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The question essentially pertains to how Gauss's lemma relates to polynomials with coefficients in a polynomial ring over the integers.

I have a polynomial $f(y) = \sum_{k=0}^D c_k y^k$ where $\forall k, c_k \in \mathbb{Z}[x_0,x_1,\dots, x_d]$ the coefficients are polynomials of other variables $x_i$ with integer coefficients. That is $c_k = \sum_i b_i \left(\prod_j x_j^{p_{i,j}}\right)$ where $b_i, p_{i,j}\in \mathbb{Z}$ where $b_i, p_{i,j}\in \mathbb{Z}$.

In my case I'm actually interested in the case where $D=3$ and $f$ is a cubic. Gauss's lemma states that when the coefficients $c_k$ are 'integers' then the rational root test should give us co-prime $p,q$ where $q|c_3$ and $p|c_0$ such that there is a root $qy-p$.

Additionally I know that $f$ is reducible.

The question is what does it mean in this case for the coefficients $c_k$ to be integers?

Intuitively if the elements $x_k$ are integers in $\mathbb{Z}[x_0,x_1,\dots, x_d]$ then I would expect the set of integers to be closed under addition and multiplication such that each $c_k$ should be an integer in the field
$\mathbb{F} = \mathbb{Z}[x_1,x_2,\dots, x_d]$. Then since $f\in \mathbb{F}[y]$ I would think that there should be a $p,q\in \mathbb{F}$ such that $(qy-p)| f$.

My apologies if the above is ridiculous, I don't have much experience working with these types of structures.

I am trying to find the factor $qy-p$ for a particular polynomial. Sympy is able to find an analytic solution for the roots of $f$ but I am surprised that there are radicals in the expression for the real root. I suppose this could be related to how the roots are found by the sympy algorithm though.

User
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    I'm not sure which version of Gauss's Lemma you are thinking of, but Gauss's Lemma does not say that "the rational root test gives" etc., in any version I am familiar with. Gauss's Lemma can be used to show that $\mathbb{Z}[x_0,\ldots,x_d]$ is a Unique Factorization Domain. The Rational Root Theorem holds for any UFD: if $p,q\in\mathbb{Z}[x_0,\ldots,x_d]$ are relatively prime, and $\frac{p}{q}$ is a root of $f$, then $p\mid c_0$ and $q\mid c_D$ in $\mathbb{Z}[x_0,\ldots,x_d]$. Gauss's Lemma works in any UFD. Note the Rational Root Test does not "give" you a root. It restricts possible roots. – Arturo Magidin Oct 11 '23 at 18:13
  • Sorry, my language is not very precise. You're saying that any rational root must take this form but that a root might not necessarily be rational? I also know that $f$ must be reducible - I had understood this to mean for the case of the cubic that this means there must be a linear factor with rational root. Is this not true? – User Oct 11 '23 at 18:23
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    I'm saying that any root of the form $\frac{p}{q}$ with $p,q\in\mathbb{Z}[x_0,\ldots,x_D]$ and such that $p$ and $q$ have no common non-unit factors in $\mathbb{Z}[x_0,\ldots,x_D]$ must satisfy that $p\mid c_0$ and $q\mid c_D$. "Fraction" is not the same thing as "rational number". Reducibility does not guarantee a root that lies in $\mathbb{Q}$, it only guarantees a root that lies in the fraction field of $\mathbb{Z}[x_0,\ldots,x_D]$. E.g., the polynomial $(y-x)^3$ is clearly reducible in $(\mathbb{Z}[x])[y]$, but has no roots in $\mathbb{Q}$. – Arturo Magidin Oct 11 '23 at 18:24
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    The linear factor $pY+q$ has $p,q\in\mathbb{Z}[x_0,\ldots,x_D]$, which gives $\frac{q}{p}$ as a root, but this need not be an element of $\mathbb{Q}$. All you know is that it is an element of $\mathbb{Z}(x_0,\ldots,x_D)$, the rational functions in $x_0,\ldots,x_D$ with coefficients in $\mathbb{Z}$ (whch is the same as $\mathbb{Q}(x_0,\ldots,x_D)$). – Arturo Magidin Oct 11 '23 at 18:28
  • Does reducibility in the case that $D=3$ guarantee the existence of a linear factor $pY+q$ with $p,q\in \mathbb{Z}[x_0, \dots, x_d]$? Or is the statement weaker if there exists $pY+q$ with $p,q\in \mathbb{Z}[x_0, \dots, x_d]$ which is a factor then $p|c_0$ and $q|c_D$? – User Oct 11 '23 at 18:33
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    Reducibility over $\mathbb{Z}[x_0,\ldots,x_D]$, yes. That means you can write it as a product of polynomials in $y$, and at least one of them has to be of degree $1$ in $y$. – Arturo Magidin Oct 11 '23 at 18:34
  • Thank you Professor. To summarize: Gauss's Lemma gives us that $\mathbb{Z}[x_0,\dots, x_d]$ is a Unique Factorization Domain (UFD). The Rational Root Theorem states that for any UFD if $p,q \in mathbb{Z}[x_0,\dots, x_d]$ are relatively prime and $\frac{p}{q}$ is a root of $f(y)$ then $p|c_0$ and $q|c_D$ in $mathbb{Z}[x_0,\dots, x_d]$. Finally assuming $f(y)$ is reducible of degree 3, then there must be a linear factor $p\cdot y +q$ where $p,q \in mathbb{Z}[x_0,\dots, x_d]$. If you'd like to submit an answer I can accept it, otherwise I can do so myself and note your contribution. – User Oct 11 '23 at 18:53
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    That is correct, once you clarify tht "reducible" means "reducible over $\mathbb{Z}[x_0,\ldots,x_d]$." – Arturo Magidin Oct 11 '23 at 18:55
  • The Rational Root Test works over any UFD (or GCD domain), e.g. see here in the linked dupe. In particular it works over the UFD $,\Bbb Z[x_0,\ldots, x_d]\ \ $ – Bill Dubuque Oct 11 '23 at 19:28

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The following is thanks to Arturo Magidin:

Gauss's Lemma gives us that $\mathbb{Z}[x_0, \dots,x_d]$ is a Unique Factorization Domain (UFD). The Rational Root Theorem states that for any UFD if $p,q \in \mathbb{Z}[x_0, \dots, x_d]$ are relatively prime and $\frac{p}{q}$ is a root of $f(y)$ then $p|c_0$ and $q|c_D$ in $\mathbb{Z}[x_0,\dots, x_d]$. Finally, assuming $f(y)$ is reducible over $\mathbb{Z}[x_0,\dots, x_d]$ and degree 3, then there must be a linear factor $p\cdot y +q$ where $p,q\in \mathbb{Z}[x_0,\dots, x_d]$.

User
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