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By a decomposition of a polynomial $F(x)$ over a field $K$ we mean writing $F(x)$ as

$$ F(x)=G(H(x)) \quad(G(x), H(x) \in K[x]), $$

which is nontrivial if $\operatorname{deg} G(x)>1$ and $\operatorname{deg} H(x)>1$. The polynomial $F(x)$ is called decomposable over $K$ if it has at least one nontrivial decomposition over $K$; otherwise it is said to be indecomposable.

Motivated by my previous question, I want to know if the decision problem of "Given an arbitrary polynomial $F(x)\in K[x]$, is it decomposable over $K$ or not?" is decidable or undecidable.

I am interested in the above question because of the famous Bilu–Tichy Theorem.

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  • $\begingroup$ The title of this question, which was identical to that of your previous question, mentions the specific field $K = \mathbb Q$ and restricts to integer polynomials, but your body doesn't. Is that intentional? (Since the Unicode letters ℤ and ℚ that you used, I guess to evade the duplicate filter, looked distractingly different on my machine, I re-worded to avoid them. I hope this was all right; but, if you didn't mean to restrict to this case, then you can change the title to something more accurate.) $\endgroup$
    – LSpice
    Commented 2 days ago
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    $\begingroup$ @LSpice: Do you mean the linked previous question, or was there a deleted question I'm not seeing? It looks to me like the titles were different - the previous question was about reducibility, rather than decomposability. $\endgroup$ Commented 2 days ago
  • $\begingroup$ @user2357112, re, yes, you are right, my mistake. Then I'm definitely not sure why the Unicode was used! $\endgroup$
    – LSpice
    Commented 2 days ago
  • $\begingroup$ @LSpice I am sorry, it was a mistake from my side because of copy-paste. I will edit my question now. $\endgroup$ Commented 7 hours ago

1 Answer 1

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Yes, this problem is decidable. Kozen and Landau gave a polynomial-time algorithm that tests if a polynomial $F$ is decomposable and produces a nontrivial decomposition $F(x) = G(H(x))$ if one exists. Their algorithm requires that $\deg(G)$ has a multiplicative inverse in $K$.

Kozen, Dexter; Landau, Susan, Polynomial decomposition algorithms, J. Symb. Comput. 7, No. 5, 445-456 (1989); errata ibid. 10, No. 5, 529 (1990). ZBL0691.68030.

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  • $\begingroup$ What happens if $\deg(G)$ is not invertible in $K$? $\endgroup$
    – aorq
    Commented 5 hours ago

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