I'm trying to see if the following polynomial is irreducible over $\mathbb{Z}[X,Y]$: $P(X,Y)=X^2Y^3+XY^2+XY+8$
Is there any simple algorithme to prove it ?
Thanks
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$P(X,1)$ is an irreducible polynomial in $X$ – ASKASK May 08 '16 at 17:36
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Thinks , but i'm really searching a general algorithm – ABRAICH Ayoub May 08 '16 at 17:38
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There is no general algorithm. – Dustan Levenstein May 08 '16 at 17:44
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@ASKASK What do you conclude from the fact that P(X,1) is an irreducible polynomial over X? – miracle173 May 09 '16 at 09:15
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$$\mathbb{Z}[X,Y] \simeq \mathbb{Z}[X][Y] \simeq \mathbb{Z}[Y][X]$$ So you can view $P(X,Y)$ as poynomial in $Y$ of degree $3$ over $\mathbb{Z}[X]$ or as polynomial in $X$ of degree $2$ over $\mathbb{Z}[Y]$(link). A polynomial of degree $2$ or $3$, that is not irreducible, must have a linear factor. So assume that $(X-p(Y))$ is a linear factor of $X^2Y^3+X(Y^2+Y)+8$ from $\mathbb{Z}[Y][X]$. $P(Y)$ must be a fraction $\frac{k}{s}$ where $k$ divisor of $8$ and $s$ is a divisor of $Y^3$ (link). These are finitely many possibilities to check. None of them gives a solution.
Check the slightly different polynomial $P(X,Y)=X^2Y^3+XY^2+XY+1$ and you will find a factor with this method, because $(X+\frac{1}{Y})$ is a linear factor. This can be shown by checking that $P(\frac{-1}{Y},Y)=0$ holds.
miracle173
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