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Let $R$ be the ring given by $R=\mathbb Z+x\mathbb Q[x]$. Then show that:

1) $R$ is an integral domain and its units are $+1$ and $-1$.

2) $x$ is not prime in $R$ and describe the quotient ring $R/(x)$.

3) Compute an ideal of $R$ which is not principal.

$R$ being a subset of $\mathbb Q[x]$ inherits non-zero divisor, unity, and commutativity. So $1$st question is done. For the third one I think the ideal $(2,{1/2}x)$ may be but can't give proper proof. Please help me for $2$nd and $3rd$ ones. Thanks in advance.

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  • $\begingroup$ $R$ doesn't inherit its units from $Q[x]$. For example, $2$ is a unit in $Q[x]$. Is it in $R$? $\endgroup$
    – user14972
    Commented Sep 12, 2017 at 1:32
  • $\begingroup$ No no its correct. I am said that only unity, commutativity and non-zero divisors can be inherits. Units can't be an inheritable property. Sir please help me for other ones. @Hurkyl $\endgroup$
    – abcdmath
    Commented Sep 12, 2017 at 1:36
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    $\begingroup$ You also said that the first question is done, but the first question says something about units. $\endgroup$
    – user14972
    Commented Sep 12, 2017 at 9:24

2 Answers 2

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This is going to sound a little strange at first, but it ends up at your solution. There is quite a bit of checking for you to do.

The set $S=\mathbb Z\times \mathbb Q/\mathbb Z$ is an Abelian group under coordinatewise addition, and it becomes a ring if you use the multiplication $(y, q+\mathbb Z)(z, p+\mathbb Z)=(yz, yp+\mathbb Z+zq+\mathbb Z)$. It's called the trivial extension of $\mathbb Q/\mathbb Z$ by $\mathbb Z$.

Now, it is not hard to prove that $S\cong R/(x)$ for the $R$ in your problem. Just look at $(x)=xR=x\mathbb Z+x^2\mathbb Q[x]$ and consider the quotient carefully.

You can check that $I=\{0\}\times \mathbb Q/\mathbb Z$ is an ideal of $S$, and furthermore $ab=0$ for any $a,b\in I$. Therefore $S$ isn't an integral domain... and that should be enough information for you to conclude that $(x)$ is not prime in $R$.

Finally, $x\mathbb Q[x]$ is your candidate for a non-finitely-generated ideal. You'll see that you have to generate everything of the form $x\mathbb Q$ using integer multiples of your generators (which won't be possible with finitely many generators.)

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  • $\begingroup$ What is the quotient will? Is it $ax+b+(x)$, where $0<a<1$ and $b$ in $Z$. Also $a$ must be in $Q$. Am I correct ? @rschwieb $\endgroup$
    – abcdmath
    Commented Sep 12, 2017 at 2:56
  • $\begingroup$ @abcdmath Right, you can find a representative of each coset with the properties you mentioned. Then you can see what the logical map is to $S$ ... $\endgroup$
    – rschwieb
    Commented Sep 12, 2017 at 20:26
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Let me give some remarks about the ring $R=\mathbb Z+x\mathbb Q[x]$. This happens to be a particular construction of the more general construction $D+xD_S[x]$, where $D$ is an integral domain and $S$ is a multiplicatively closed subset of $D$ such that $0\notin S$.

In general, there is no a "good" relationship between $D$ and $D+xD_S[x]$, except when $S=D\setminus \{0\}$. In this case there are many interesting properties related to both $D$ and $D+xD_S[x]=D+K[x]$ (here $K$ is the fraction field of $D$).

For example, in this paper by Costa and Zafrullah it's proved the following remarkable result:

Theorem 1: $D+xK[x]$ is a bézout domain iff $D$ is a bézout domain.

Proof: This is corollary 4.13 in the paper cited above.

Now, in your case, $D=\Bbb Z$ is a bézout domain (it's even a PID), so by the above theorem $\mathbb Z+x\mathbb Q[x]$ is also a bézout domain.

This explains why in order to produce a non principal ideal you need to take a non finitely generated ideal, e.g., the ideal given in rschwieb's answer: $x\Bbb Q[x]$. This also justifies why your example, namely $(2,1/2x)$ won't work, since it's a finitely generated ideal.

On the other hand, rschwieb already answered what is the form of $R/(x)$ and from there he proved that $x$ is not prime, but there is another way to proof the later result.

It's enough to show that $x$ is not irreducible and indeed, in the ring $\mathbb Z+x\mathbb Q[x]$ we have the non-trivial factorization of $x$: $$x=\Bigl(\frac{1}{2}x\Bigr)2.$$ This shows that $x$ is not irreducible in $\mathbb Z+x\mathbb Q[x]$.

More generally, Costa and Zafrullah proved in their paper above a nice result about the structure of the prime ideals of $D+xK[x]$. We have the following:

Theorem 2: The nonzero prime ideals of $D+xK[x]$ are the ideals $P+xK[x]$ where $P$ is a prime ideal of $D$, and the principal ideals $f(x)(D+xK[x])$, where $f(X)$ is irreducible in $K[x]$ and $f(0)=1$.

Proof: This is theorem 4.21 in the paper cited above.

So by the theorem above we can conclude again that $x$ is not prime in $\mathbb Z+x\mathbb Q[x]$ by noticing that $x$ is irreducible in $\Bbb Q[x]$, but $x(0)=0$.

There has been many researching in commutative algebra about the ring $D+xD_S[x]$ and similar constructions. If you want to explore and deepen these constructions I recommend you the following articles:

  1. Costa, D.; Mott, J.L.; Zafrullah, M. The Construction $D+XD_S[X]$. Journal of Algebra. 53 (1978), p. 423-439.

  2. Zafrullah M. The $D+XD_S[X]$ construction from GCD domains. Journal of Pure and Applied Algebra. 50 (1988), p. 93-107.

  3. Jackson, T.; Zafrullah, M. Examples in modern algebra with which students can play. Problems, Resources, and Issues in Mathematics Undergraduate Studies. 6.4 (1996), p. 351-354.

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