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A polynomial $f\in k[x]$ is irreducible if :

1. $\deg(f)\geq 1$

2. If $f=gh$ then $\deg(g)=0$ or $\deg(h)=0$

First of all, this definition confuses me. For example, Let $f=2x^2+2$ and the field be $\mathbb{R}$. By this definition f is irreducible because $f=2(x^2+1)$ and $\deg(2)=0$. Is this correct? I would have thought $f$ is reducible since it can be divided by two..

Anyway, on with the proof:

If $f$ is reducible then $f=gh$ where $g,h\in k[x]$.

There are two cases, either $\deg(g)=\deg(h)=1$ or $\deg(g)=2 \text{ and } \deg(h)=0$. If the first case is true then this implies $f$ has roots. If the second case is true then either $\deg(g)=0 \text{ or } \deg(h)=0$ which would imply that f is irreducible according to the above definition?

Am I missing something or should the second criterion of irreducibility be:

If$~~f=gh \text{ then }\deg(g)=-\infty \text{ or } \deg(h)=-\infty$

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    $\begingroup$ We are working over a field $k$. The polynomial $2(x^2+1)$ is irreducible over the reals. Reducible means splits in a non-trivial way. After all, we could also say that $x^2+1=\frac{1}{42}(42x^2+42)$. $\endgroup$ – André Nicolas Nov 25 '15 at 21:22
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It is much simpler to consider Euclidean division for polynomials: for any $\alpha \in k$, divide $f$ by $x-\alpha$: $$f(x)=q(x)(x-\alpha) +r$$ Furthermore, $r=f(\alpha)$, hence

$\alpha$ is a root of $f$ if and only if $f(x)$ is divisible by $x-\alpha$.

Thus $f(x) $ has a root if and only if $f(x)$ is divisible by some $x-\alpha$, which implies $f$ is reducible.

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The second part of the definition has an implied universal quantor: "For all polynomials g and h such that f=g.h, at least one of g or h has degree 0 (i.e., is constant)"

This is really just a definition. A polynomial is irreducible if and only if it cannot be divided by any nonconstant polynomial. Division by a constant such as 2 does not count.

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  • $\begingroup$ Ok thanks for clearling this up! So my second case cannot happen because then this contradicts the assumption that f is reducible, thus only the first case remains. $\endgroup$ – NormalsNotFar Nov 25 '15 at 21:30

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