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I have the following definition of irreducible polynomials, given by my professor:

If $p = f * g$ , where $f, g \in F[x]$ then $degree(f) = 0 \ \ or \ \ degree(g) = 0$

And then the following sentence: it is always possible to represent a polynomial as $p = f*g$, where $deg(f) = 0$ (like this: $p = a * (a^{-1}*p)$)

I am confused, need your help. If every polynomial can be represented as such product with $deg(f) = 0$, given the definition of irreducible polynomials above, doesn't it mean every polynomial is irreducible? (I know, this is false).

F.e., I have a polynomial f = (x+2)(x-5). Well, I can represent it as $a*(a^{-1}*(x+2)(x-5)$ and say that it's irreducible as $deg(a) = 0$.

Do you see what I mean, where my problem is? Would you please help me to understand, what my prof. wanted to tell us?

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    $\begingroup$ Irreducible means that the only way to write it as a product is that one of the factors is a unit. You can of course write everything as "unit times whatever", but irreducibles, you can write in only that way as a product. $\endgroup$ – Daniel Fischer Nov 8 '13 at 20:10
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    $\begingroup$ Try to work out some examples to get a handle on the definition. For example, try to prove that $x^3 - 1$ is not irreducible in $\mathbb{R}[x]$. However, you can try to show that $x^2 + 1$ is irreducible in $\mathbb{R}[x]$. Play around with these examples to make sure you see what part of the definition is being used. $\endgroup$ – tylerc0816 Nov 8 '13 at 20:17
  • $\begingroup$ @Daniel, that means we talk about writing them exactly as a product of two polynomials.. I guess I see now what I overlooked. $\endgroup$ – spike Nov 8 '13 at 20:21
  • $\begingroup$ @tylerco816, $x^3 - 1 = (x-1)(x^2+ab+b^2)$. So we can represent it as a product of exactly two polynomials and non of them is of degree 0. But x-1 must have a zero point, as well as $(x^2+ab+b^2)$. So in $\Bbb R$ they both have. Am I going in the right direction? Or I could say $x^3 = 1$, i.e. $x=1$ is a zero point for $x^3 -1$... And I don't know how to continue. $\endgroup$ – spike Nov 8 '13 at 20:26
  • $\begingroup$ Got it.. that means I can factor it and get what I've just written above. So we have a product of two polynomials and non of them is of degree zeor. :) $\endgroup$ – spike Nov 8 '13 at 20:32
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The confusion was resolved in comments:

  • Every polynomial can be factored in a trivial way, i.e., with some of factors being units.
  • An irreducible polynomial is one that can be factored only in a trivial way.
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