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# Understanding Gauss' Lemma and Irreducible Polynomials in Fields watch

1. I have a question I am trying to find the answer to:

Factorise the following polynomials into irreducibles in .

(Hint: use the Gauss Lemma).

So I have answered the question and it is correct, but it seems so arduous a task that I am sure there must be a better way. Also I do not understand where or why Gauss' Lemma has anything to do with it. (This may be a lack of understanding of Gauss' Lemma itself).

So here's what I did:

is a root with

This implies tat it must be split into two quadratics as there are no roots., and this is where it may get a bit hazy.

Then am unsure how to go about finding a, b, c and d as this will take so much toil especially when working in so I think this is where Gauss' Lemma kicks in, but I don't know and I don't know the effect it has on what I am doing. Therefore in my ignorance I continued to try and calculate a, b, c and d.

I said:

Try

and I got lucky I think that I found this on the first go. Giving my answer:

Right my issue is although I have an answer I assumed b and d would be integers. I also don't know where I used Gauss' Lemma if at all. Another issue is that I lucked out that I chose b = 1, d = 2 first, I seem to have been very lucky in this question, and I do not want to throw away easy marks in the exam because I do not understand what Gauss' Lemma is. Can anyone tell me how to work out this question more efficiently or maybe (if I have used it or not) where Gauss' lemma slots into play.

Thank you for reading this even if you do not have an answer I realise it is a long post. All help will be rewarded
2. You used Gauss' lemma when you assumed b and d are integers, and also that the leading coefficients of the two quadratics that you have factorised the polynomial into are 2 and 1. Gauss' lemma tells us that an integer polynomial is irreducible over the rationals iff it is irreducible over the integers. This is essentially stating that we can factorise an integer polynomial into two polynomials (of degree at least one) with rational coefficients iff we can factorise it into two polynomials (of degree at least one) with integer coefficients. Thus, we may assume b and d to be integers and that the leading coefficients of the quadratics are 2 and 1.

The condition easiest to work with, as you found, was that bd = 2. So then it's just choosing which choice of b and d you want. By this I mean it's essentially trial and improvement/error. Pick one pair, say (b,d) = (2,1). See whether this 'works': if it does, then we're done, if not then let's try (b,d) = (-1,-2). And rinse and repeat. You were lucky that you chose (b,d) = (1,2) first, but if you had picked a 'wrong' pair for (b,d), then it wouldn't take you long (say a couple of minutes) to realise that this pair wouldn't work. And there are only four possible pairs for (b,d), so it's not too much to be asking you to find the one which works.

Other than guessing a pair for (b,d) you could go about trying to solve the four equations simultaneously just by directly making substitutions like a = -2-2c into the other equations and such, and solving for a, b, c and d like that. But things like that are often messy...

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Updated: March 25, 2011
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