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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 $R/M$be the field. Then every element from $R/M$ (except 0) have an inverse element. Let $x+M \in R/M$. $\left(\exists y \in R \right)$ so that $(x+M)(y+M)=1+M$. Then $xy+M=1+M$. $(\exists m^{'},m^{''} \in M)$ so that $xy+m^{'}=1+m^{'#39;}$ . Because$m^{'}-m^{''}=m \in M$ we have $1-xy=m \in M$. So we have $1-xy \in M$. 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 $M'$ and pick $x\in M'\setminus M$. Show that if $x$ has an inverse in $R/M$ according to your conclusion, then $xR=R$ contradicting that $M'$ 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 $xR=R$ contradicting that $M'$ 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 $x\in M'$ and $M'\ne R$ is a proper ideal, then $xR\subseteq M'R=M#39;\subset R$. The problem is to show that $xR=R$. But this is possible if there exists a $y$ such that $xy+m_1=1+m_2$ for some $m_1,m_2\in M$.
 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 $1-xy \in M$ (which is same thing as $xy-1 \in M$ ). Because $x \notin M$ it's $xy \in (x)$ for every $y \in R$. Hence we have $1 \in (x)+M$. But, there is only one ideal which can contain a unit. That is whole ring $R$. So we have $(x)+M=R$, which means $M$ is maximal ideal.
January 4th, 2014, 06:44 AM   #7
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Joined: Dec 2013
From: Russia

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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 $x \notin M$ it's $xy \in (x)$ for every $y \in R$.
$xy\in (x)$ regardless of whether $x\in M$. I assume $(x)$ is the principal ideal generated by $x$.

Quote:
 Originally Posted by limes5 So we have $(x)+M=R$, which means $M$ is maximal ideal.
It is here that we assume that $x\notin M$. Then the smallest ideal containing $x$ and $M$ must also contain $(x)+M$ and therefore coincide with $R$.

My reasoning was the following, continuing from my last post. Multiply $xy+m_1=1+m_2$ by an arbitrary $z\in R$ to get $x(yz)+m_1z=z+m_2z$. Then $x(yz)+m_1'=z+m_2#39;$ for some $m_1',m_2'\in M$, i.e., $x(yz)=z+(m_2'-m_1#39" />. Actually, now I am not sure that $x(yz)$ with various $z$ ranges over all elements of $R$ because $m_1',m_2'$ also depend on $z$. However, $m_2'-m_1'\in M$, so $x(yz)$ has a representative in each element of $R/M$ and is not restricted to $M'$.

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).

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