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  
Senior Member Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108  Re: Maximal ideal Quote:
Quote:
Quote:
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) = (xy) + 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 nonzero 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 preimage 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). 

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