Show that the polynomial $x^3-xy^2+y+1$ is irreducible in $\mathbb{Q}[x,y]$ I know that $\mathbb{Q}[y][x]=\mathbb{Q}[y,x]$; thus, we can see this polynomial as a polynomial with variable $x$ and with coefficients in $\mathbb{Q}[y]$. Someone explains to me that we could use Gauss's Lemma to reduce to proving irreducibility over the $\mathbb{Q}(x)[y]$ where $\mathbb{Q}(x)$ is the fraction field and after apply Eisenstein for the prime $y+1$. I am blocked for a while on this problem. 
Could anyone solve this problem?
 A: The hint, as you write it, is a bit messy: at some point you seem to be willing to use Eisenstein's criterion on the polynomial ring $R[x]$, with $R=\Bbb Q[y]$, but then it's mixed up with another hint, which is "try to show that it is irreducible in $\Bbb Q(x)[y]$". However, the latter does not use Eisenstein's criterion. In fact, if $F$ is a field, no polynomial in $F[t]$ satisfies the hypothesis of Eisenstein's criterion (exercise for the reader).
I'll go with the second hint, the way I see it.
Since the coefficients in $\Bbb Q[x]$ of your polynomial have no common factor, by Gauss' lemma your polynomial is irreducible in $\Bbb Q[x,y]$ if and only if it is irreducible in $\Bbb Q(x)[y]$.
However, you can easily decide whether $-xy^2+y+x^3+1$ is irreducible in $\Bbb Q(x)[y]$ by using the $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ formula for the roots.
A: Looking for a factorization in $\mathbb Q[x][y]$, $-xy^2+y+1+x^3 = (P_1(x)y+P_2(x))(P_3(x)y+P_4(x))$ where the $P_i$ are in $\mathbb Q[x]$.
Expanding yields $$P_1(x)P_3(x)=-x\\P_2(x)P_3(x)+P_1(x)P_4(x)=1\\P_2(x)P_4(x)=1+x^3=(x+1)(x^2-x+1)$$
The second equality implies $\deg P_2P_3=\deg P_1P_4$ , and combined with the others yield $(\deg P_1,\deg P_2,\deg P_3,\deg P_4)\in \{(1,2,0,1),(0,1,1,2)\}$
If $P_1(x)=\alpha x, P_3(x)=-\frac{1}{\alpha}, P_2(x)=\beta (x^2-x+1), P_4(x)=\frac{1}{\beta}(x+1)$, second equation gives $\beta =0$, a contradiction.
If $P_1(x)=\alpha,P_2(x)=\beta(x+1),P_3(x)=-\frac{1}{\alpha}x,P_4(x)=\frac{1}{\beta}(x^2-x+1)$ second equation gives $\beta =0$, a contradiction.
