<p>Lets say we define a reducible multivariate <span class="math-container">$f \in \mathbb{F}[x_1,...,x_n]$</span> to be partionable by <span class="math-container">$y_1,...,y_r \in \mathbb{F}[x_1,...,x_n]$</span> iff
<span class="math-container">\begin{equation}
f(x_1,,,,.x_n) = f_1(y_1)\cdot f_2(y_2) \cdot \cdot \cdot f_r(y_r)
\end{equation}</span>
where <span class="math-container">$f_i(y_i) \in \mathbb{F}[y_i]$</span>.
Can we always partition a polynomial by algebraically independent elements in a non-trivial way (i.e. not partitioning f by itself)? Can you also link references to work done on this problem?</p>

CC BY-SA 4.0
2023-05-18T10:40:01.337Z
I may be missing something, but isn’t the answer obviously *no*, e.g. for irreducible polynomials, or for polynomials that are already univariate?
Emil Jeřábek
CC BY-SA 4.0
2023-05-18T10:54:20.827Z
The question isn't very well framed. You are right; for irreducible polynomials and univariate, we can't do anything better than trivial. The interesting case would be multivariate and reducible polynomials. We can try to partition by irreducible elements, but I don't think they would be algebraically independent.
Rishabh Kothary
CC BY-SA 4.0
2023-05-18T11:23:43.567Z
Since $\mathbb F[x_1,\dots,x_n]$ is a UFD, you can’t do better than what the prime factorization gives. E.g., $(x+y)(x+y+1)$ is multivariate and reducible, but its only nontrivial partition is to $x+y$ and $x+y+1$, which are not algebraically independent.
Emil Jeřábek
CC BY-SA 4.0
2023-05-18T13:59:51.663Z
But in the above example if we consider $x+y$ as an algebraic variable the polynomial reduces to a univariate. In general can we collect the different factors that can be expressed as univariates in some variable
Rishabh Kothary
CC BY-SA 4.0
2023-05-18T14:12:17.660Z
This makes no sense. *Every* polynomial $f$ reduces to a univariate polynomial in some variable (namely, take $f$ itself as an algebraic variable).
Emil Jeřábek
CC BY-SA 4.0
2023-05-18T14:17:02.853Z
You are right my question is not well defined. I am unable to clearly describe what I am truly getting at. When I have a clearer picture I'll add a new question or edit this question.The discussion was quite helpful though :)
Rishabh Kothary
CC BY-SA 4.0
2023-05-18T14:17:36.457Z
Look, I have no idea what you want to use the property in the question for, but there is no sensible way in which it would “always” hold. With a bit of algebra, you can easily characterize *which* polynomials $f$ have such a nontrivial partition: exactly those that have an irreducible factor $g$ such that $g$ is algebraically independent from $f$.
Emil Jeřábek
CC BY-SA 4.0
2023-05-18T14:19:29.610Z
Probably you are right thanks
Rishabh Kothary
CC BY-SA 4.0
2023-05-19T19:37:49.730Z
Unless I am misunderstanding something, the univariate case of your question seems similar to [polynomial decomposition](https://en.wikipedia.org/wiki/Polynomial_decomposition).
Robert Andrews
<polynomials><algebraic-complexity>
Can every reducible multivariate polynomial be partitioned into product of univariate polynomials of algebraically independent elements?
Rishabh Kothary
# Can every reducible multivariate polynomial be partitioned into product of univariate polynomials of algebraically independent elements?

(Asked by: username_0 on 2023-05-18)

Lets say we define a reducible multivariate $f \in \mathbb{F}[x\_1,...,x\_n]$ to be partionable by $y\_1,...,y\_r \in \mathbb{F}[x\_1,...,x\_n]$ iff
\begin{equation}
f(x\_1,,,,.x\_n) = f\_1(y\_1)\cdot f\_2(y\_2) \cdot \cdot \cdot f\_r(y\_r)
\end{equation}
where $f\_i(y\_i) \in \mathbb{F}[y\_i]$.
Can we always partition a polynomial by algebraically independent elements in a non-trivial way (i.e. not partitioning f by itself)? Can you also link references to work done on this problem?

**Tags:** polynomials, algebraic-complexity
> **Comments:**
> **username_1**: I may be missing something, but isn’t the answer obviously *no*, e.g. for irreducible polynomials, or for polynomials that are already univariate?  
> **username_0**: The question isn't very well framed. You are right; for irreducible polynomials and univariate, we can't do anything better than trivial. The interesting case would be multivariate and reducible polynomials. We can try to partition by irreducible elements, but I don't think they would be algebraically independent.  
> **username_1**: Since $\mathbb F[x_1,\dots,x_n]$ is a UFD, you can’t do better than what the prime factorization gives. E.g., $(x+y)(x+y+1)$ is multivariate and reducible, but its only nontrivial partition is to $x+y$ and $x+y+1$, which are not algebraically independent.  
> **username_0**: But in the above example if we consider $x+y$ as an algebraic variable the polynomial reduces to a univariate. In general can we collect the different factors that can be expressed as univariates in some variable  
> **username_1**: This makes no sense. *Every* polynomial $f$ reduces to a univariate polynomial in some variable (namely, take $f$ itself as an algebraic variable).  
> **username_0**: You are right my question is not well defined. I am unable to clearly describe what I am truly getting at. When I have a clearer picture I'll add a new question or edit this question.The discussion was quite helpful though :)  
> **username_1**: Look, I have no idea what you want to use the property in the question for, but there is no sensible way in which it would “always” hold. With a bit of algebra, you can easily characterize *which* polynomials $f$ have such a nontrivial partition: exactly those that have an irreducible factor $g$ such that $g$ is algebraically independent from $f$.  
> **username_0**: Probably you are right thanks  
> **username_2**: Unless I am misunderstanding something, the univariate case of your question seems similar to [polynomial decomposition](https://en.wikipedia.org/wiki/Polynomial_decomposition).  

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