With arbitrary presentations of the ring allowed, this problem has as a corner case the word problem for groups (<a href="http://en.wikipedia.org/wiki/Word_problem_for_groups">http://en.wikipedia.org/wiki/Word_problem_for_groups</a>). <div>
We take the ring to be K = CG, the group algebra over C of a group G. Then take the two elements in K to be the images under the natural inclusion of G in CG of two elements of G.<div><br></div><div>Regards,</div><div>Michael<br>
<br><div class="gmail_quote">On Sat, Jul 10, 2010 at 10:09 PM, Roman Beslik <span dir="ltr"><<a href="mailto:beroal@ukr.net">beroal@ukr.net</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;">
Hi.<div class="im"><br>
On 10.07.10 21:40, Grigory Sarnitskiy wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
I'm not very familiar with algebra and I have a question.<br>
<br>
Imagine we have ring K. We also have two expressions formed by elements from K and binary operations (+) (*) from K.<br>
</blockquote></div>
In what follows I assume "elements from K" ==> "variables"<div class="im"><br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
Can we decide weather these two expressions are equivalent? If there is such an algorithm, where can I find something in Haskell about it?<br>
</blockquote></div>
Using distributivity of ring you convert an expression to a normal form. "A normal form" is "a sum of products". If normal forms are equal (up to associativity and commutativity of ring), expressions are equivalent. I am not aware whether Haskell has a library.<br>
<br>
-- <br>
Best regards,<br><font color="#888888">
Roman Beslik.</font><div><div></div><div class="h5"><br>
<br>
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