Hi Let be a ring, such that . Suppose that is finite. Prove that We can see that , so if , it's over. Any idea?
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Originally Posted by clic-clac Hi Let be a ring, such that . Suppose that is finite. Prove that We can see that , so if , it's over. Any idea? we have: hence, since is finite, there exist such that which gives us: now multiply both sides of (1) from the right by and note that then you'll get:
Last edited by NonCommAlg; November 16th 2008 at 01:02 PM.
Thanks for your answer. I don't see why there exists such that . The conclusion is still the same with for a but there is something I'm missing...
Originally Posted by clic-clac Thanks for your answer. I don't see why there exists such that . The conclusion is still the same with for a but there is something I'm missing... you're right, thanks! it was just one of those really weird mistakes that i make sometimes! it's fixed now!
You're welcomed. But for me the problem is that doing this, we assume that every element of is equal to a , . Why can we say that?
Originally Posted by clic-clac we assume that every element of is equal to a , . Why can we say that? i didn't say that! what we have is this: and we're given that is finite. so must also be finite and thus we can find two elements of which are equal.
Last edited by NonCommAlg; November 16th 2008 at 01:51 PM.
Great NonCommAlg, I finally got it! Thank you.
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