show that a quotient ring is in integral domain

**frankmelody** · September 5th 2009, 06:39 PM

Let A be an integral domain, and a,b are elements in A. Let B=A[x]/(ax+b), where (ax+b) is ideal generated by ax+b. Suppose $(a)\cap(b)=(ab)$ , show that B is an integral domain.

I think it suffices to show that (ax+b) is prime under the assumption, and hence need to show that if f(x)g(x) is in (ax+b), then either f(x) or g(x) is in (ax+b).

thank you for help!

**NonCommAlg** · September 6th 2009, 02:31 AM

Originally Posted by frankmelody

Let A be an integral domain, and a,b are elements in A. Let B=A[x]/(ax+b), where (ax+b) is ideal generated by ax+b. Suppose $(a)\cap(b)=(ab)$ , show that B is an integral domain.

I think it suffices to show that (ax+b) is prime under the assumption, and hence need to show that if f(x)g(x) is in (ax+b), then either f(x) or g(x) is in (ax+b).

thank you for help!

first of all we also need to assume that $a \neq 0, \ b \neq 0.$ i've been trying for a couple of hours now to find a simpler proof for this problem but i haven't been able to do so!

so, i'm just going to give

you my "not very simple" proof. we need a lemma:

Lemma: suppose $d \mid a^n c$ and $d \mid b^n c,$ for some $c,d \in A$ and integer $n \geq 0.$ then $d \mid c.$

Proof: by induction on $n.$ if $n=0$ or $c=0,$ there is nothing to prove. so suppose the claim is true for $n-1$ and $d \mid a^nc, \ d \mid b^n c,$ where $c \neq 0.$ so we have $a^n c = rd, \ b^n c = sd,$ for some $r,s \in A,$

which gives us $a^ns = b^nr \in <a> \cap <b> = <ab>.$ hence $a^ns=abu, \ b^nr=abv,$ for some $u,v \in A.$ therefore $a^{n-1}s=bu$ and $b^{n-1}r=av.$ hence by the induction hypothesis $b \mid s$ and $a \mid r.$

so $s = bt, \ r = az,$ for some $t,z \in A.$ but then $a^n c = rd = azd, \ b^n c = sd = btd,$ and so $a^{n-1}c = zd, \ b^{n-1}c = td,$ and we're done by the induction hypothesis. $\Box$

now suppose $f(x)g(x) \in <ax + b>,$ for some $f(x),g(x) \in A[x].$ we want to prove that either $f(x) \in <ax + b>$ or $g(x) \in <ax+b>.$ obviously we may assume that both $f(x),g(x)$ are non-

zero. let $Q$ be the field of fractions of $A.$ clearly in $Q$ we have $f(-b/a)g(-b/a)=0.$ so we may assume that $f(-b/a)=0.$ so there exists $h(x) \in Q[x]$ such that $f(x)=(ax+b)h(x),$ which we

will call it (1). let $f(x)=c_0x^n + \cdots + c_n$ and $h(x)=\frac{d_0}{d}x^{n-1} + \cdots + \frac{d_{n-1}}{d},$ where $c_j, d_j , d \in A$ and $d \neq 0.$ so (1) becomes: $dc_0x^n + \cdots + dc_n = (ax+b)(d_0x^{n-1} + \cdots + d_{n-1}).$ call this (2).

the claim is that $d \mid d_j,$ for all $j,$ which will complete the solution because then $h(x) \in A[x]$ and thus by (1): $f(x) \in <ax+b>.$ here's the proof of this claim:

from (2) we have: $dc_0=ad_0, \ dc_n = bd_{n-1}, \ dc_j = ad_j + b d_{j-1}, \ 1 \leq j \leq n-1.$ call this one (3). now, ignoring $dc_n = bd_{n-1}$ in (3), the rest of the relations can be written as $CX=dY,$ where:

$C$ is an $n \times n$ lower triangular matrix with $a$ on the main diagonal, $X$ an $n \times 1$ vector with the entries $d_0, \cdots , d_{n-1},$ and $Y$ an $n \times 1$ vector with the entries $c_0, \cdots , c_{n-1}.$ multiplying $CX=dY$

from the left by $\text{adj}(C)$ gives us $a^n X = \det(C) X = d \cdot \text{adj}(C)Y$ and so $d \mid a^n d_j,$ for all $j.$ similarly if we ignore $dc_0=ad_0$ in (3) and write the remaining relations in terms of matrices, we will get

$d \mid b^n d_j,$ for all $j.$ therefore $d \mid d_j,$ for all $j,$ by the lemma. $\Box$

**frankmelody** · September 6th 2009, 07:35 AM

hi, thank you for your help! I think most of your solution is correct, only the last part of your proof has some problems. Since A is only an integral domain, adj(C) need not to exist in A. But still, I can get d|d_j under your idea, just by considering each equation one by one. I can solve the problem now, thank you so much!

**NonCommAlg** · September 6th 2009, 07:54 AM

Originally Posted by frankmelody

... Since A is only an integral domain, adj(C) need not to exist in A.

that is not true! actually for the adjoint matrix of a square matrix C over a ring A to exist, we even don't need A to be integral domain!! we only need A to be commutative with identity.

i suggest you take a look at a standard textbook in linear algebra or just google it!

**frankmelody** · September 6th 2009, 08:01 AM

oh~yep! sorry for my mistake! we do not need division in the denifition of the adjoint matrix, thank you!

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