A subring of the reals???

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tom.young84
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A subring of the reals???

Postby tom.young84 » Tue Jan 26, 2010 1:44 am UTC

Given this set:

Q[a+b[imath]\sqrt[3]2[/imath]+c[imath]\sqrt[3]4[/imath] : a,b [imath]\in[/imath]Q}

I am able to show this is closed under addition, has additive inverse, and has additive identity. I am stuck proving that this is closed under multiplication.

As a side question, are the above four things sufficient to show that this is indeed a subring. We have not "learned" that showing that subtraction and multiplication are closed is sufficient to demonstrate something is a subring.

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Re: A subring of the reals???

Postby Qoppa » Tue Jan 26, 2010 1:53 am UTC

Remember: [imath]a^xb^x=(ab)^x[/imath]. That includes cases where [imath]x=1/3[/imath] (like you have). With this in mind, just multiply two generic elements and see that the product can be massaged into the desired form.

Code: Select all

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):0;%>O(x,l)??<_='['/7;{return!(x%(_-11))
?x??'l:x^(1+ ++l);}??>main(){t=&O;w=a();}

tom.young84
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Re: A subring of the reals???

Postby tom.young84 » Tue Jan 26, 2010 2:00 am UTC

Cool.

Finally, is it just four axioms that demonstrate that a subset is a subring?

Suffusion of Yellow
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Re: A subring of the reals???

Postby Suffusion of Yellow » Tue Jan 26, 2010 3:00 am UTC

tom.young84 wrote:Cool.

Finally, is it just four axioms that demonstrate that a subset is a subring?


Also it has to be non-empty, but that's trivial here.

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Re: A subring of the reals???

Postby skeptical scientist » Tue Jan 26, 2010 3:01 am UTC

tom.young84 wrote:Finally, is it just four axioms that demonstrate that a subset is a subring?

Well, since a subring is a subset of a ring, you can deduce most of the ring axioms because they hold in the larger structure. You really only need to check that contains the zero element and is closed under addition, multiplication, and taking additive inverses. If these three properties hold, all of the ring axioms can be shown to hold as well, and that's the real definition of a subring: a subset of a ring which is itself a ring under the inherited operations of addition and multiplication.

Suffusion of Yellow wrote:Also it has to be non-empty, but that's trivial here.

If it contains the additive identity/zero element, it is certainly non-empty.
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tom.young84
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Re: A subring of the reals???

Postby tom.young84 » Tue Jan 26, 2010 4:49 am UTC

This is just to make sure I did multiplication right:

ea+(af+eb)[imath]\sqrt[3]2[/imath]+(ag+bf+ec)[imath]\sqrt[3]4[/imath]+(bg+cf)[imath]\sqrt[3]8[/imath]+cg[imath]\sqrt[3]16[/imath]

Additionally...this looks like it will be a pain to prove it is a field

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Re: A subring of the reals???

Postby antonfire » Tue Jan 26, 2010 6:17 am UTC

This depends on your definitions, but if your rings are required to have an identity element, you also need to check that your subring contains 1.

Note that it's not enough to show that the subring has an identity. If your rings are required to have identity, {(n,0) | n in Z} is generally not considered to be a subring of Z x Z.
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tom.young84
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Re: A subring of the reals???

Postby tom.young84 » Tue Jan 26, 2010 6:20 am UTC

Don't I have to computer a solution to the equation ax=1.

As such, won't this inverse be obnoxious since there is a lot of book keeping.

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Re: A subring of the reals???

Postby t0rajir0u » Tue Jan 26, 2010 6:24 am UTC

It will be if you insist on writing everything out explicitly, but there are very general ways to prove statements like this that require much less work.

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Re: A subring of the reals???

Postby stephentyrone » Tue Jan 26, 2010 6:26 am UTC

tom.young84 wrote:Additionally...this looks like it will be a pain to prove it is a field


When did fields come into this? I thought you just had to show it was a subring?
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Re: A subring of the reals???

Postby tom.young84 » Tue Jan 26, 2010 6:34 am UTC

It was attached to the question. I was consumed by showing the explicit details of how multiplication is closed.

What are the general ways to show this is a field? I believed I calculated what the inverse would look like, but if there is one that requires far less work I would like to know for future reference.

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Re: A subring of the reals???

Postby antonfire » Tue Jan 26, 2010 6:55 am UTC

Hint: It is a finite-dimensional vector space over Q, so injective linear maps from itself to itself are also surjective.
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Re: A subring of the reals???

Postby edahl » Tue Jan 26, 2010 8:17 am UTC

I think the slickest way of checking if the subset E of a ring R is a subring is the following: For all x, y in E, x-y is in E, and xy is in E. If you want a ring with identity, substitute the last check with xy^(-1) edit: multiplication isn't necessarily a group. You should verify that this accounts for all the axioms of a ring, which yes you are correct about.
Last edited by edahl on Tue Jan 26, 2010 4:22 pm UTC, edited 1 time in total.

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Re: A subring of the reals???

Postby skeptical scientist » Tue Jan 26, 2010 2:41 pm UTC

edahl wrote:If you want a ring with identity, substitute the last check with xy^(-1).

There are things which are subrings which don't satisfy that last requirement. E.g. F[x2] ≤ F[x].
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Re: A subring of the reals???

Postby t0rajir0u » Tue Jan 26, 2010 2:54 pm UTC

You show that something is a field by verifying the field axioms. Anyway, I never understood this [imath]x - y[/imath] condition. It's usually just as hard to verify as showing closure under addition and additive inverse, and either way it doesn't take much time.

What I meant by my comment is that there are really slick ways to show closure under addition, multiplication, and division, but from the look of the OP it sounds like you haven't learned about quotients yet, so you should wait for that.

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Re: A subring of the reals???

Postby tom.young84 » Tue Jan 26, 2010 3:01 pm UTC

We have just finished up our summary of subrings. So I'm not sure of all the slick tricks.

Additionally, I have computed by hand what the solution to ax=1 would be (the condition to verify this is a field according to Introduction to Algebra). It's ugly and I'm not sure if I did it right:
[math]\frac{a-b[imath]\sqrt[3]2[/imath]-c[imath]\sqrt[3]4[/imath]}{a2-b[imath]\sqrt[3]4[/imath]-2bc[imath]\sqrt[3]8[/imath]-c2[imath]\sqrt[3]16[/imath]}[/math]

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Re: A subring of the reals???

Postby Cleverbeans » Tue Jan 26, 2010 3:33 pm UTC

Quick simplification - what's the cube root of 8?
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Re: A subring of the reals???

Postby tom.young84 » Tue Jan 26, 2010 3:37 pm UTC

2 !

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Re: A subring of the reals???

Postby skeptical scientist » Tue Jan 26, 2010 3:38 pm UTC

tom.young84 wrote:We have just finished up our summary of subrings. So I'm not sure of all the slick tricks.

Additionally, I have computed by hand what the solution to ax=1 would be (the condition to verify this is a field according to Introduction to Algebra). It's ugly and I'm not sure if I did it right:
[math]\frac{a-b[imath]\sqrt[3]2[/imath]-c[imath]\sqrt[3]4[/imath]}{a2-b[imath]\sqrt[3]4[/imath]-2bc[imath]\sqrt[3]8[/imath]-c2[imath]\sqrt[3]16[/imath]}[/math]

That doesn't look like an element of your subring. I mean, it might be, but how would you know since it doesn't have the correct form?
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Re: A subring of the reals???

Postby Cleverbeans » Tue Jan 26, 2010 3:39 pm UTC

Then you'll want to simplify the cube root of 16 as well.
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Re: A subring of the reals???

Postby edahl » Tue Jan 26, 2010 4:30 pm UTC

t0rajir0u wrote:You show that something is a field by verifying the field axioms. Anyway, I never understood this [imath]x - y[/imath] condition. It's usually just as hard to verify as showing closure under addition and additive inverse, and either way it doesn't take much time.

What I meant by my comment is that there are really slick ways to show closure under addition, multiplication, and division, but from the look of the OP it sounds like you haven't learned about quotients yet, so you should wait for that.


It's just a cool little thing you can do as an exercise (I'm taking abstract algebra now, and I thought it was a bit cool), so it's not something to pay too much attention to.

And yes, I did err somewhat in my post up there, so don't listen to that crap!


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