Irreducibility Question (Rationals, Eisenstein!)

  • Thread starter mattmns
  • Start date
  • #1
1,090
6
Hello. I have this question from the book:

Prove that the polynomial [tex]x^4 + 2x^2 + 2[/tex] is irreducible in Q[x]. (Q being the rational fancy Q)
---

So I used Eisenstein Criterion (because we are dealing with the rationals) and said that the coefficents are: 1 (for x^4), 2(for x^2), 2 with the last 2 coefficients being the important ones. So I then said that lets use 2 as the prime, and then 2 | 2 (the middle coefficient), and 4 (2^2) does not divide 2 (the last coefficient), so [tex]x^4 + 2x^2 + 2[/tex] is not reducible in Q[x].

That seems quite acceptable to me, but the book has no examples, did I do the problem correctly? Danke!
 
Last edited:

Answers and Replies

  • #2
shmoe
Science Advisor
Homework Helper
1,992
1
All the coefficients are important, 2 divides the x^3 and x coefficients trivially. More importantly, there's a condition about the lead coefficient that you haven't mentioned...
 
  • #3
1,090
6
Thanks, I did not think about those. However, I am not sure what you are talking about with the leading coefficient (I am guessing you are talking about the 1 in front of x^4).

In my book the definition of The Eisenstein Criterion is:

Let [tex]f(x) = x^n + a_{1}x^{n-1} + .. + a_{n}[/tex] be a polynomial with integer coefficients. Suppose that there is some prime p such that [tex]p|a_{1}, p|a_{2}, ..., p|a_{n}[/tex], but p^2 does not divide [tex]a_{n}[/tex] Then f(x) is irreducible in Q[x].



Hmm... I just looked in another book, and on wikipedia, at the definition of The Eisenstein Criterion and they both mention a property about the first coefficient (1 for x^4); specifically: p cannot divide the first coefficient(1 for x^4). Which looks true too. Not sure why one of my books says that and the other does not. Thanks
 
Last edited:
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
966
Your book is assuming a "monic" polynomial- the "leading coefficient" (the coefficient of the highest power) is 1 and no prime divides that so it isn't necessary to mention it. Since factoring out a constant doesn affect irreducibility, the two forms are equivalent.
 
  • #5
1,090
6
HallsofIvy said:
Your book is assuming a "monic" polynomial- the "leading coefficient" (the coefficient of the highest power) is 1 and no prime divides that so it isn't necessary to mention it. Since factoring out a constant doesn affect irreducibility, the two forms are equivalent.
You are absolutely correct, I missed that part! Thanks.
 

Related Threads on Irreducibility Question (Rationals, Eisenstein!)

Replies
0
Views
810
Replies
2
Views
3K
  • Last Post
Replies
1
Views
568
Replies
4
Views
550
Replies
11
Views
1K
Replies
1
Views
2K
Replies
1
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
3
Views
957
  • Last Post
Replies
1
Views
2K
Top