# Smallest subring

• Jan 21st 2009, 11:24 AM
HTale
Smallest subring
Hi guys,

I'm scheduled to see my lecturer about this problem, but he isn't available until next week, and I feel that I am likely to get a clearer explanation here.

This is for a Galois Theory course, and he states the following theorem,

Let $(S, +, \times)$ be a subring of $(R, +, \times)$ and fix $r \in R$. Then the smallest subring of $(R,+,\times)$ that contains both $S$ and $r$ is equal to

$\displaystyle \{ \sum^{i=n}_{i=0} s_ir^i : s_i \in S, n \in \mathbb{N}\}$

and then he writes this subring as $S[r]$.

Then, he writes an example, $S[r] = \mathbb{Q}[i]$, and concludes that

$\mathbb{Q}[i] = \displaystyle \{ r_0 + ir_1 : r_0, r_1 \in \mathbb{Q}\}$

This is fine. The problem is when he introduces subrings of two elements. He claims

$\mathbb{Q}[i, \sqrt{2}] = \displaystyle \{ r_0 + ir_1 + r_2\sqrt{2} + r_3\sqrt{2}i : r_0, r_1, r_2, r_3 \in \mathbb{Q}\}$

I don't understand how he has got to here. The second claim is

$\mathbb{Q}[i, \sqrt{2}] = \mathbb{Q}[i, \sqrt{2} + i]$

Intiutively, I think I have to prove $\mathbb{Q}[i, \sqrt{2}] \subseteq \mathbb{Q}[i, \sqrt{2} + i]$ and $\mathbb{Q}[i, \sqrt{2}+i] \subseteq \mathbb{Q}[i, \sqrt{2} ]$. The first I think is clear - I can't think of how to prove the second.

Sorry about the length of this question. Any help is much appreciated.

HTale.
• Jan 21st 2009, 12:30 PM
Opalg
Quote:

Originally Posted by HTale
Let $(S, +, \times)$ be a subring of $(R, +, \times)$ and fix $r \in R$. Then the smallest subring of $(R,+,\times)$ that contains both $S$ and $r$ is equal to

$\displaystyle \{ \sum^{i=n}_{i=0} s_ir^i : s_i \in S, n \in \mathbb{N}\}$

If a subring of R contains S and r then it has to contain all the powers of r, also their products by elements of S, and also any finite sum of such elements. So it has to contain all elements of the form $\sum s_ir^i$ (*).

On the other hand, provided that R is a commutative ring (the result wouldn't be true in a noncommutative ring, but if this is from a Galois theory course, presumably there is an underlying assumption that all rings will be commutative), the set of all elements of the form (*) is closed under addition and multiplication, and contains the identity element (since S does). Therefore it is a subring, and clearly it is the smallest subring containing S and r.

Quote:

Originally Posted by HTale
He claims

$\mathbb{Q}[i, \sqrt{2}] = \displaystyle \{ r_0 + ir_1 + r_2\sqrt{2} + r_3\sqrt{2}i : r_0, r_1, r_2, r_3 \in \mathbb{Q}\}$

I don't understand how he has got to here.

Similar sort of argument. Clearly $\mathbb{Q}[i, \sqrt{2}]$ has to contain all elements of that form. Conversely (with a bit more work) you can see that the set of all such elements is closed under multiplication. (It is obviously closed under addition.)

Quote:

Originally Posted by HTale
The second claim is

$\mathbb{Q}[i, \sqrt{2}] = \mathbb{Q}[i, \sqrt{2} + i]$

Intiutively, I think I have to prove $\mathbb{Q}[i, \sqrt{2}] \subseteq \mathbb{Q}[i, \sqrt{2} + i]$ and $\mathbb{Q}[i, \sqrt{2}+i] \subseteq \mathbb{Q}[i, \sqrt{2} ]$. The first I think is clear - I can't think of how to prove the second.

Yes you can. Just use the fact that √2 = (√2 + i) – i.
• Jan 21st 2009, 12:59 PM
HTale
Quote:

Originally Posted by Opalg
Yes you can. Just use the fact that √2 = (√2 + i) – i.

Thanks Opalg for your response. So, correct me if I'm wrong, what you're saying here is that I can write $\sqrt{2} \in \mathbb{Q}[i, \sqrt{2}]$, as a linear combination of some elements of $\mathbb{Q}[i, \sqrt{2}+i]$, to conclude that $\mathbb{Q}[i, \sqrt{2}+i] \subseteq \mathbb{Q}[i, \sqrt{2}]$?
• Jan 21st 2009, 01:58 PM
ThePerfectHacker
Quote:

Originally Posted by HTale
Hi guys,

I'm scheduled to see my lecturer about this problem, but he isn't available until next week, and I feel that I am likely to get a clearer explanation here.

This is for a Galois Theory course, and he states the following theorem,

Let $(S, +, \times)$ be a subring of $(R, +, \times)$ and fix $r \in R$. Then the smallest subring of $(R,+,\times)$ that contains both $S$ and $r$ is equal to

$\displaystyle \{ \sum^{i=n}_{i=0} s_ir^i : s_i \in S, n \in \mathbb{N}\}$

and then he writes this subring as $S[r]$.

Then, he writes an example, $S[r] = \mathbb{Q}[i]$, and concludes that

$\mathbb{Q}[i] = \displaystyle \{ r_0 + ir_1 : r_0, r_1 \in \mathbb{Q}\}$

This is fine. The problem is when he introduces subrings of two elements. He claims

Just remember this. If you have $S[r]$ then it means it must contain $r$, but if it contains $r$ then it contains, $r,r^2,r^3,r^4,r^5,...$ because $S[r]$ is a subring so it is closed. And so that is why it must contains elements of the form $a+br+cr^2+dr^3+...$.

EDIT: I did not see Opalg's post.
• Jan 22nd 2009, 12:23 AM
Opalg
Quote:

Originally Posted by HTale
Thanks Opalg for your response. So, correct me if I'm wrong, what you're saying here is that I can write $\sqrt{2} \in \mathbb{Q}[i, \sqrt{2}]$, as a linear combination of some elements of $\mathbb{Q}[i, \sqrt{2}+i]$, to conclude that $\mathbb{Q}[i, \sqrt{2}+i] \subseteq \mathbb{Q}[i, \sqrt{2}]$?

Yes. (Cool)