Smallest subring

**HTale** · January 21st 2009, 11:24 AM

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.

**Opalg** · January 21st 2009, 12:30 PM

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.

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.)

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.

**HTale** · January 21st 2009, 12:59 PM

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

**ThePerfectHacker** · January 21st 2009, 01:58 PM

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.

**Opalg** · January 22nd 2009, 12:23 AM

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.

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