# Show that $A[X]/(aX+b)$ is an integral domain

Let $A$ be an integral domain, $a$ and $b \in A-\{0\}$, and let $B = A[X]/(aX+b)$. Show that, if $Aa \cap Ab=Aab$, then $B$ is an integral domain.

My attempt at proof (following a hint). Denote by $K$ the field of fractions of $A$. Let $\phi: A[X] \to K$, with $\phi(X)=-b/a$ and $\phi(y)=y, y \in A$.

Then $\phi(aX+b)=-b+b=0$. If $p(X) \in A[X]$ and $p(X) \not \in (aX+b)$, then $p(X)=q(X)(aX+b)+r$, $r \in A$. Therefore, $\phi(p(X))=\phi(r)=r$, so $p(X) \in \ker(\phi) \iff r=0 \iff p(X) \in (aX+b)$.

Hence $\ker(\phi)=(aX+b)$; by the first isomorphism theorem, $A[X]/(aX+b)$ must be isomorphic to Im$(\phi)$. In conclusion $B$ is isomorphic to a subfield of $K$, so it's an integral domain.

The problem is, I don't know where I used the condition $Aa \cap Ab=Aab$. Am I doing something wrong?

• Be careful where you consider your principal ideals. Note that $X^2 + X$ is not in the ideal generated by $2X + 2$ in the ring of integral polynomials. – quid Feb 8 '15 at 20:32
• Why should $\operatorname{img} ϕ ⊂ K$ be a subfield? (Of course, the conclusion that it’s integral would still hold.) – k.stm Feb 8 '15 at 20:33
• I thought that resulted from the isomorphism theorem, but looking back at it, I'm actually not sure. – odnerpmocon Feb 8 '15 at 20:35
• @quid: True, but can you explain how it applies here? – odnerpmocon Feb 8 '15 at 20:37
• @odnerpmocon What quid was getting at was that if the ideal is in $K[x]$ instead of $A[x]$ you will come to difference conclusions, since different rings have different workings. Also, the division algorithm only works if you have a Euclidean domain, in the case $A$ is not a field, then no such $q,r$ are guaranteed to exist, which undermines that entire approach! – Adam Hughes Feb 8 '15 at 21:41

It suffices to prove $$\,f = ax+b\,$$ is prime in $$\,A[x],\,$$ which follows as below.

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

Proof $$\$$ Assume $$\,f\mid gg'\,$$ for $$\,g,g'\! \in D[x].\,$$ $$\,f = a(x + \frac{b}a)= a\bar f$$ in $$\,\bar D = D[a^{-1}]\,$$ where $$\,\bar f\,$$ is prime, hence $$\,\bar f\mid g\,$$ or $$\,\bar f\mid g'\,;\,$$ wlog $$\,\bar f\mid g,\,$$ so $$\, g = \bar f \bar h,\ \bar h \in \bar D[x].\,$$ Scaling by $$a^n$$ for big enough $$\,n\,$$ yields $$\, a^n g = f h,\ h\in D[x],\,$$ so $$\,f\mid a^n g\,\Rightarrow\,f\mid g,\,$$ by iterating below Lemma. Hence $$\,f\,$$ is prime $$\ \$$ QED

Lemma $$\,\ f\mid ag\,\Rightarrow\,f\mid(f\!-\!ax)g=bg\,\Rightarrow\,f\mid ag,bg\,\Rightarrow\,a,b\mid abg/f\,\color{#c00}\Rightarrow\,ab\mid abg/f\,\Rightarrow\,f\mid g$$

Remark  If localizations like $$\,D[a^{-1}]\,\cong\, D[\,t]/(at\!-\!1)\,$$ are unfamiliar then we can eliminate it by instead using the nonmonic form of the polynomial division algorithm, namely $$\, a^n g = q\, f + r\,$$ where $$\,a\,$$ is the lead coef of $$\,f,\,$$ which has a simple inductive proof. I presented it here in local form so to show how this Theorem generalizes to localizations. The results there generalize as below. Above we used $$\, S^{-1}D = D[a^{-1}],\,$$ whereas there we used the $$\,S^{-1}D = K =$$ full fraction field.

Theorem $$\,\ f\,$$ is prime in $$\,D[x] \!\iff\! f\,$$ is prime in $$\,S^{-1} D[x]\,$$ and $$\,f\,$$ is $$S$$-superprimitive.

where $$\,f\,$$ is $$\,S$$-superprimitive if it satisfies one of the following equivalent properties

$$(1)\quad c\mid gf\,\Rightarrow\, c\mid g\quad$$ for all $$\ c\in S,\ g\in D[x]$$

$$(2)\quad c\mid df\,\Rightarrow\, c\mid d\quad$$ for all $$\ c\in S,\ d\in D$$

$$(3)\quad f\mid g\,$$ in $$\,S^{-1} D[x]\,\Rightarrow\, f\mid g\,$$ in $$\,D[x]\quad$$ for all $$\, g\in D[x]$$

$$(4)\quad f\mid cg\,\Rightarrow\,f\mid g\quad$$ for all $$\ c\in S,\ g\in D[x]$$

$$(5)\quad gf\in D[x]\,\Rightarrow\, g \in D[x]\quad$$ for all $$\,g\in S^{-1}D[x]$$

You write:

If $p(X) \in A[X]$ and $p(X) \not \in (aX+b)$, then $p(X)=q(X)(aX+b)+r, r \in A$.

Yet how do you actually justify it? One cannot divide by a polynomial in an integral domain in general; one needs the leading coefficient is invertible (or some other assumption).

So, let us work over the quotient field instead. We can define the $\phi$ just as well over $K[x]$ and then we get that the kernel for the map defined on $K[X]$ is the ideal generated by $aX + b$ in $K[X]$.

Now, restrict the map to $A[X]$. Then the kernel is the intersection of $A[X]$ with the ideal generated by $aX+b$ in $K[X]$. In general this is not the same as the ideal generated by $aX+ b$ in $A[X]$ but here you can then use your assumption.

• Could you explain how I could use the assumption? – odnerpmocon Feb 9 '15 at 1:59
• The linked to proposed duplicate question contains various arguments along these lines. – quid Feb 9 '15 at 11:25