Irreducible polynomials in two and four variables I am trying to show that the polynomials :
1) $X^2+Y^2-1$
2) $XT-YZ$
are irreducible in $\mathbb Q[X,Y]$ and $\mathbb Q[X,Y,Z,T]$ respectively.
For 1) I know that $\mathbb  Q[X,Y]=(\mathbb Q[X])[Y]$ and $X^2+Y^2-1=(X-1)(X+1)+Y^2$ but this didn't help me, I would appreciate any hints or suggestions for both polynomials. Thanks in advance.
 A: I think this should work for #1: Look at $f(X,Y) = X^{2}+Y^{2}-1$ as a polynomial in $(\mathbb{Q}[X])[Y]$. Then as you noted, $f(X,Y) = Y^{2} + (X-1)(X+1)$. Applying Eisenstein's criterion to the prime ideal $(X-1)$ shows $f(X,Y)$ is irreducible in $(\mathbb{Q}[X])[Y]$, hence is irreducible $\mathbb{Q}[X,Y]$.  
As observed in the comments by the OP, one can view $f(X,Y, Z,T) = XT-YZ$ as a polynomial in $(\mathbb{Q}[X, Y, Z])[T]$, whence we apply Eisenstein's criterion to the prime ideal $(Z)$ in $\mathbb{Q}[X, Y, Z]$, which shows the desired conclusion. 
A: For 1st problem ,I have tried something check whether it works or not
Writing the polynomial as a product of two 1 degree polynomials in $x$ and $y$(which has to be the case in order for the polynomial to be reducible)
$x^2+y^2-1=(a_1x+b_1y+c_1)(a_2x+b_2y+c_2)$
Expanding the product on the R.H.S. we have $a_1a_2=1;b_1b_2=1;a_1b_2+b_1a_2=0$
which is not possible as $a_1$and $a_2$ must be of same sign ;$b_1$and $b_2$ must be of same sign for two equations to hold which clearly contradicts the 3rd equation .Thus polynomial not reducible
Please comment on whether this technique works
