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Let $C$ be the field of complex numbers. Polynomial $f \in K[x,y]$ is called closed or non-decomposible if $C[f]$ is algebraically closed (definition for $f \in K[x1,...,xn]$ is the same).

Theorem. Next conditions are equivalent: (1) $f$ is closed;

(2) there does not exist any $F \in K[t]$ such that $f = F(h)$ for some $h \in K[x,y]$;

(3) $f + a$ is irreducible for all except finitely many $a \in C$.

(4) there exist $a \in C$ such that $f + a$ is irreducible. (end of the theorem)

It is clear, if $f$ is irreducible then $f$ is closed.

It is easy to see that if $f$ and $g$ are irreducible, then $fg$ is closed.

Question: Assume $f$ and $g$ are closed. Is it true that $fg$ is closed?

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You don't mean $C[f]$ is algebraically closed, just algebraically closed in $K[x,y]$, I hope. Similarly, (2) above, I hope you mean $\deg F>1$. Thirdly, it is not true that if $f,g$ are irreducible, then $fg$ is closed, since one can take $f=g=x$, but $x^2=fg$ is not closed. –  Mohan Jan 25 '13 at 17:16
    
If we take the definition of closed to be (2) with $\deg F > 1$, then we get counterexamples whenever $f$ is closed and $g=f$, since then $fg=F(f)$ with $F=t^2$. –  Michael Zieve Jan 25 '13 at 19:25
    
I am sorry, I forgot to write that $f$ and $g$ are algebraically independent! But anyway, it is not true (and easy), an example is: $f=x^{2} y$, $g = y (x+y)^{2}$ and then $fg = (xy(x+y))^{2}$. –  Andriy Regeta Jan 28 '13 at 12:03
    
Mohan, yes, I mean $C[f]$ is algebraically closed in $C[f,g]$ (I mean $C=K$) and in $(2)$, of source (sorry, I did not write it), I mean $deg F > 1$. –  Andriy Regeta Jan 28 '13 at 12:09

1 Answer 1

up vote 0 down vote accepted

As I mentioned above in the comment, the answer to my question (very easy question) is negative!

P.S. Of course I meant that $f$ and $g$ are algebraically independent!

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