Proving that $(ac - (b^2 + 1))$ is irreducible in $k[a,b,c].$
Here is my trial:
I usually work by degrees and taking the lower degree of my polynomial, but the thing here is that I have two variables of the same degree $a,c$ and even they are of degree $1.$ So I am a little bit confused in my case here.
Assume that $f(a,b,c) = ac - (b^2 + 1) = ac - b^2 -1 = g(a,b,c)h(a,b,c),$ we will look at the ring $k[a,b,c]$ as $k[a,b][c]$ i.e., a polynomial in $c$ whose coefficients come from $k[a,b].$ In this ring $f(a,b,c)$ is irreducible because $a,b^2 + 1$ are units and $f(a,b,c)$ is a linear polynomial. So either $g$ or $h$is a unit in $k[a,b][c].$
Without loss of generality, suppose $g(a,b,c)$ is a unit in $k[a,b][c]$. But then the only units there are elements of $k[a,b]$, so, $g(a,b,c) = e(a,b)$ for some polynomial $e(a,b) \in k[a,b].$
So, $f(a,b,c) = e(a,b)h(a,b,c).$ But then this means that $e(a,b)$ is a common divisor of $a,b^2 +1,$ but they are polynomials of different variables.
But then I do not know how to complete. Any help will be greatly appreciated!
I think it is not clear in my mind what should I do with this $e(a,b).$ Should I just say ok, from here $g$ is a unit in $k[x,y,z]$ and I am done? I know that I would have by this that $h(a,b,c)$ is a linear polynomial in $c, $ but how that say that $f$ is irreducible in $k[a,b,c]$?
EDIT:
A quick question, should I write $k[a,b][c]$ or $k(a,b)[c]$? which is more accurate?
Note that we can write $h(a,b,c)=h_n(a,b)c^n + \ldots + h_0(a,b)$ for some $n$ where each $h_i(a,b)$ are polynomials in $a,b$. From $ac-(b^2+1)=h(a,b,c)e(a,b)$ we have that $$h_1(a,b)e(a,b)=a, \quad h_0(a,b)e(a,b)=b^2+1 \ \text{ and } h_i(a,b)=0, \forall i\ge 2$$
Now use similar argument as you did with $g$, we see that $h_1(a,b)e(a,b)=a$ implies $h_1$ or $e$ is constant. If $e$ is constant then we are done. If $h_1(a,b)=k$ is constant then $e(a,b)=\dfrac{a}{k}$, hence, $h_0(a,b)\dfrac{a}{k}=b^2+1$ (contradiction).