# If $a,b$ is an $R$-sequence, then $ax-b$ is prime (Eisenbud, Exercise 10.4)

This is the exercise mentioned above:

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

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thanks, I'll do soon! –  user149240 May 16 at 21:38

Here's the general statement: Let $R$ be any ring, and $a, b \in R$ a weak regular sequence, i.e. $a$ is a nonzerodivisor and $(a) : (b) = (a)$. Then the $R$-algebra map $\phi : R[x] \to R[\frac{1}{a}]$, $x \mapsto \frac{b}{a}$, has $\ker \phi = (ax - b)$. The proof consists of 3 steps:

1) Certainly $(ax - b) \subseteq \ker \phi$. If $(ax - b) \subsetneq \ker \phi$, then there exists $g \in \ker \phi \setminus (ax - b)$, of minimal degree with respect to not being contained in $(ax - b)$. Since $a$ is a nonzerodivisor, $\ker \phi$ contains no constants, so $g(x) = c_nx^n + \ldots + c_0$, where $n \ge 1$.

2) $c_n \not \in (a)$. If not, then $d := \frac{c_nb}{a}$ is a well-defined element of $R$ (since $a$ nonzerodivisor), and $0 = g(\frac{b}{a}) = c_n(\frac{b}{a})^n + c_{n-1}(\frac{b}{a})^{n-1} + \ldots + c_0 = (d + c_{n-1})(\frac{b}{a})^{n-1} + \ldots + c_0$. Thus $h(x) := (d + c_{n-1})x^{n-1} + \ldots + c_0$ is in $\ker \phi$ of strictly smaller degree than $g$ (since $n \ge 1$), so $h(x) \in (ax - b)$, and thus $g = h + c_nx^n - dx^{n-1} = h + \frac{c_n}{a}x^{n-1}(ax - b) \in (ax - b)$.

3) However, $0 = c_n(\frac{b}{a})^n + \ldots + c_0 \implies -c_nb^n = c_{n-1}b^{n-1}a + \ldots + c_0a^n \in (a)$, and $(a) : (b) = (a) \implies (a) : (b^n) = (a)$, so $c_n \in (a) : (b^n) = (a)$, in contradiction to (2).

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@user26857: I suppose you're right about homework questions. On the other hand I personally don't like to give hints. I feel that the details for this particular problem are sufficiently non-obvious though (for instance, the OP still wasn't able to solve it based on the hint in your answer) –  zcn May 17 at 7:45
I'm sorry, But I didn't understand a little step. You define $d:=(c_n\cdot b)/a$. Generally, it is not invertible even if it's not a $0$-divisor. So, are you working in a sort of localization? which one? –  user149240 May 18 at 15:49
@user149240: Here I'm assuming that $c_n \in (a)$ for contradiction, and thus writing $c_n = ra$ for some unique $r \in R$ (unique because $a$ is a nonzerodivisor). Then $d$ is really just $rb$ –  zcn May 18 at 17:52
sorry, stupid question! –  user149240 May 18 at 18:15
@user149240: No worries! –  zcn May 18 at 18:16

Hint $\$ The following theorem applies.

Theorem $\$ Suppose that $\,D\,$ is a domain, and $\,0\ne a,b \in D\,$ satisfy $\,a,b\mid e\, \color{#c00}\Rightarrow\, ab\mid e\,$ for all $\,e \in D.\,$ Then $\, f = ax+b\,$ is prime in $\,D[x].$

$\quad$ Lemma $\,\ d\mid cf\ \Rightarrow\, d\mid c,\$ for all $\ 0\ne c,d\in D,\$ i.e. $\,f\$ is superprimitive.

$\quad$ Proof $\,\ d\mid c(ax\!+\!b)\, \Rightarrow\, d\mid ac,bc\,\Rightarrow\, a,b\mid abc/d\,\color{#c00}\Rightarrow\, ab\mid abc/d \,\Rightarrow\, d\mid c\ \$ QED

Proof $\$ If $\,f\,$ is not prime, $\,f\mid hk,\,\ f\nmid h, k\,$ for some $\,h,k \in D[x].\,$ Let $\,K\,$ be the fraction field of $\,D.\,$ Since $\,K[x]\,$ is a UFD, $\,f\,$ irreducible $\,\Rightarrow\, f\,$ prime, so $\,f\mid hk\, \Rightarrow\ f\mid h\,$ or $\,f\mid k.\,$ Wlog let $\,f\mid h\,$ in $\,K[x],\,$ so $\, h/f = g/d,\,$ for some $\,g \in D[x],\,\ 0\ne d \in D,\,\ d\nmid g\,$ in $\,D[x].\,$ Also $\,d\nmid f\$ by $\,c = 1\,$ in the Lemma. Hence $\,fg = dh\,$ implies $\,f\,$ is a zero-divisor in $\,(D/d)[x].\,$ Hence by McCoy's theorem $\,cf = 0\,$ for some $\,0\ne c\in D/d,\$ so $\ d\mid cf,\,\, d\nmid c,\,$ contra Lemma. $\$ QED

More generally the above proof is easily extended to this

Theorem $\,\ f\,$ is prime in $\,D[x] \!\iff\! f\,$ is irreducible in $\,K[x]\,$ and $\,f\,$ is superprimitive.

For further results on superprimitivity and Gauss' Lemma see e.g. Hwa Tsang Tang, Gauss' Lemma, Proc. Amer. Math. Soc. 35 (1972) 372-376 (summary of her thesis under Kaplansky).

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$R$ is an integral domain, and therefore has a quotient field $K$. Now just to illustrate take the case where $a=1$ then if $x-b|p(x)r(x)$ we have $p(b)r(b)=0$ and so one of them is zero, say $p(b)=0$ and then $x-b|p(x)$ in $R[x]$.

Now consider the general case if $ax-b|p(x)r(x)$ then in $K[x]$ we have

$x-\frac{b}{a}|p(x)r(x)$ 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 $n>1$ 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.

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