Proving a polynomial over Q[x] is irreducible

  • #1
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.
 

Answers and Replies

  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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
 
  • #6
No, b could be negative. Try substituting in the other direction.
 
  • #7
well i'm saying that c^2 can't equal a negative number in Q, it would imply an i somewhere
 
  • #8
Yes, but if b is negative, then -1/b will be positive, which is not an issue.
 
  • #9
hmm...so I'm not sure what you mean then.
 
  • #10
By substitute in the other direction, did you mean:

a=1/c and b=1/d

d/c+c/d=0

This is impossible
 
  • #11
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
 
  • #12
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.
 
  • #13
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.
 
  • #14
But c and d can't be zero because ac=1 and bd=1
 
  • #17
Office_Shredder
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Science Advisor
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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
 
  • #18
284
3
You can use Eisenstein and the fact f(x) is irreducible if and only if f(x+1) is irreducible.
 

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