Show the polynomial $p(x,y) =y^3+xy^2+(x^2+x)y+(4x^3+2x)$ is irreducible in $\mathbb{Z}[x, y]$

Question

I've thought about $p(x,y)$ being in $\mathbb{Z}[x][y]$ and tried to use Eisenstein's criterion, but I haven't had any luck. The best I've got is, we can suppose $p(x,y)$ is reducible and write it as $$p(x,y)=(y+a)(y^2+by+c) = y^3 + (a+b)y^2 + (ab+c)y + ac$$ for some $a,b,c \in \mathbb{Z}[x]$. Comparing coefficients we have \begin{align} &a+b=x \\ &ab+c = x^2+x \\ &ac = 4x^3+2x = 2x(2x^2+1) \end{align} From the last equation we have $a = 2x$ and $c=2x^2+1$ or alternatively $a = 2x^2+1$ and $c=2x$.

If $a=2x$, $c=2x^2+1$ then from the first equation, $2x+b=x$ so $b=-x$. From the second equation, $-2x^2+x-1=x^2+x$, but this doesn't hold for all $x$...

If $a=2x^2+1$, $c=2x$ then from the first equation, $2x^2+1+b=x$ so $b=-2x^2+x-1$. From the second equation, $(2x^2+1)(-2x^2+x-1) = x^2+x$, but again the l.h.s. and r.h.s. don't match since without even computing we see the l.h.s. is degree $4$.

Is there a better way to prove this?

Answer 1

Rewrite: $$p(x,y)= y^3+xy^2+x(x+1)y+2x(2x^2+1), $$ and as $p(x,y)\in \mathbf Z[x][y]$, we can use the generalisation of Eisenstein's criterion directly:

$(x)$ is a prime ideal in the U.F.D. $\mathbf Z[x]$, and

• $x$ divides divides all coefficients, but the leading coefficient;
• $x^2$ does not divide the constant term. Hence $p(x,y)$ is irreducible in $\mathbf Z[x][y]\simeq\mathbf Z[x,y]$.

Answer 2

$$p(3,y)=y^3+3y^2+18y+114$$

Now use Eisenstein's criterion with $p=3$.

Answer 3

Take $x=3$ and use Eisenstein for $p=3$.

Answer 4

In $\mathbb{Z}[x][y]$, a generalization of Eisenstein's criterion will apply, since $\mathbb{Z}[x]$ is a UFD. Now, apply this generalization using the irreducible element $x \in \mathbb{Z}[x]$.

As an alternative: if $p$ were reducible, then its terms of highest total degree $4x^3 + x^2 y + x y^2 + y^3$ would have to be a product of the highest-degree terms of the factors. Now suppose the highest-degree terms of the factors are $ax^2 + bxy + cy^2$ and $dx + ey$, respectively. Then $(ax^2 + bxy + cy^2) (dx + ey) = 4x^3 + x^2 y + x y^2 + y^3$; now if we substitute $x := t$ and $y := 1$, we get $(at^2 + bt + c) (dt + e) = 4t^3 + t^2 + t + 1$, so $4 t^3 + t^2 + t + 1$ would have a rational root $-\frac{e}{d}$. But then, $d \mid 4$ and $e \mid 1$, and from here it is easy to get a contradiction.

The only other possibility would be if one of the factors were constant; however, then this constant would have to be $\pm 1$ since no other integer divides $p(x, y)$.