Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$K,L$ are fields, $K\subseteq L$. $f,g \in K[x]$. Suppose that $f,g$ are relatively prime as elements of $K[x]$. Prove they remain relatively prime in $L[x]$.

I've tried everything I can think of. I feel like working with the contrapositive may be helpful but that's just a feeling.

share|cite|improve this question

Hint: Note that Bezout's identity holds for polynomial rings in one variable over a field, since such rings are principal ideal domains (PIDs):

$$f,g\in F[x]\text{ relatively prime }\iff \exists a,b\in F[x]\text{ such that }af+bg=1.$$

Use this both with $F=K$ and $F=L$.

share|cite|improve this answer
I still fail to see how I connect the two fields – Tim anderson Nov 18 '11 at 13:44
@Tim: If $f$ and $g$ are relatively prime as elements of $K[x]$, there are $a,b\in K[x]$ such that $af+bg=1$. But $K[x]\subseteq L[x]$, so $a,b,f,g\in L[x]$, so the fact that $af+bg=1$ tells us that... – Zev Chonoles Nov 18 '11 at 14:38
Rephrase @ZevChonoles's comment: If $f$ and $g$ are relatively in $K[x]$ $\Rightarrow$ $\exists a,b\in K[x]$ such that $af+bg=1\in K[x]$ $\Rightarrow$ $\exists a,b\in L[x]$ such that $af+bg=1\in L[x]$ $\Rightarrow$ $f$ and $g$ are relatively prime in $L[x]$. – bfhaha Sep 27 '15 at 13:47

Zev's answer is in some sense the canonical one, but here is another point of view, which is less elegant, but perhaps more intuitive.

We can embed $L$ into its algebraic closure $\overline{L}$; the algebraic closure of $K$ in $\overline{L}$ is then an algebraic closure of $K$.

Now $f$ and $g$ coprime in $K[x]$ means that they have distinct roots in $\overline{K}$. But these are also the roots of $f$ and $g$ in $\overline{L}$, and so $f$ and $g$ have distinct roots in $\overline{L}$. Thus $f$ and $g$ are coprime in $L[x]$ as well.

share|cite|improve this answer
Shouldn't this imply that being separable is same as being square-free? I am not sure why the following Wikepedia website says the otherwise: – Gil Jul 23 '15 at 6:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.