisfies this! :confused:

I think there is a theorem that says if g(x+1) is irreducible then g(x) is irreducible. So in this case

g(x+1) = x4 + 8x3 + 18 x2 + 16 x + 17

But again I cannot find a p to satisfiy Eisenstein’s irreducibility criterion... So why does the method fail? Then what other method can one use to establish g(x)'s irreducibility?
 
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  • #2
You could use:
[tex] (g\mod p) \text{ is irreducible in } (\mathbb{Z}/p\mathbb{Z})[X] \Rightarrow
g \text{ is irreducible in } \mathbb{Z}[X] \Rightarrow
g \text{ is irreducible in } \mathbb{Q}[X]. [/tex]

where p is prime
 
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  • #3
g(x+2) works for Eisenstein.
 
  • #4
losiu99 said:
g(x+2) works for Eisenstein.

Are you sure?

[tex]g(x+2)= x^4 + 12 x^3 + 48 x^2 + 87 x + 67[/tex]

2 does not divide 87 and 3 does not divide 67. So there are no primes to use for Eisenstein...

You could use:
[tex] (g\mod p) \text{ is irreducible in } (\mathbb{Z}/p\mathbb{Z})[X] \Rightarrow g \text{ is irreducible in } \mathbb{Z}[X] \Rightarrowg \text{ is irreducible in } \mathbb{Q}[X]. [/tex]

where p is prime

I haven't seen this before but there s a similar test in my book called "Mod p irreducibility test", where p is a prime, here we must reduce the polynomial in Zp and see if the reduce polynomial is irreducible over Zp, if it is, it imples irreducibility over Q for the original polynomial. So here's what I've done:

Let g(x)=x4+4x3+7x+5. Then over [tex]\mathbb{Z}_2[/tex], we have [tex]\bar{g}(x)=x^4+x+1[/tex]. If we show that [tex]\bar{g}(x)[/tex] is irreducible over [tex]\mathbb{Z}_2[/tex] that implies that g(x) is irreducible over [tex]\mathbb{Q}[/tex].

But how do we show this?? We surely can't use Eisenstein here... :confused:
 
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  • #5
roam said:
Let g(x)=x4+4x3+7x+5. Then over [tex]\mathbb{Z}_2[/tex], we have [tex]\bar{g}(x)=x^4+x+1[/tex]. If we show that [tex]\bar{g}(x)[/tex] is irreducible over [tex]\mathbb{Z}_2[/tex] that implies that g(x) is irreducible over [tex]\mathbb{Q}[/tex].

But how do we show this?? We surely can't use Eisenstein here... :confused:

In [tex]\mathbb{Z}_2[/tex] all coefficients are either 1 or 0, so ...
 
  • #6
gerben said:
In [tex]\mathbb{Z}_2[/tex] all coefficients are either 1 or 0, so ...

What do you mean?? I know that [tex]\bar{g}(0)=1[/tex] and [tex]\bar{g}(1)=1[/tex] in [tex]\mathbb{Z}_2[/tex], so there are no zeros. But we can't use this to conclude that it is irreducible because this test is only for degrees 2 and 3, so I can't use it when the polynomial has degree 4...

And for the Eisenstein we need a prime that divides 1 and doesn not divide 1 at the same time which is impossible!
 
  • #7
roam said:
What do you mean?? I know that [tex]\bar{g}(0)=1[/tex] and [tex]\bar{g}(1)=1[/tex] in [tex]\mathbb{Z}_2[/tex], so there are no zeros. But we can't use this to conclude that it is irreducible because this test is only for degrees 2 and 3, so I can't use it when the polynomial has degree 4...

And for the Eisenstein we need a prime that divides 1 and doesn not divide 1 at the same time which is impossible!

You have [tex]\bar{g}(x)=x^4+x+1[/tex]. You cannot divide out a factor x, so if there were some factor it would have to be of the form (x^n+q) for some polynomial q and some integer n. Any coefficient in these polynomials is 1, use this to argue that you must have [tex](x^n+q_1)(x^n+q_2) \neq x^4+x+1[/tex] for all [tex]q_1,q_2\in \mathbb{Z}_2[X][/tex].
 
  • #8
gerben said:
You have [tex]\bar{g}(x)=x^4+x+1[/tex]. You cannot divide out a factor x, so if there were some factor it would have to be of the form (x^n+q) for some polynomial q and some integer n. Any coefficient in these polynomials is 1, use this to argue that you must have [tex](x^n+q_1)(x^n+q_2) \neq x^4+x+1[/tex] for all [tex]q_1,q_2\in \mathbb{Z}_2[X][/tex].

I think it's better to use xn and xm because they can be different as long as n+m=4.

[tex](x^n+q_1)(x^m+q_2) = x^{n+m} + q_2 x^n + q_1 x^m +q_1q_2[/tex]

In order for all coefficients to be 1 or 0, q1=q2=1 OR 0. Then we have:

[tex](x^n+1)(x^m+1) = x^{n+m} + x^n + x^m + 1[/tex] ...(1)

or

[tex](x^n+0)(x^m+0) = x^{n+m}[/tex] ...(2)

The polynomial (2) is not equal to [tex]\bar{g}(x)=x^4+x+1[/tex], and in polynomial (2) regardless of our choice for m and n, we will never end up with [tex]\bar{g}(x)[/tex]:

n,m = 4,0 [tex]\implies x^4+x^4+1[/tex]
n,m = 3,1 [tex]\implies x^4 + x^3 + x^2 +1[/tex]
n,m = 2,2 [tex]\implies x^4 + x^2 + x^2 +1[/tex]

Is this correct now?
 
  • #9
losiu99 said:
g(x+2) works for Eisenstein.
Sorry, I must have made some terrible mistake :blushing:
 
  • #10
roam said:
I think it's better to use xn and xm because they can be different as long as n+m=4.
Yes, I should have done that

roam said:
In order for all coefficients to be 1 or 0, q1=q2=1 OR 0. Then we have:

[tex](x^n+1)(x^m+1) = x^{n+m} + x^n + x^m + 1[/tex] ...(1)
or
[tex](x^n+0)(x^m+0) = x^{n+m}[/tex] ...(2)

No, q1 and q2 could be any polynomial just all the coefficients in those polynomials are either 0 or 1, i.e. all the ai in [tex]q_1 = \sum a_i X^i [/tex] are either 0 or 1 and all the bi in [tex]q_2 = \sum b_i X^i [/tex] are either 0 or 1.

roam said:
The polynomial (2) is not equal to [tex]\bar{g}(x)=x^4+x+1[/tex], and in polynomial (1) regardless of our choice for m and n, we will never end up with [tex]\bar{g}(x)[/tex]:

n,m = 4,0 [tex]\implies x^4+x^4+1[/tex]
n,m = 3,1 [tex]\implies x^4 + x^3 + x^2 +1[/tex]
n,m = 2,2 [tex]\implies x^4 + x^2 + x^2 +1[/tex]

Is this correct now?

You can use your idea like this:

if n,m = 4,0 then we have [tex] (x^4 + q_1) (1+q_2) =...[/tex] which cannot be equal to [tex]\bar{g}[/tex] because...
if n,m = 3,1 then we have [tex] (x^3 + q_1) (x+q_2) =... [/tex] which cannot be equal to [tex]\bar{g}[/tex] because...
if n,m = 2,2 then we have [tex] (x^2 + q_1) (x^2+q_2) =...[/tex] which cannot be equal to [tex]\bar{g}[/tex] because...

(better not use "\implies" here, whenever you use "\implies" you need to make really sure that whenever the thing on the left of the arrow is true that then also the thing on the right is true, and it is really easy to forget something on the left)
 
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  • #11
Thank you so much;

We have [tex](x^n+q_1)(x^m+q_2) = x^{n+m} + q_2x^n + q_1x^m + q_1q_2[/tex]

if n,m = 4,0 then we have [tex] (x^4 + q_1) (1+q_2) = x^4+q_2x^4+q_1+q_1q_2[/tex] which cannot be equal to [tex]\bar{g}[/tex] because regardless of what value (1 or 0) we choose for the coefficients we will never get [tex]\bar{g}[/tex].

if n,m = 3,1 then we have [tex] (x^3 + q_1) (x+q_2) =x^4+q_2x^3+q_1x+q_1q_2[/tex] which cannot be equal to [tex]\bar{g}[/tex] because no possible choice of coefficients can make it equal to [tex]\bar{g}[/tex].

if n,m = 2,2 then we have [tex] (x^2 + q_1) (x^2+q_2) =x^4+q_2x^2+q_1x^2+q_1q_2[/tex] which cannot be equal to [tex]\bar{g}[/tex] because whatver value we pick for [tex]q_1[/tex] and [tex]q_2[/tex] we will never end up with [tex]\bar{g}[/tex].

This is okay now I guess?
 

Related to Eisenstein's Irreducibility Criterion

1. What is Eisenstein's Irreducibility Criterion?

Eisenstein's Irreducibility Criterion is a mathematical test used to determine whether a polynomial is irreducible, meaning it cannot be factored into polynomials with smaller degrees. It is named after mathematician Ferdinand Eisenstein.

2. How does Eisenstein's Irreducibility Criterion work?

Eisenstein's Irreducibility Criterion states that if a polynomial has integer coefficients and there exists a prime number that divides all coefficients except the leading coefficient, and the prime number squared does not divide the constant term, then the polynomial is irreducible.

3. What is the significance of Eisenstein's Irreducibility Criterion?

Eisenstein's Irreducibility Criterion is a powerful tool in determining whether a polynomial is irreducible. It is especially useful in fields such as algebraic number theory and algebraic geometry.

4. Can Eisenstein's Irreducibility Criterion be applied to polynomials with non-integer coefficients?

No, Eisenstein's Irreducibility Criterion only applies to polynomials with integer coefficients. If the coefficients are not integers, then a different method must be used to determine irreducibility.

5. What are some examples of polynomials that can be proven irreducible using Eisenstein's Irreducibility Criterion?

An example of a polynomial that can be proven irreducible using Eisenstein's Irreducibility Criterion is x^4 + 3x^3 + 6x^2 + 9x + 12. Another example is x^5 + 5x^3 + 10x + 15.

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