2
$\begingroup$

I have two questions, just the first one remains unsolved for me.

  1. Is there an example of a commutative unital ring $B$, a unital subring $A \subsetneq B$ and a non-zero ideal $I$ of $B$ such that $I \subseteq A$?

  2. Let $A \subset B$ be two commutative rings. Assume that any proper ideal of $A$ is also an ideal of $B$. Do we have $A=B$ ?

I noticed that $2)$ is wrong if we assume that any proper ideal of $B$ is included in (thus an ideal of) $A$; take $B$ to be a field. Actually, I've just noticed that $2)$ is wrong: take $A$ to be any field.

For $1)$ I thought to $B=\Bbb Z \times \Bbb Z \supsetneq A = \{(x,x),x\in \Bbb Z\}$. But it didn't work.

Thank you!

  • 0
    must the unit be the same for the two rings? – 2017-01-10
  • 0
    @JorgeFernándezHidalgo : yes, sorry. I should have said "subring" (which means same unit element for me). – 2017-01-10
  • 1
    but $A$ is not a subring – 2017-01-10
  • 0
    I've found it : try $I=A=\{p(x)\in B\mid p(0)=0\} \subset B=\Bbb R[x]$. :-) – 2017-01-10
  • 0
    Actually my comment above is wrong (I've taken it from http://math.stackexchange.com/q/11378 but there "subring" doesn't have to contain 1). Here also (https://crazyproject.wordpress.com/2010/09/04/determine-whether-a-subring-is-an-ideal/) there is an answer : take $B$ the ring of all functions from the closed interval [0,1] to $\mathbb{R}$, and $I=A$ the set of all functions f(x) such that f(q) = 0 for all $q \in \mathbb{Q} \cap [0,1]$. But A doesn't contain 1. – 2017-01-11
  • 3
    If $A=\mathbb Z+X\mathbb Q[X]$ and $B=\mathbb Q[X]$, then $X\mathbb Q[X]$ is a common ideal. – 2017-01-11
  • 0
    @AlexM. : If $I$ is included in $A$, then $I$ is an ideal of $A$, since $I$ is an ideal of the ring $B$, or am I wrong? – 2017-01-11
  • 0
    @Alphonse yes, $I$ is certainly an ideal of $A$ in this case. – 2017-01-11
  • 1
    @user26857 : Thank you very much for the example $I = X \Bbb Q[X] \subset A=\Bbb Z+X\Bbb Q[X] \subset B=\Bbb Q[X]$, this is really helpful ! – 2017-01-11
  • 0
    @Watson I've lost the point of the parallel conversation, but your link it's about the idealization of a module. Is this related to the present question? – 2017-01-11
  • 0
    @Alphonse yeah haha I moticed it was incorrect but you deleted too soon. plz dont delete interesting question in the future – 2017-01-11
  • 1
    @Alphonse: In fact, **user26857**'s example gives us the following general scheme: if $S \le B$ are two unital commutative rings, and if $I \subset B$ is a proper ideal, then let $A = S + I$. – 2017-01-11
  • 0
    @JorgeFernándezHidalgo : yes, sorry, I will avoid this. – 2017-01-11

1 Answers 1

1

Let $R, S$ be commutative unital rings and let $S' \le S$ be a commutative unital subring. Let $A = R \times S'$ and $B = R \times S$. Clearly $A \le B$ and $(1,1) \in A$. If $I \subset R$ is a proper ideal, then $I \times \{0\} \subset A$ is a proper ideal too.

(Alternatively, following @user26857's excellent suggestion, one may start with $B$ and consider a unital subring $S$; then $A = S + I$ is a unital subring as desired.)