Rings Generated by Elements - Lovett, Example 5.2.1 .... ....

[Math Amateur]

I am reading Stephen Lovett's book, "Abstract Algebra: Structures and Applications" and am currently focused on Chapter 5 ...

I need some help with Example 5.2.1 in Section 5.2: Rings Generated by Elements ...
View attachment 6407
In the Introduction to Section 5.2.1 (see text above) Lovett writes:

" ... ... \(\displaystyle R\) denotes the smallest (by inclusion) subring of \(\displaystyle A\) that contains both \(\displaystyle R\) and \(\displaystyle S\) ... ... "Then, a bit later, in Example 5.2.1 concerning the ring \(\displaystyle \mathbb{Z} [ \frac{1}{2} ]\) Lovett writes:" ... ... It is not hard to show that the set\(\displaystyle \{ \frac{k}{ 2^n} \ | \ k.n \in \mathbb{Z} \}\)is a subring of \(\displaystyle \mathbb{Q}\). Hence, this set is precisely the ring \(\displaystyle \mathbb{Z} [ \frac{1}{2} ]\) ... ... ... "BUT ...

How has Lovett actually shown that the set \(\displaystyle \{ \frac{k}{ 2^n} \ | \ k,n \in \mathbb{Z} \}\) as a subring of \(\displaystyle \mathbb{Q}\) is actually (precisely in Lovett's words) the ring \(\displaystyle \mathbb{Z} [ \frac{1}{2} ]\) ... ... ?

... ... according to his introduction which I quoted Lovett says that the ring \(\displaystyle \mathbb{Z} [ \frac{1}{2} ]\) is the smallest (by inclusion) subring of \(\displaystyle \mathbb{Q}\) that contains \(\displaystyle \mathbb{Z}\) and \(\displaystyle \frac{1}{2}\) ... ...Can someone please explain to me exactly how Lovett has demonstrated this ... ...

... and ... if Lovett has not clearly proved this can someone please demonstrate a proof ...Just one further clarification ... is Lovett here dealing with ring extensions ... ... ?

Hope someone can help ...

Peter

 

[johng1]

Let R be any subring of the rationals Q that contains both 1/2 and the integers. Let n be any positive integer; induct on n to show ${1\over 2^n}\in R$. (Inductive step: ${1\over 2^n}\in R$ implies ${1\over 2}{1\over 2^n}\in R$ since R is closed under products. Also if n is any non-positive integer, ${1\over 2^n}=2^{-n}$ is an integer and so belongs to R. Again since R is closed under products, $k{1\over 2^n}\in R$. So the given subring is a subset of R.

 

[Math Amateur]

Let R be any subring of the rationals Q that contains both 1/2 and the integers. Let n be any positive integer; induct on n to show ${1\over 2^n}\in R$. (Inductive step: ${1\over 2^n}\in R$ implies ${1\over 2}{1\over 2^n}\in R$ since R is closed under products. Also if n is any non-positive integer, ${1\over 2^n}=2^{-n}$ is an integer and so belongs to R. Again since R is closed under products, $k{1\over 2^n}\in R$. So the given subring is a subset of R.

Thanks for the help johng ...

But i am still a bit puzzled ... I think we have to show that \(\displaystyle \{ \frac{k}{ 2^n} \ | \ k.n \in \mathbb{Z} \}\) is actually the smallest subring of \(\displaystyle \mathbb{Q}\) that contains both \(\displaystyle \frac{1}{2}\) and \(\displaystyle \mathbb{Z}\) ... ... thus making it equal to or identified with \(\displaystyle \mathbb{Z} [ \frac{1}{2} ]\) ... ...

How do we accomplish this explicitly and rigorously ...

Can you help further ...

Peter

 

[johng1]

You have to understand what is meant by the smallest element of a partially ordered set S. This is an element m of S such that $m\leq s$ for any $s\in S$. For example let S be the set of positive even integers ordered by $\leq$. Then 2 is the smallest element of S. How do you prove this?

Let s be any positive even integer. Then s=2q for some $q\geq 1$, and so $2\leq 2q=s$.

Then the above applies directly to the partially ordered set S of subrings of Q that contain 1/2 and the integers (the order on S is $\subseteq$). In my previous post, I said let R be any subring that contains 1/2 and the integers; i.e. let R be any element of S. I proved
$$\{{k\over 2^n}\,: k,n\in Z\}\subseteq R\}$$

 

[HallsofIvy]

Suppose $$\{\frac{k}{2^n}|k,n\in Z\}$$ were not the smallest ring containing $$\frac{1}{2}$$ and all integers. That is, there exist some ring, S, that is smaller. Then there must be some integers, k and n, such that $$\frac{k}{2^n}$$ that is not in S. But, as johng showed, since $$\frac{1}{2}$$ is in the set, by the multiplication is closed, and induction on n, $$\frac{1}{2^n}$$ is in the set. And then, since k is an integer $$\frac{k}{2^n}$$ is in the set, contradicting the statement that $$\frac{k}{2^n}$$ is not in the set.

 

[Math Amateur]

Suppose $$\{\frac{k}{2^n}|k,n\in Z\}$$ were not the smallest ring containing $$\frac{1}{2}$$ and all integers. That is that, the exist some ring, S, that is smaller. Then there must be some integers, k and n, such that $$\frac{k}{2^n}$$ that is not in S. But, as johng showed, since $$\frac{1}{2}$$ is in the set, by the multiplication is closed, and induction on n, $$\frac{1}{2^n}$$ is in the set. And then, since multiplication is closed and the integer k is closed, $$\frac{k}{2^n}$$ is in the set, contradicting the statement that $$\frac{k}{2^n}$$ is not in the set.

Well! ... that is a thoroughly convincing proof!

Thanks HallsofIvy ... really helpful ...

Peter