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

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My statement of Eisenstein's criterion is the following:

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

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

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


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

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

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

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

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

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


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

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

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



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

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

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

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

DonAntonio


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

2+i 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.
 
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