TITLE: If $a,b$ is an $R$-sequence, then $ax-b$ is prime (Eisenbud, Exercise 10.4)
QUESTION [6 upvotes]: This is the exercise mentioned above:

Let $a,b$ be regular sequence over a domain $R$. Prove that $ax-b$ is a prime of $R[x]$.

Thank you for your answer!

REPLY [3 votes]: $R$ is an integral domain, and therefore has a quotient field $K$. Now just to illustrate take the case where $a=1$ then if $x-b\mid p(x)r(x)$ we have 
$p(b)r(b)=0$ and so one of them is zero, say $p(b)=0$ and then $x-b\mid p(x)$ in $R[x]$.
Now consider the general case if $ax-b\mid p(x)r(x)$ then in $K[x]$ we have 
$x-\frac{b}{a}\mid p(x)r(x)$ so by the above argument we have say, $p(\frac{b}{a})=0$.
I claim in general that there are natural number $n$, $q(x)\in R[x]$ and $r \in R$ such that $a^np(x)=(ax-b)q(x)+r$. The proof is by induction on the degree on $p$. Let $cx^k$ be the leading term of $p(x)$ consider $ap(x)-cx^{k-1}(ax-b)$ this has lower degree and applying induction to this gives our result immediately.  
It is now clear to see that if $p(\frac{b}{a})=0$ then $r$ is zero.
Assume now that $n$ is minimal such that 
$$a^np(x)=(ax-b)q(x)$$ with $q(x) \in R[x]$.
If $n=0$ we are done so for $n\ge1$ we derive a contradiction by showing that $q$ is divisible by $a$.
Now if we look in the ring $R/(a)$ we have 
$bq(x)=0$ and since $b$ is not a zero divisor in $R/(a)$ this means that $q(x)=0$ in 
$R/(a)$ and therefore $q$ is divisible by $a$. 
Therefore we have in fact,
$$p(x)=(ax-b)q(x)$$ verifying that $ax-b$ is prime.