Proving a polynomial over Q[x] is irreducible

[chaotixmonjuish]

Show that the polynomial x^2+1 is irreducible in Q[x].

Hint: If not, it must factor as (ax+b)(cx+d) with a,b,c,d in Q. Show that this is impossible.

So I got this far:

ac =1

ad+bc=0

bd=1

I'm not sure how to go further than this.

 

[Wretchosoft]

Hmm, I don't understand why they restrict to just Q, as it's fairly easy to show that it's irreducible over R as well.

Hint: Square.

 

[chaotixmonjuish]

Well the question restricts it to Q[x].

I asked my prof if just saying it factors to (x-i)(x+i)...however he said it needs to be general. Hence why I have the ac, ac+db, and bd. I'm not sure how to 'generalize' the proof.

 

[Wretchosoft]

Looking at the equation, the core issue is that squares can't be negative in Q, right? So, you have a set of mixed products. Try to mix them together in a way that generates an absurdity.

 

[chaotixmonjuish]

So does this work:

ad=-bc
a=-bc/d

sub into the first equation

-bc^2=1
c^2=-1/b^ This is impossible in Q

 

[Wretchosoft]

No, b could be negative. Try substituting in the other direction.

 

[chaotixmonjuish]

well I'm saying that c^2 can't equal a negative number in Q, it would imply an i somewhere

 

[Wretchosoft]

Yes, but if b is negative, then -1/b will be positive, which is not an issue.

 

[chaotixmonjuish]

hmm...so I'm not sure what you mean then.

 

[chaotixmonjuish]

By substitute in the other direction, did you mean:

a=1/c and b=1/d

d/c+c/d=0

This is impossible

 

[Wretchosoft]

That works, though you did it differently from what I had in mind. I'd like to see you complete the reasoning though.

Here's what I was hinting at: (ad+bc)^2=0

(ad)^2 + 2abcd + (bc)^2 = 0

(ad)^2 + (bc)^2 = -2

 

[chaotixmonjuish]

Well to play with my idea more:

d^2+c^2/cd=0

d^2+c^2=0

This is not possible...to be honest I'm not sure how to follow the logic of yours.

 

[Wretchosoft]

Well to play with my idea more:

d^2+c^2/cd=0

d^2+c^2=0

This is not possible...to be honest I'm not sure how to follow the logic of yours.

What if c and d are 0?

In moving from the second to the third line, I replaced acbd with 1*1.

 

[chaotixmonjuish]

But c and d can't be zero because ac=1 and bd=1

 

[Wretchosoft]

But c and d can't be zero because ac=1 and bd=1

Well then, it seems our work here is done.

 

[chaotixmonjuish]

Thanks!

 

[Office_Shredder]

You can basically show how it's just split into x-i and x+i.


If x² + 1 =a(x-b)(x-c) then b, c are roots of the equation, so in particular b² = -1. This is impossible, hence it can't factor

 

[Focus]

You can use Eisenstein and the fact f(x) is irreducible if and only if f(x+1) is irreducible.