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Prove: $x^2+1$ is irreducible over $\mathbb{Q}$

Proof: Since $x^2+1=(x-i)(x+i)$, so $x^2+1$ is irreducible over $\mathbb{Q}$.

Is it right?

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You have to argue that $\mathbb Q[X]$ is a unique factorization domain, too. –  zarathustra Aug 8 '13 at 12:15
    
You need more argument here. After all, $x^2+1=(x-k)(x+k)$ in the quaternions, does that make it irreducible over $\mathbb{C}$? –  MartianInvader Aug 8 '13 at 16:41

5 Answers 5

You are reducing it outside of $\mathbb{Q}$. How about

$$x^2 + 1 = (x-a)(x-b) = x^2 - (a+b)x + ab, \quad a,b \in \mathbb{Q}?$$

Then, you have $a = -b$ and $ab = 1$, so $0 \le a^2 = -1$, which doesn't have a solution in $\mathbb{Q}$.

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I must upvote; I was typing the almost exact copy of this :) –  Marra Aug 8 '13 at 12:16
    
Thank you. It happens quite often to me as well. :-) –  Vedran Šego Aug 8 '13 at 12:17
    
But why make it so complicated? The polynomial is of degree $2$, so you just need to show it has no roots. –  Tobias Kildetoft Aug 8 '13 at 12:18
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Of course, we may notice that $x^2 \ge 0$, so $x^2 + 1 > 0$, but that won't help the OP for something like $x^2 - 2 = 0$, so I went for a bit more general approach, with no "tricks". –  Vedran Šego Aug 8 '13 at 12:20
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How can the above be more general, when it still only deals with degree $2$, where "no roots" is equivalent to "irreducible"? –  Tobias Kildetoft Aug 8 '13 at 12:28

You could also show that, should it be reductible, is has roots; therefore there must exist an $x\in\mathbb{Q}$ such that $x^2+1=0$. But that means that $x²=-1$, which is impossible since $x\in\mathbb{Q}$ implies that $x^2\geq 0$.

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Your approach can be made right.

  • If $x^2 + 1$ was reducible over $\mathbb{Q}$, then there are monic polynomials $f,g$ over $\mathbb{Q}$ such that $fg = x^2 + 1$, and neither $f$ nor $g$ are constants
  • If $x^2 + 1$ was reducible over $\mathbb{Q}$, then there are monic, linear polynomials $f,g$ over $\mathbb{Q}$ such that $fg = x^2 + 1$
  • If $x^2 + 1$ was reducible over $\mathbb{Q}$, then there are monic, linear polynomials $f,g$ over $\mathbb{C}$ such that $fg = x^2 + 1$ and $f,g$ both have coefficients in $\mathbb{Q}$

However, we know there is only one such factorization over $\mathbb{C}$, so we can plug in the only possible factorization (up to swapping factors) into this statement:

  • If $x^2 + 1$ was reducible over $\mathbb{Q}$, then $x+i$ and $x-i$ both have coefficients in $\mathbb{Q}$.

This last statement is false, so the original must be as well.

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Hi, the first three items' descriptions are the same? –  forlorn Aug 8 '13 at 12:55
    
@mathdummy: Not quite! They are each slightly different statements, each following from the previous one. I find when i'm confused, it helps to be exceedingly clear about what facts I'm trying to invoke to prove things, and why those facts should be true given previously known information. Of course, one could be more compact if you just wrote the "then ... " portion of each bullet; I was feeling particularly verbose when I wrote my post. (and sometimes it helps to make each individual statement self contained, rather than relying on previous ones, when doing proof by contradiction or cases) –  Hurkyl Aug 8 '13 at 12:56
    
I think our answers are equivalent but you did write your answer a few minutes before me ... Anyway, your answer is nicely explained in detail so +1! (And, in general, your contributions to this site are excellent, needless to say! It's great that we have a contributor of your calibre in the area of abstract algebra.) –  Amitesh Datta Aug 8 '13 at 22:28

The polynomial $f(x)$ is irreducible if and only if $f(x+1)$ is irreducible. But in your case, $$f(x+1) = (x+1)^2 +1 = x^2 + 2x + 2$$ is irreducible by Eisenstein's criterion (with $p=2$.)

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+1 It's great that we have so many alternative answers to this question! (Also, it's important to keep in mind that sometimes, Eisenstein's criterion might not be applicable to the original $f(x)$ but is applicable to $f(x+1)$; an excellent trick to remember when proving certain polynomials are irreducible.) Edit: Looks like my daily vote limit has been transgressed. I will upvote in one hour though! –  Amitesh Datta Aug 8 '13 at 22:56

The answers provided by Gustavo and Vedran are great! In addition, let me indicate how one can fix your argument: you've factored $x^2+1\in \mathbb{C}[x]$ via the inclusion $\mathbb{Q}[x]\subseteq \mathbb{C}[x]$ of rings. If $x^2+1$ were reducible in $\mathbb{Q}[x]$, then it'd be reducible in $\mathbb{C}[x]$. By virtue of the fact that $\mathbb{C}[x]$ is a UFD, any reduction of $x^2+1\in \mathbb{C}[x]$ into irreducibles must be the reduction $(x-i)(x+i)$ (up to multiplication by scalars and rearrangement). However, this is impossible as $i\not\in \mathbb{Q}$. (Check to ensure you understand the details; the key point is that $\mathbb{C}[x]$ is a UFD!)

Of course, this solution is indirect and doesn't elucidate the key point. However, I wanted to mention this because it's based on the original argument you had in mind. (See Hurkyl's great solution for an explanation along similar lines!)

I hope this helps!

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I believe you mean $x^2+1$ here everywhere you use $x^2-1$; the latter polynomial is eminently reducible in $\mathbb{Q}[x]$. :-) –  Steven Stadnicki Aug 8 '13 at 22:48
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Dear @Steven, thanks so much for this correction! Yes, in light of your comment I was thinking, in fact, that $x^2-1$ is reducible over any ring! That shows what difference a sign can make in mathematics. :-) –  Amitesh Datta Aug 8 '13 at 22:52

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