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I would like to verify the validity of the following line of thought:

Let $K \subset E$ be a field extension. Let $K[x]$ be the polynomial ring over $x$ and denote $K(x)$ its field of quotients (the rational functions over $K$).

Now assume $f(x)=g(x)h(x)$ with $f(x),g(x) \in K[x]$ and $h(x) \in E[x]$ and all $f$, $g$, $h$ are different than zero. Then $\frac{f(x)}{g(x)} = h(x)$ in $E(x)$. So $h(x) \in K(x)$. But $h(x)$ is a polynomial, so $h(x) \in K[x]$.

Is there any other simpler argument than that?


Let me be more specific. Let $x^p - a \in K[x]$, with $p$ being prime and $char(K)=p$. Let $b$ be the unique root of $x^p - a$ in its splitting field $K(b)$. Then $x^p - a=(x-b)^p \in K(b)[x].$ Let $m_b(x) \in K[x]$ be the minimal polynomial of $b$ over $K$. Then $m_b(x)$ divides $(x-b)^p$ and so $m_b(x)=(x-b)^l$ and $l$ is the smallest positive integer for which $(x-b)^l \in K[x]$. Let $p = lq+r$ where $0 \ge <l$. Then $x^p-a=(m_b(x))^q (x-b)^r$. Can i conclude that $(x-b)^r \in K[x]$ and so $r=0$ by the minimality of $l$?


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If you have shown elsewhere that $K(x)\cap E[x]=K[x]$ (you can form the intersection inside $E(x)$), then this is ok. – Jyrki Lahtonen Dec 5 '11 at 21:45
i see, that's a subtle point – Manos Dec 5 '11 at 21:50
@Jyrki: i added a specific instance of my question. What do you think in that case? – Manos Dec 6 '11 at 0:40
up vote 4 down vote accepted

HINT $\ $ It follows from the uniqueness of the quotient (and remainder) in the division algorithm (which is the same in $\rm\:K[x]\:$ and $\rm\:E[x]\:,\:$ using the polynomial degree as the Euclidean valuation). Thus since dividing $\rm\:f\:$ by $\rm\:g\:$ in $\rm\:E[x]\:$ leaves remainder $0$, by uniqueness, the remainder must also be $0$ in $\rm\:K[x]\:,\:$ i.e. $\rm\:g\ |\ f\ $ in $\rm\:E[x]\:$ $\:\Rightarrow\:$ $\rm\:g\ |\ 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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This is very good. Thanks a lot. – Manos Dec 6 '11 at 2:11

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