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May 10th, 2017, 06:17 PM   #1
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Why is $2x^2 + 4$ reducible over Z but not Q?

We can rewrite $2x^2 +4$ as $2(x^2 + 2)$ so wouldnt this polynomial be reducible over both $\mathbb{Z}$ and $\mathbb{Q}$?
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May 11th, 2017, 06:22 PM   #2
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Quote:
Originally Posted by Diehardwalnut View Post
We can rewrite $2x^2 +4$ as $2(x^2 + 2)$ so wouldnt this polynomial be reducible over both $\mathbb{Z}$ and $\mathbb{Q}$?
Actually your polynomial is not factorable over the reals. Thus factorization fails over $\mathbb{Z} , \mathbb{Q}$ and $\mathbb{R}$. Factoring out a constant (i.e. 2) doesn't count.

It is factorable over the complex numbers

$2x^2 + 4 = 2(x - i \sqrt{2})( x + i \sqrt{2}) $

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May 12th, 2017, 01:40 AM   #3
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Equation $\displaystyle \; \; x^2 + y^2 =0 \; \; \: \; (x-iy)(x+iy)=0$
$\displaystyle \begin{cases} x-iy=0 \\ x+iy=0 \end{cases}$
$\displaystyle x=y=0$
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