Is Eisenstein's Criterion Applicable to Polynomials in the Gaussian Integers?

  • Thread starter Ted123
  • Start date
  • Tags
    Rings
In summary, Eisenstein's criterion states that if a polynomial f(x) = a_0 + a_1x + ... + a_n x^n \in R[x] satisfies certain conditions in an integral domain R, then it is irreducible. This criterion can be applied in the Gaussian integers R=\mathbb{Z}[i] to determine if a polynomial is irreducible. To use this criterion, one must check that the coefficients of the polynomial satisfy certain conditions involving a prime ideal P of R. An example of using this criterion in the Gaussian integers is given, where the polynomial x^2+(1+i) is shown to be irreducible. However, the polynomial x^2-1 is not irreducible in \math
  • #1
Ted123
446
0
My statement of Eisenstein's criterion is the following:

Let [itex]R[/itex] be an integral domain, [itex]P[/itex] be prime ideal of [itex]R[/itex] and [itex]f(x) = a_0 + a_1x + ... + a_n x^n \in R[x][/itex].

Suppose
(1) [itex]a_0 , a_1 , ... , a_{n-1} \in P[/itex]
(2) [itex]a_0 \in P[/itex] but [itex]a_0 \not\in P^2[/itex]
(3) [itex]a_n \not\in P[/itex]

Then [itex]f[/itex] has no divisors of degree [itex]d[/itex] such that [itex]1\leqslant d \leqslant n-1[/itex]. In particular if [itex]f[/itex] is primitive and (1)-(3) hold then [itex]f[/itex] is irreducible.


I would like to see an example of how we can use this criterion in the Gaussian integers [itex]R= \mathbb{Z}[/itex].

I know [itex]1+i[/itex] is a Gaussian prime so can I anyone give me a quick example of a polynomial with coefficients in [itex]\mathbb{Z}[/itex] and how to use this criterion?

I know how to use it when [itex]R=\mathbb{Z}[/itex], for example to show [itex]f(x)= x^2 -1 \in \mathbb{Z}[x][/itex] is irreducible we just check that for a prime [itex]p[/itex]: [itex]p | a_n[/itex], [itex]p | a_i[/itex] for all [itex]i<n[/itex] and [itex]p^2 \not | a_0[/itex] . I'm confused as to how to use the version of it above for a polynomial in [itex]\mathbb{Z}[x][/itex].
 
Last edited:
Physics news on Phys.org
  • #2
Ted123 said:
My statement of Eisenstein's criterion is the following:

Let [itex]R[/itex] be an integral domain, [itex]P[/itex] be prime ideal of [itex]R[/itex] and [itex]f(x) = a_0 + a_1x + ... + a_n x^n \in R[x][/itex].

Suppose
(1) [itex]a_0 , a_1 , ... , a_{n-1} \in P[/itex]
(2) [itex]a_0 \in P[/itex] but [itex]a_0 \not\in P^2[/itex]
(3) [itex]a_n \not\in P[/itex]

Then [itex]f[/itex] has no divisors of degree [itex]d[/itex] such that [itex]1\leqslant d \leqslant n-1[/itex]. In particular if [itex]f[/itex] is primitive and (1)-(3) hold then [itex]f[/itex] is irreducible.


I would like to see an example of how we can use this criterion in the Gaussian integers [itex]R= \mathbb{Z}[/itex].

I know [itex]1+i[/itex] is a Gaussian prime so can I anyone give me a quick example of a polynomial with coefficients in [itex]\mathbb{Z}[/itex] and how to use this criterion?

I know how to use it when [itex]R=\mathbb{Z}[/itex], for example to show [itex]f(x)= x^2 -1 \in \mathbb{Z}[x][/itex] is irreducible we just check that for a prime [itex]p[/itex]: [itex]p | a_n[/itex], [itex]p | a_i[/itex] for all [itex]i<n[/itex] and [itex]p^2 \not | a_0[/itex] . I'm confused as to how to use the version of it above for a polynomial in [itex]\mathbb{Z}[x][/itex].



What about the polynomial [itex]\,x^2+(1+i)\in\left(\mathbb{Z}\right)[x]\,[/itex] ?

BTW, in your example, [itex]\,x^2-1\in\mathbb{Z}[x]\,[/itex] is reducible...;>)

DonAntonio
 
  • #3
DonAntonio said:
What about the polynomial [itex]\,x^2+(1+i)\in\left(\mathbb{Z}\right)[x]\,[/itex] ?

BTW, in your example, [itex]\,x^2-1\in\mathbb{Z}[x]\,[/itex] is reducible...;>)

DonAntonio


Is the polynomial [tex]f(x) = x^7 + (3-i)x^2 + (3+4i)x + (4+2i) \in \mathbb{Z}[x][/tex] irreducible?

[itex]2+i[/itex] is a Gauassian prime isn't it? And 2+i does not divide 1, 2+i | 3-i , 2+i | 3+4i , 2+i | 4+2i and (2+i)^2 = 3+4i which does not divide 4+2i.
 
Last edited:

FAQ: Is Eisenstein's Criterion Applicable to Polynomials in the Gaussian Integers?

What is the Eisenstein criterion?

The Eisenstein criterion is a test used to determine if a polynomial with integer coefficients is irreducible over the rational numbers. It states that if a polynomial is of the form anxn + an-1xn-1 + ... + a1x + a0, where all coefficients except a0 are divisible by a prime number p and a0 is not divisible by p2, then the polynomial is irreducible over the rational numbers.

How is the Eisenstein criterion used?

The Eisenstein criterion is used to determine if a polynomial is irreducible over the rational numbers without having to perform polynomial long division or factoring. It is a quick and efficient way to determine the irreducibility of a polynomial, which is important in many areas of mathematics and science.

What is the significance of the Eisenstein criterion?

The Eisenstein criterion is significant because it provides a necessary and sufficient condition for the irreducibility of a polynomial over the rational numbers. This means that if the criterion is satisfied, the polynomial is definitely irreducible, and if it is not satisfied, the polynomial is definitely reducible.

What are the limitations of the Eisenstein criterion?

The Eisenstein criterion can only be applied to polynomials with integer coefficients. It cannot be used for polynomials with coefficients in other fields, such as the complex numbers. Additionally, the criterion only applies to a specific form of polynomials, so it cannot be used for all polynomials.

Are there any extensions of the Eisenstein criterion?

Yes, there are extensions of the Eisenstein criterion for polynomials with coefficients in other fields, such as the complex numbers. These extensions involve using different primes and conditions. Additionally, there are other tests for irreducibility, such as the Berlekamp's criterion and the Perron's criterion, which can be used for polynomials that do not satisfy the Eisenstein criterion.

Similar threads

Replies
4
Views
2K
Replies
16
Views
3K
Replies
5
Views
1K
Replies
1
Views
1K
Replies
1
Views
2K
Replies
2
Views
20K
Back
Top