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Hypothesis:

$E$ is a field extension of $F$.

$p, q \in F[x]$ s.t. $p$ and $q$ are not relatively prime over $E$.

Goal:

Show that $p$ and $q$ are not relatively prime over $F$.

Attempt:

  1. Since $p$ and $q$ are not relatively prime over $E$, we have that $\gcd(p,q) \ne 1$ in $E[x]$.

  2. Then let $d(x)$ be the greatest common divisor of $p,q$ in $E[x]$ which from (1) satisfies $\deg(d(x)) \ge 1$.

And from here I'm not sure how to proceed. Since $E$ is just a field extension over $F$, we don't even know if there exist roots of $d(x)$ in $E$, much less $F$. If necessary, we can extend $E$ to some field $K$ s.t. $K$ possesses all of the roots of $d(x), p(x)$ and $q(x)$ -- but I'm not sure how this helps us.

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marked as duplicate by user26857, Davide Giraudo, egreg, Austin Mohr, Norbert Dec 29 '13 at 22:11

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  • $\begingroup$ I would suggest proving it indirectly. If $p,q$ are relatively prime over $F$, then they are also relatively prime over $E$. Straightforward since $F[x]$ is a PID. $\endgroup$ – Daniel Fischer Oct 20 '13 at 23:09
  • $\begingroup$ I am assuming OP meant p and q are NOT relatively prime over F. Statement as stated is clearly false (take p=q non unit in F for instance). $\endgroup$ – user24503 Oct 21 '13 at 2:17
  • $\begingroup$ Souparna: Yes -- that's what I meant (and sorry for that very problematic typo). I just edited the problem statement. Daniel: I thought about it a bit and can't see how $F[x]$ as a PID helps us here. Thanks for the hint though. $\endgroup$ – user1770201 Oct 21 '13 at 2:25
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“Not relatively prime over $E$ $\Rightarrow$ not relatively prime over $F$” is the same as “Relatively prime over $F$ $\Rightarrow$ relatively prime over $E$.”

So let’s suppose that $p,q\in F[x]$ and relatively prime over $F$. Then there are $A,B\in F[x]$ with $Ap+Bq=1$. But since $F[x]\subset E[x]$, you’re done.

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  1. Suppose for sake of contradiction that $p$ and $q$ were not relatively prime over $F$.

  2. Then $\gcd(p(x),q(x)) = d(x)$ s.t. $\deg(d(x)) \ge 1$ for $d(x) \in F[x]$.

  3. But then $d(x) \in F[x] \implies d(x) \in E[x]$ since $E$ is an extension of $F$.

  4. Then in $E$ we have that $\gcd(p(x),q(x)) = d(x)$ for $\deg(d(x)) \ge 1$, contradicting our assumption that $p$ and $q$ are not relatively prime over $E$.

  5. Then $p$ and $q$ must be relatively prime over $F$ as desired.

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  • $\begingroup$ But you’ve proved the converse of what was asked. $\endgroup$ – Lubin Oct 22 '13 at 0:11
  • $\begingroup$ Yes -- indeed. Strike this. $\endgroup$ – user1770201 Oct 22 '13 at 0:17

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