# If a polynomial has a rational root is it automatically reducible?

This seems obvious, but I just can't crack it.

Let K be a field and $F(X) \in K[X]$ be a polynomial. Does $F(a)=0$ for some $a\in K$ imply that F(X) is reducible. Clearly, by the fundamental theorem of algebra, $F(X) = (X-a)G(X)$ for some $G(X) \in \bar{K}[X]$, but does it follow that actually $G(X) \in K[X]$?

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The answer to the question: NO. A polynomial of degree 1 may have a root but still be irreducible. So: state it more carefully. –  GEdgar Apr 5 '12 at 12:08
The title of the question does not really reflect what is being asked. –  lhf Apr 5 '12 at 12:25
@lhf Yes, it does. In this context "rational" is a common abbreviation for $K$-rational. It is also the better formulation to use in titles, where there is no other reference to $K$. –  Alex B. Apr 5 '12 at 12:28
It's obvious that $G(X) \in K(X)$ if you think to think about that. What can you do with that information? –  Hurkyl Apr 5 '12 at 21:25

Yes: if $a\in K$ then $G \in K[X]$ because the Euclidean division algorithm of $F$ by $X-a$ works entirely over $K$.

Indeed, you get $F(X)=(X-a)Q(X)$ with $Q\in K[X]$. But $(X-a)G(X)=F(X)=(X-a)Q(X)$ over $\bar K$ and you can cancel $X-a$ to conclude $G=Q \in K[X]$.

The fundamental theorem of algebra plays no role here. The Euclidean division algorithm works over any field (and over any commutative ring when the divisor is monic).

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Until you make rigorous the term "works entirely", this cannot be considered a valid proof. If you make it rigorous, you will see that it amounts to the uniqueness theorem that I mention. –  Bill Dubuque Apr 5 '12 at 13:47
@Bill, your uniqueness approach is certainly elegant but my point is that polynomial division is an algorithm that can be performed using the field operations and so never gets outside $K$. –  lhf Apr 5 '12 at 13:50
But the existence of one (algorithmically derived) quotient and remainder is not enough to deduce the uniqueness, There are also algorithms for factoring integers, but this does not imply that such factorizations are unique. –  Bill Dubuque Apr 5 '12 at 13:58
@Bill, I see what you mean now. I was answering the first part of the question, that $F$ has a $X-a$ factor over $K$. The last part of the question does mention $\bar K$ and I have addressed that in my edited answer. So, yes, we're saying the same thing, though in different ways. –  lhf Apr 5 '12 at 14:14
@Dustan Confusion seems to stem from different interpretations of the question. I read it as asking that if we somehow find a factorization $\rm\:f = (x-a)\:g\$ for $\rm\:g\in \hat K[x]\:$ over some extension of the coefficient ring $\rm\:\hat K\supset K,\:$ then does this necessarily yield a factorization over $\rm\:K,\:$ i.e. is $\rm\:g\in K[x]$? Perhaps you interpret it more simply as: must there exist $\rm\:g\in K[x],\:$ vs. $\rm\:g\in \hat K[x]\:$ $\Rightarrow$ $\rm\:g\in K[x].$ $\$ $\$ $\$ $\$ –  Bill Dubuque Apr 5 '12 at 20:41

Note $\$ There seems to be varying interpretations of the question. I read it as asking that if we somehow find a factorization $\rm\:f=(x−a)\:g\:$ for $\rm\:g\in \hat K[x]\:$ over some extension of the coefficient ring $\rm\:\hat K\supset K,\:$ then does this necessarily yield a factorization over $\rm K,\:$ i.e. is $\rm\:g\in K[x]$? Some readers interpreted the question more simply as: must there exist $\rm\:g\in K[x],\:$ vs. $\rm\:g\in \hat K[x]\: \Rightarrow\: g\in K[x].$

Hint $\$ It follows from the uniqueness of the quotient (and remainder) in the division algorithm (which is the same in $\rm\:K[x]\:$ and $\rm\:\bar K[x],\:$ using the polynomial degree as the Euclidean valuation). Thus since dividing $\rm\:f\:$ by $\rm\:x-a\:$ in $\rm\:\bar K[x]\:$ leaves remainder $0$, by uniqueness, the remainder must also be $0$ in $\rm K[x]\:,\:$ i.e. $\rm\:x-a\ |\ f\$ in $\rm \bar K[x]\:$ $\:\Rightarrow\:$ $\rm\:x-a\ |\ f\$ in $\rm K[x]\:.\:$

This is but one of many examples of the power of uniqueness theorems for proving equalities.

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