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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.2012-04-05
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    The title of the question does not really reflect what is being asked.2012-04-05
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    @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 $K$.2012-04-05
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    It's obvious that $G(X) \in K(X)$ if you think to think about that. What can you do with that information?2012-04-05

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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.2012-04-05
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    @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 $K$.2012-04-05
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    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.2012-04-05
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    @Bill, sure, but where is uniqueness needed here if you don't appeal to $\bar K$?2012-04-05
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    I don't know what you mean by your last comment. Could you please expand your answer to a *complete rigorous* proof so that I can understand what you have in mind.2012-04-05
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    Uniqueness is not necessary at all; using the division algorithm write $F(X) = q(X) (X-a) + r(X)$, where $r(x)$ must have degree less than X-a, ie, it must be constant. But then $F(a) = r(a)$, so that constant must be 0.2012-04-05
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    @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.2012-04-05
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    Can you explain what the issue with my argument is, then?2012-04-05
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    @Dustan Comments above refer to the first version of the answer (= first sentence of current answer). That version, and your comment, do not yield a proof that if $\rm\:f = (x-a)\:g\:$ for $\rm\:g\in \bar K[x]\:$ then $\rm\:g\in K[x].\:$ This does *not* follow from the fact that applying the division algorithm in $\rm\:\bar K[x]\:$ is the same as applying it in $\rm\:K[x],\:$ so yields a quotient $\rm\:g \in K[x].$2012-04-05
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    Only if one knows that such quotients are *unique* can one rule out the existence of multiple solutions for $\rm\:g.\:$ In fact if we generalize from $\rm\: K\subset \bar K$ to an arbitrary ring (vs. domain) extension then the claim fails, i.e. we can have $\rm\:f = gh = \bar g h\:$ for $\rm\:g\ne \bar g\:$ (but not for *monic* $\rm\:h\:$ e.g. $\rm\ x-a,\:$ which are not zero-divisors). The uniqueness of the quotient $\rm\:g\:$ is equivalent to the divisor $\rm\:h\:$ (here $\rm\:x-a$) being a non-zero-divisor. This is not a consequence of the *existence* (and persistence) of the quotient.2012-04-05
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    @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. $\rm\:g\in \hat K[x]\:$ $\Rightarrow$ $\rm\:g\in K[x].$ $\ $ $\ $ $\ $ $\ $2012-04-05
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    That makes sense, Thanks.2012-04-06
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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.