Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
thanks, I'll do soon! –  user149240 May 16 '14 at 21:38

3 Answers 3

up vote 1 down vote accepted

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).

share|improve this answer
@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 '14 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 '14 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 '14 at 17:52
sorry, stupid question! –  user149240 May 18 '14 at 18:15
@user149240: No worries! –  zcn May 18 '14 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).

share|improve this answer

$R$ 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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.