I have the following:

Let A be a ring and B $\subset$ A a subring of A, and X $\subset$ A a subset of A. Show that exists a subring B[X] of A such that X $\subset$ B[X] and B $\subset$ B[X] and any other subring of A that contains B and X, contains B[X]

I'm not sure if what I'm thinking is right, but I wanted to extend X to a ring X', and then define B[X]:={ bx$\in$ A / b$\in$B and x$\in$X' }

X' should be defined as the set generated by all of the elements of X. Is this right?

  • $\begingroup$ Yes, you should precise what does mean "generated by $X$", but if you mean generated multiplicatively by $X$ it's ok, i.e all the product $x_1 \dots x_n$ for $x_i \in X$. In fact the notation is very suggestive : it is exactly all the polynomial with coefficients in $B$, and variables the elements of $X$. $\endgroup$ – Nicolas Hemelsoet Oct 11 '17 at 21:02
  • $\begingroup$ Since the intersection of subrings is again a subring, you can just take the intersection of all subrings that contain $B$ and $X$ (which is a non-empty set of subrings, as it contains $A$.) This trivially fulfills the condition $\endgroup$ – Lukas Heger Oct 11 '17 at 21:21
  • $\begingroup$ Thank you both! I liked better MatheiBoulomenos solution. I think it's simpler for this case, but it's really good to know what the answer is going to be, leaving the abstract definition he gave. $\endgroup$ – Dani Seidler Oct 11 '17 at 22:01

In this answer, I am making the assumption that $A$ is unital; that is, that there is an element $1_A \in A$ such that $1_A a = a1_A = a$ for all $a \in A$. This assumption is, I believe, in accord with current the current usage of the word "ring", the recently coined term "rng" being reserved for entities satisfying all the ring axioms except the existence of a multiplicative unit; so, for example, $\Bbb Z$ is a ring, but $2\Bbb Z$ is a rng. And of course, since $B \subset A$ is a subring, we assume it inherits the multitplicative identity of $A$ so that $1_A$ is the unit element of $B$. I'm pretty sure, though I have not worked through the details, that this result is correct even in the even $B$ is merely a rng, but I believe the demonstration in that case to be somewhat more delicate; so here, I'll stick with unital rings.

Having said these things,

The way I would attack this problem is to consider the set $S$ defined by

$S = \{ \sum_1^n \prod_1^m y_j \mid m, n \in \Bbb N, y_j \ \in B \cup X \}; \tag 1$

that is, $S$ is the set of all finite sums of finite products of elements selected from $B$ and $X$. It is clear that $B \subset S$ and $X \subset S$; we show that in fact $S$ is a subring of $A$. Since $0 \in B$, it is clear that $0 \in S$; furthermore, since the sum of any two sums of finite products of elements from $B$ and $X$ is again a sum of such products, it is also clear that $S$ is closed under the "$+$" operation of $A$. The additive inverse of any such sum $s$ may be had by multiplying it by $-1_A$, which we have assumed is also an element of $B$. Thus, since the other properties of ring addition are inherited from $A$, $S$ forms an additive subgroup of $A$. Furthermore, it is easy to see that the product of two elements of $S$ may again be expressed as a finite sum of finite products of elements taken from $B$ and $X$; and again, since axiomatic properties of the multiplication operation in $A$ will be inherited by elements of $S$, it ($S$) will obey all the usual ring axioms for multiplication, including the distributive axioms and so forth. So $S$ is indeed a subring of $A$.

If $R$ is a subring of $A$ such that $B \subset R$ and $X \subset R$, and $m, n \in \Bbb N$ and $y_j \in B \cup X$, it is easy to see that for $s \in S$,

$s = \sum_1^n \prod_1^m y_j \in R, \tag 2$

since (as has in fact been made clear already in the above), $s$ is formed from the $y_i$ via a finite number of the ring operations $+$ and $\cdot$; thus $S \subset R$ for every subring $R$ containing $B$ and $X$. In fact we have

$S \subset \bigcap_{B, X \subset R \subset A; \; R \; \text{a subring}}R, \tag 3$

and since $S$ is a subring satisfying $B, X \subset R \subset A$, we see that it actually occurs on the right of (3), whence in fact

$S = \bigcap_{B, X \subset R \subset A; \; R \; \text{a subring}}R. \tag 4$

(4) shows that every subring R of $A$ with $B, X \subset R$ contains $S$; thus we may take

$S = B[X]. \tag 5$

There is no need to extend $X$ to a subring $X'$; the process of taking finite sums of finite products accomplishes this extension, and in any event would be "executed" in so extending $X$. In any event, one would need to take sums of the products $bx \in A$ occurring in the OP's definition of $B[X]$ since in general, the set of such products will not be additively closed in and of itself. Finally, defining $X'$ to be the set of elements generated from $X$, for example via the finite sums of finite products procedure, may not produce a subring; it is not clear that we will necessarily have a multiplicative unit in $X'$; perhaps $X'$ will be a rng, but I leave that discovery to my readers.

It is worth noting that the finite sums of finite products construction is similar to taking the product of a set of ideals; the idea in each case is to obtain an object closed under addition etc.


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