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In Charles Pinter's Abstract Algebra, he presents the factorization theorem for polynomials. That is,

Every polynomial $a(x)$ of positive degree in $F[x]$ can be written as a product \begin{equation*} a( x) =kp_{1}( x) p_{2}( x) \cdots p_{r}( x) \end{equation*} where $k$ is a constant and $p_1(x),...,p_r(x)$ are monic irreducible polynomials of $F[x]$.

But I am confused by the next paragraph,

If this were not true, we could choose a polynomial $a(x)$ of lowest degree among those which cannot be factored into irreducibles. Then $a(x)$ is reducible, so $a(x)=b(x)c(x)$ where $b(x)$ and $c(x)$ have lower degree than $a(x)$. But this means that $b(x)$ and $c(x)$ can be factored into irreducibles, and therefore $a(x)$ can also.

My issue is if we choose a polynomial $a(x)$ among those which cannot be factored into irreducibles then how is $a(x)$ reducible? We just said choose one that isn't reducible. Moreover, I don't understand how we immediately know that $b(x)$ and $c(x)$ are reducible as well. All we said was that they were of lower degree, not that they were reducible.

This leads to the more general question, how can every polynomial of positive degree be factored into irreducibles? For example, take any one of these irreducible factors $p_r(x)$, by definition, $p_r(x)$ cannot be written as a product of irreducibles since this would mean that it is reducible and thus couldn't be an irreducible factor of $a(x)$. It seems to me that we are saying that even irreducible polynomials can be written as a product of irreducibles. I can't seem to locate where my misunderstanding is.

jinks908
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  • I'm having some trouble with the theorem as stated. If $F$ is $\Bbb{Z}$, $x^2 - 2$ is irreducible and surely can't be written as a product of monic polynomials in $\Bbb{Z}[x]$. So the theorem would not hold, unless I'm missing some context. The same happens in $\Bbb{R}[x]$ with $x^2 + 2$. Is $F$ an algebraically closed field? – Niki Di Giano Apr 24 '22 at 01:06
  • @NikiDiGiano Sorry, I should have specified. Pinter uses $F$ to specifically represent a field. So in this case, $F \neq \Bbb{Z}$ – jinks908 Apr 24 '22 at 02:28
  • Here, an irreducible is considered to be a "product" of irreducibles (a product of elements from a set containing only one element). Such singleton (and empty) products often prove handing for induction & recursion. This is explained at length in the analogous proof for integers here in the linked dupe. – Bill Dubuque Apr 24 '22 at 10:21
  • Note also that the proof is clearer when presented in non-contradiction form as in the linked dupe. – Bill Dubuque Apr 24 '22 at 10:27
  • ^^^ typo: "handing" -> "handy" – Bill Dubuque Apr 24 '22 at 10:38
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    @BillDubuque Yes, that proof is much more clear and elaborate. Thanks! – jinks908 Apr 24 '22 at 12:24

1 Answers1

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The point you need to understand is that factoring a polynomial into a product of a single term is allowed. Indeed, this is the only option for irreducible monic polynomials.

In this way, if $a(x)$ were irreducible, then writing $a(x)$ would be a product of irreducible factors (which is assumed not to exist), so $a(x)$ must be reducible. As for $b(x)$ and $c(x)$, they are not necessarily reducible, and could just be a product each of one term.

Theo Bendit
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  • But this sounds like every polynomial is reducible then. What constitutes an irreducible polynomial if we can always find some product to represent it? – jinks908 Apr 24 '22 at 00:39
  • An irreducible polynomial cannot be written as a product of two polynomials of lesser degree. This includes the $3$ or more case as well. All polynomials, reducible or not, can be written trivially as a product of one polynomial. – Theo Bendit Apr 24 '22 at 03:26