Check the proof of:$x^2+1$ is irreducible over $\Bbb Q$ 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? 
 A: Your approach can be made right.


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*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.
A: 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!
A: 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 $\mathbb{Q}$.
A: 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 $x^2\geq 0$.
A: 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 $p=2$.)
