
#1
Oct910, 10:30 PM

P: 884

I have a question about the application of Eisenstein’s Criterion. I want to show that
g(x)=x^{4}+4x^{3}+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! 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) = x^{4} + 8x^{3} + 18 x^{2} + 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? 



#2
Oct910, 11:24 PM

P: 540

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 



#3
Oct1010, 02:31 AM

P: 134

g(x+2) works for Eisenstein.




#4
Oct1110, 11:32 PM

P: 884

Eisenstein's Irreducibility Criterion[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... Let g(x)=x^{4}+4x^{3}+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... 



#5
Oct1210, 12:30 AM

P: 540





#6
Oct1210, 01:23 AM

P: 884

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



#7
Oct1210, 04:36 AM

P: 540





#8
Oct1210, 07:45 PM

P: 884

[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, q_{1}=q_{2}=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
Oct1310, 04:37 AM

P: 134





#10
Oct1310, 07:04 PM

P: 540

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) 



#11
Oct1410, 04:32 AM

P: 884

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? 


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