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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?

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    $\begingroup$ 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. $\endgroup$ – quid Feb 8 '15 at 20:32
  • $\begingroup$ Why should $\operatorname{img} ϕ ⊂ K$ be a subfield? (Of course, the conclusion that it’s integral would still hold.) $\endgroup$ – k.stm Feb 8 '15 at 20:33
  • $\begingroup$ I thought that resulted from the isomorphism theorem, but looking back at it, I'm actually not sure. $\endgroup$ – odnerpmocon Feb 8 '15 at 20:35
  • $\begingroup$ @quid: True, but can you explain how it applies here? $\endgroup$ – odnerpmocon Feb 8 '15 at 20:37
  • $\begingroup$ @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! $\endgroup$ – Adam Hughes Feb 8 '15 at 21:41
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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 $\,g\in S^{-1}D[x]$

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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.

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

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