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Given $f(x),g(x) \in \mathbb{Z}[x]$, two irreducible polynomials, is there a polynomial $h(x) \in \mathbb{Z}[x]$ coprime to $f(x)$ such that $f(x) + g(x)h(x)$ is reducible over $\mathbb{Z}[x]$ with factors not congruent to $1 \mod g(x)$?

If so, what is the best way to find one such $h(x) \in \mathbb{Z}[x]$?

Let degree of $f(x),g(x)$ be bounded by $n$ (with degree of $g(x)$ strictly less than degree of $f(x)$) and maximum size of coefficients of $f(x),g(x)$ be $2^{s}$. Is there a $(ns)^{c}$ or $(n2^{s})^{c}$ algorithm to find such an $h(x)$ (if such a polynomial exists) where $c \in \mathbb{N}$ is fixed?

In general, is there a way to define the probability that $f(x) + g(x)h(x)$ is reducible into factors not congruent to $1 \mod g(x)$? It is likely this probability asymptotically goes to zero in terms of $n$ and $s$. However how fast does it go to zero? Does it go to zero asymptotically as $\frac{1}{(ns)^{c_{1}}}$ or $\frac{1}{(n2^{s})^{c_{1}}}$ for some positive constant $c_{1}$?

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I suppose you do not want $h$ to be a multiple of $f$, then? In general, can you prove the existence a (non-trivial) $h$ that does the job? –  René Jun 28 '13 at 22:54
    
Thankyou corrected. –  J.A Jun 28 '13 at 22:56
    
Good question! There should exist. I do not have a proof. May be I will change the question. –  J.A Jun 28 '13 at 23:01

1 Answer 1

Let $h(x)=f(x)(f(x)+g(x))+1$, obviously coprime to $f(x)$. Then $$ f(x)+g(x)h(x) = (f(x)+g(x))(1+f(x)g(x)). $$

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I think $h=1+fg+f$ if your displayed formula is to hold. –  Venkataramana Jun 29 '13 at 3:05
    
not exactly what I was thinking. However with polynomials there is always a sneaky way. –  J.A Jun 29 '13 at 7:27

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