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 January 3rd, 2014, 08:03 AM #1 Senior Member   Joined: Feb 2013 Posts: 153 Thanks: 0 Maximal ideal Hi guys. I'm trying to prove next assertion: Let R be commutative ring with unity, and M ideal of that ring so that R/M is field. Prove that M is maximal ideal. I need help at the end of the proof. Here is my try: Let be the field. Then every element from (except 0) have an inverse element. Let . so that . Then . so that . Because we have . So we have . How can I now conclude that M is maximal ideal? Thanks! January 3rd, 2014, 09:26 AM #2 Senior Member   Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 Re: Maximal ideal I am not a specialist in this, but you can try the following. Assume that there is a larger proper ideal and pick . Show that if has an inverse in according to your conclusion, then contradicting that is a proper ideal. January 3rd, 2014, 11:35 AM #3 Senior Member   Joined: Feb 2013 Posts: 153 Thanks: 0 Re: Maximal ideal Evgeny, I'm not sure what I have to do... How contradicting that is proper ideal? January 3rd, 2014, 11:57 AM #4 Senior Member   Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 Re: Maximal ideal If and is a proper ideal, then . The problem is to show that . But this is possible if there exists a such that for some . January 3rd, 2014, 11:07 PM #5 Senior Member   Joined: Feb 2013 Posts: 153 Thanks: 0 Re: Maximal ideal As you said, I'm not a specialist in this... I dont understand. Can you explain this more detailed please. January 4th, 2014, 03:32 AM #6 Senior Member   Joined: Feb 2013 Posts: 153 Thanks: 0 Re: Maximal ideal Is following conclusion correct? We have (which is same thing as ). Because it's for every . Hence we have . But, there is only one ideal which can contain a unit. That is whole ring . So we have , which means is maximal ideal. January 4th, 2014, 06:44 AM   #7
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Re: Maximal ideal

Quote:
 Originally Posted by limes5 Is following conclusion correct?
Pretty much, I believe.

Quote:
 Originally Posted by limes5 Because it's for every .
regardless of whether . I assume is the principal ideal generated by .

Quote:
 Originally Posted by limes5 So we have , which means is maximal ideal.
It is here that we assume that . Then the smallest ideal containing and must also contain and therefore coincide with .

My reasoning was the following, continuing from my last post. Multiply by an arbitrary to get . Then for some , i.e., " />. Actually, now I am not sure that with various ranges over all elements of because also depend on . However, , so has a representative in each element of and is not restricted to .

Your solution seems clearer and more concise. January 6th, 2014, 03:43 PM #8 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: Maximal ideal Here is what I consider a simpler approach: Suppose we have an ideal M < M' in R. Consider the image in R/M of M' under the canonical ring homomorphism R-->R/M: this is an ideal of R/M, because: Given x,y in M', we have: (x+M) - (y+M) = (x-y) + M, which is in the image of M', since M' is an ideal (and thus x - y is in M'). Given any element a+M in R/M, and x in M', we have: (a+M)(x+M) = ax + M, and ax is in M', so M' is an ideal of R/M. But R/M is a field, and has only two ideals: R/M, and M (= (0) in R/M), because all non-zero elements are units. Since the containment M < M' is proper, we have x in M' with x + M not equal to M. Thus the image of M' under the surjection is all of R/M, which means that M', being the pre-image of the image, is all of R, so M is maximal. (In other words: use the morphisms, don't waste time investigating "certain elements". This is the key lesson of the fundamental homomorphism theorem: a lattice isomorphism between ideals of R/M and ideals of R that contain M). Tags ideal, maximal ,

### r has only two maximal ideal

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