# Irreducibility Question (Rationals, Eisenstein!)

1. Dec 1, 2005

### mattmns

Hello. I have this question from the book:

Prove that the polynomial $$x^4 + 2x^2 + 2$$ is irreducible in Q[x]. (Q being the rational fancy Q)
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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 $$x^4 + 2x^2 + 2$$ 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: Dec 1, 2005
2. Dec 1, 2005

### shmoe

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. Dec 1, 2005

### mattmns

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 $$f(x) = x^n + a_{1}x^{n-1} + .. + a_{n}$$ be a polynomial with integer coefficients. Suppose that there is some prime p such that $$p|a_{1}, p|a_{2}, ..., p|a_{n}$$, but p^2 does not divide $$a_{n}$$ 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: Dec 1, 2005
4. Dec 2, 2005

### HallsofIvy

Staff Emeritus
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. Dec 2, 2005

### mattmns

You are absolutely correct, I missed that part! Thanks.