Hey there! Sign in to join this conversationNew here? Join for free
 You are Here: Home >< Maths

# Minimal polynomial of sqrt2 + sqrt3 Watch

1. Can someone please show how to prove it is irreducible.

2. If it isn't irreducible, it must contain a linear or quadratic factor.
3. (Original post by constellarknight)
If it isn't irreducible, it must contain a linear or quadratic factor.
I know but how do you prove that it doesn't?

I know that we can simply find the roots of the polynomial and say that they are not in Q, because this is a quartic.
4. Let the polynomial equal (a+bx+x^2)(c+dx+x^2) and deduce a contradiction. But actually, there's a much faster way: 2 divides -10, but does not divide 1, and 2^2 does not divide 1, so by Eisenstein's criterion, the polynomial is immediately irreducible over the rational numbers.
5. (Original post by constellarknight)
Let the polynomial equal (a+bx+x^2)(c+dx+x^2) and deduce a contradiction. But actually, there's a much faster way: 2 divides -10, but does not divide 1, and 2^2 does not divide 1, so by Eisenstein's criterion, the polynomial is immediately irreducible over the rational numbers.
Please can we not use any criterion because we have not come across that so we it is useless to us at the moment...
Is this OK?

6. Yes, that's fine. Eisenstein's is faster though.
From Wikipedia:
To prove the validity of the criterion, suppose Q satisfies the criterion for the prime number p, but that it is nevertheless reducible in Q[x], from which we wish to obtain a contradiction. From Gauss' lemma it follows that Q is reducible in Z[x] as well, and in fact can be written as the product Q = GH of two non-constant polynomials G, H (in case Q is not primitive, one applies the lemma to the primitive polynomial Q/c (where the integer c is the content of Q) to obtain a decomposition for it, and multiplies c into one of the factors to obtain a decomposition for Q). Now reduce Q = GH modulo p to obtain a decomposition in (Z/pZ)[x]. But by hypothesis this reduction for Q leaves its leading term, of the form axn for a non-zero constant aZ/pZ, as the only nonzero term. But then necessarily the reductions modulo p of G and H also make all non-leading terms vanish (and cannot make their leading terms vanish), since no other decompositions of axn are possible in (Z/pZ)[x], which is a unique factorization domain. In particular the constant terms of G and H vanish in the reduction, so they are divisible by p, but then the constant term of Q, which is their product, is divisible by p2, contrary to the hypothesis, and one has a contradiction.
7. (Original post by constellarknight)
Yes, that's fine. Eisenstein's is faster though.
From Wikipedia:
To prove the validity of the criterion, suppose Q satisfies the criterion for the prime number p, but that it is nevertheless reducible in Q[x], from which we wish to obtain a contradiction. From Gauss' lemma it follows that Q is reducible in Z[x] as well, and in fact can be written as the product Q = GH of two non-constant polynomials G, H (in case Q is not primitive, one applies the lemma to the primitive polynomial Q/c (where the integer c is the content of Q) to obtain a decomposition for it, and multiplies c into one of the factors to obtain a decomposition for Q). Now reduce Q = GH modulo p to obtain a decomposition in (Z/pZ)[x]. But by hypothesis this reduction for Q leaves its leading term, of the form axn for a non-zero constant aZ/pZ, as the only nonzero term. But then necessarily the reductions modulo p of G and H also make all non-leading terms vanish (and cannot make their leading terms vanish), since no other decompositions of axn are possible in (Z/pZ)[x], which is a unique factorization domain. In particular the constant terms of G and H vanish in the reduction, so they are divisible by p, but then the constant term of Q, which is their product, is divisible by p2, contrary to the hypothesis, and one has a contradiction.
I see. Hopefully we will come to this in lectures soon.

Can I ask, the factorisation I got, I got it off wolfram alpha, how would you come to that factorisation?
When i solve it for x^2, I get x^2 = 5 +/- i
8. Then set (a+bi)^2=5+i (or 5-i) and solve for a and b.
9. (Original post by cooldudeman)
Please can we not use any criterion because we have not come across that so we it is useless to us at the moment...
Have you come across Gauss's lemma yet?
10. (Original post by Gregorius)
Have you come across Gauss's lemma yet?
Nope I dont think we will. I don't know why but our lecturer seems to want us to do everything from scratch, well at the moment anyway...

Posted from TSR Mobile

Reply
Submit reply
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 21, 2016
Today on TSR

### Best unis for graduate salaries

Half of the top 10 aren't even RG...

### My alarm went off in an exam!

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.