The subring Z[p/q] of Q

**Jumping** · December 4th 2010, 10:15 PM

Hello. I am working with the subring Z[p/q] of the ring of integers Q. Among other things, I want to show that Z[1/2], as a Z-module, cannot be generated by an element a + b/2 in Z[1/2]. Otherwise, if we take the element 1/2 in Z[1/2] then for some m in Z we can write

1/2 = ma +mb/2.

So this implies a = 0 and b = 1 so Z[1/2] is generated by 1/2. But then, for 1 + 1/2 in Z[1/2],

1 + 1/2 = m/2

for some m in Z is impossible. This is where I am starting to get confused. Couldn't we just add the left hand side of the last equation and then say m had to of been 3?

**tonio** · December 5th 2010, 07:33 AM

Originally Posted by Jumping

Hello. I am working with the subring Z[p/q] of the ring of integers Q. Among other things, I want to show that Z[1/2], as a Z-module, cannot be generated by an element a + b/2 in Z[1/2].

Do you mean here $\displaystyle{a+\frac{b}{2}\,,\,\,a,b\in\mathbb{Z} }$ ?

Otherwise, if we take the element 1/2 in Z[1/2] then for some m in Z we can write

1/2 = ma +mb/2.

So this implies a = 0 and b = 1

If the answer to my question above is yes then no: it doesn't imply this. For example, take $a=2\,,\,b=-3$ , so

the question is: what do you really mean here?

Tonio

so Z[1/2] is generated by 1/2. But then, for 1 + 1/2 in Z[1/2],

1 + 1/2 = m/2

for some m in Z is impossible. This is where I am starting to get confused. Couldn't we just add the left hand side of the last equation and then say m had to of been 3?

.

**Jumping** · December 5th 2010, 08:43 AM

Thank you for the help Tonio. My goal is to show that the subring Z[1/2] of Q, when considered as a Z-module, cannot be finitely generated. I figure I ought to first show that Z[1/2] cannot be generated by a single element from Z[1/2].

I think my confusion starts when I was writing on my scratch paper that

Z[1/2] = {a + b*(1/2) : a, b are in Z}.

But, Z[1/2] is a subring of Q and so it also has a multiplicative structure. For example 1/2 is in Z[1/2] and so (1/2)(1/2) = 1/4 is also in Z[1/2]. But I don't see how I can produce an element 1/4 in Z[1/2] by my above description of Z[1/2].

So I think what I am really trying to do is show that if a is any element of Z[1/2], then I can find some element b in Z[1/2] so that it is impossible to write

b = m*a

for any m in Z. This would then show that the Z-module Z[1/2] cannot be generated by any single element a in Z[1/2].

I will keep working on this. Thanks again Tonio.

**tonio** · December 5th 2010, 12:10 PM

Originally Posted by Jumping

Thank you for the help Tonio. My goal is to show that the subring Z[1/2] of Q, when considered as a Z-module, cannot be finitely generated. I figure I ought to first show that Z[1/2] cannot be generated by a single element from Z[1/2].

I think my confusion starts when I was writing on my scratch paper that

Z[1/2] = {a + b*(1/2) : a, b are in Z}.

This, of course, is false. We can define $\displaystyle{\mathbb{Z}\left[\frac{1}{2}\right]:=\{f(1/2)\,;\,\,f(x)\in\mathbb{Z}[x]\}$ , so for example, with $f(x)=x^4$ , we get

$\displaystyle{f\left(\frac{1}{2}\right)=\frac{1}{1 6}\in\mathbb{Z}\left[\frac{1}{2}\right]-\left\{a+b\cdot\frac{1}{2}\,;\,\,a,b\in\mathbb{Z}\ right\}}$

But, Z[1/2] is a subring of Q and so it also has a multiplicative structure. For example 1/2 is in Z[1/2] and so (1/2)(1/2) = 1/4 is also in Z[1/2]. But I don't see how I can produce an element 1/4 in Z[1/2] by my above description of Z[1/2].

So I think what I am really trying to do is show that if a is any element of Z[1/2], then I can find some element b in Z[1/2] so that it is impossible to write

b = m*a

for any m in Z. This would then show that the Z-module Z[1/2] cannot be generated by any single element a in Z[1/2].

I will keep working on this. Thanks again Tonio.

Do you know what an integral element is? And do you know that $\mathbb{Z}$ is an integrally

closed domain?

If you know the above then it is inmediate that $\mathbb{Z}\left[\frac{1}{2}\right]$ cannot be a f.g. $\mathbb{Z}-$ module,

lest $\frac{1}{2}$ is an integral non-integer rational number, which of course is absurd.

If you still don't know the above, then you'll have to work (slightly) harder: suppose the above ring is a f.g.

module over the integers, so that $\displaystyle{\mathbb{Z}\left[\frac{1}{2}\right]=\left\{\sum\limits^r_{k=1}m_kf_k(1/2)\,,\,\,m_k\in\mathbb{Z}\,,\,\,f_k(x)\in\mathbb{Z }[x]\right\}}$ , and let

$n:=\max\{\deg f_1,\,\deg f_2,\ldots,\deg f_r\}$ . Well, now I let you to prove the

easy fact that $\displaystyle{\frac{1}{2^{n+1}}}$ cannot be generated by the above "generators"...

Tonio

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