Subring of R

• Feb 10th 2010, 01:56 PM
zhupolongjoe
Subring of R
Let S be the set of all elements of the form a+(b*cubed root of 2)+(c*cubed root of 4) with a,b,c rationals. Show that S is a subring of R.

My book does not require multiplicative identity.

I think I can handle showing the zero element is there, and closure under addition, but I'm having trouble with closure under multiplication and additive identity. Thanks.
• Feb 10th 2010, 03:08 PM
Drexel28
Quote:

Originally Posted by zhupolongjoe
Let S be the set of all elements of the form a+(b*cubed root of 2)+(c*cubed root of 4) with a,b,c rationals. Show that S is a subring of R.

My book does not require multiplicative identity.

I think I can handle showing the zero element is there, and closure under addition, but I'm having trouble with closure under multiplication and additive identity. Thanks.

I'm assuming that this says $S=\left\{a+b\sqrt[3]{2}+c\sqrt[3]{4}:a,b,c\in\mathbb{Q}\right\}$. Closure under mult. is just a lot of work. But, what do you mean by additive identity? Isn't the zero element the additive identity?
• Feb 10th 2010, 03:17 PM
zhupolongjoe
Yes, that is the way it looks. Sorry, I meat additive inverse, not additive identity.
• Feb 10th 2010, 03:19 PM
Drexel28
Quote:

Originally Posted by zhupolongjoe
Yes, that is the way it looks. Sorry, I meat additive inverse, not additive identity.

$a+b\sqrt[3]{2}+c\sqrt[3]{4}+-a+-b\sqrt[3]{2}+-c\sqrt[3]{4}=0$?
• Feb 10th 2010, 03:47 PM
zhupolongjoe
I didn't think it would be that easy lol, but thanks...the multiplication is just troublesome.
• Feb 10th 2010, 03:52 PM
Drexel28
Quote:

Originally Posted by zhupolongjoe
I didn't think it would be that easy lol, but thanks...the multiplication is just troublesome.

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