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Been asked to show this is true with hints $R/I$ field $\Longleftrightarrow$ $I$ is maximal and $R/I$ integral domain $\Longleftrightarrow$ $I$ prime.

Can you check this please, I have had a ten month break from university mathematics and am returning shortly so need to know I'm on the right track). How do you format TeX on this site as well?

Let $R/I$ be a field (so neccessarily $I \neq R$) and $J$ be an ideal of $R$ where $I$ is a subset of $J$. If $I$ is not equal to $J$ then there exists an $x \in J\backslash I$ ( $J$ minus the set $I$) so the coset of $x$, $I + x$, is non-zero hence there is an element $a \in R$ such that $(I+a)(I+x) = (I+1)$ so $I + (ax - 1) = I$ hence $ax - 1$ is in $I$. But $x \in J$ and $J$ is an ideal so $ax \in J$ and as $I \subseteq J$, $ax - (ax-1) = 1 \in J$; this implies $J = R$.

Now, consider $I$ maximal and $(I+a) \neq I$. This means $a \notin I$ so consider the ideal $I + <a>$ (the principal ideal generated by $a$) which obviously contains $I$. It follows that the ideal must be $R$ so there exist $i \in I$, $r \in R$ such that the following holds; $$ i + ra = 1 $$

as $1 \in R$. Thus $I + (i+ra) = I + 1$, so $I + ra = I + 1 = (I+a)(I+r)$ hence $I+a \in R/I$ does indeed have an inverse if $a$ is non zero.

Let $R/I$ be an integral domain. Neccesarily $I \neq R$. Let $xy \in I$. Then $(I+xy) = I = (I+x)(I+y)$ so as $R/I$ is an integral domain this implies $x$ or $y$ is in $I$, hence $I$ is prime.

Now, let $I$ be prime. Suppose $(I+x)(I+y) = I$. Then $I+xy = I$ hence $xy$ is in $I$ so $x \in I$ or $y \in I$ so $I+x$ or $I+y$ is $I$ (which is zero).

A field is an integral domain thus

$I$ maximal implies $I$ is prime.

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Welcome to MSE =) To format TeX here just put what you want to code in math between \$ $cash \, signs \, like \, this$ \$. –  Patrick Da Silva Nov 29 '11 at 19:13
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4 Answers

up vote 2 down vote accepted

Yes, your proofs seem fine. You can make the proof of $R/I$ a field $\Leftrightarrow$ $I$ maximal a bit more compact by using the following propositions:

Proposition 1: If $R$ is commutative with identity then $R$ is a field if and only if the only ideals of $R$ are $0$ and $R$.

Proposition 2: Let $R$ be commutative with identity, and let $I$ be an ideal of $R$. Then every ideal of $R/I$ is of the form $J/I$ where $J$ is an ideal of $R$ and $I\subseteq J$. Conversely, if $J$ is an ideal of $R$ and $I\subseteq J$, then $J/I$ is an ideal of $R/I$.

The proofs are fairly basic (and canonical exercises), and I suggest giving them a try.

With the propositions above in hand, suppose $R/I$ is a field. Then the only ideals of $R/I$ are $I$ (ie $0+I$) and $R/I$. Therefore the only ideals of $R$ that contain $I$ are $I$ and $R$. Therefore $I$ is maximal.

On the other hand, suppose $I$ is maximal, and consider an ideal of $R/I$. This ideal must be of the form $J/I$ for $I\subseteq J \subseteq R$. However, as $I$ is maximal, we have either $J=I$ or $J=R$. Thus, the only ideals of $R/I$ are $0+I$ and $R/I$, hence $R/I$ is a field.

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Propsotion 1; If R is a field and R $diamond_arrow$I is not the zero ideal then there is a nonzero element x in I hence 1 is in I (R is a field so the multiplicative inverse of x exists in R so $x.x^{-1} = 1$ will be in I hence I = R. Assume now the only ideals of R are {0} and R. Let x be non-zero in R. Then $<x>$ is itself an Ideal of R hence contains 1 so there is a multiplicative inverse for x in R. Cheers Jack, just trying TeXing and getting used to basic arguments once more. –  Adam Nov 29 '11 at 20:36
    
I am confused by the notation of proposition 2 - is J/I the quotient ring formed from J and I? I have not seen such a cosntruction before, but I guess this is what you mean. –  Adam Nov 29 '11 at 20:36
    
Thanks for your help! –  Adam Nov 29 '11 at 20:37
    
@Adam - Yes, if $I\subseteq J$, then $J/I=\{j+i\,\vert\,j\in J,i\in I\}$ is exactly what you think it is: cosets of $R/I$ whose representatives lie in $J$. There are plenty of reasonably good tutorials on LaTeX (FYI, if you're googling anything with "LaTeX" in it and you need SFW results, then choose your search query carefully!), and you can always hit the edit button on any question or answer to see the LaTeX markup that the author used. –  user5137 Nov 29 '11 at 20:41
    
@Adam - You're most welcome. Since you're new to MSE, I also recommend that you consider accepting the answer to your question that you found the most helpful (and yes, it's okay if you don't mark my answer as accepted--no hard feelings. :) –  user5137 Nov 29 '11 at 20:43
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This is a prototypical example of modular reduction - which serves to simplify and/or reduce to canonical form. Here, reducing mod $\rm\:I\:,\:$ the problem simplifies to $(0)$ maximal $\implies (0)$ prime, i.e. field $\implies$ domain, a simple "canonical" form of this implication - packaged in a form for optimal reuse. It's important to bring this structure to the fore, and master these techniques in simple contexts such as here, since it will be much more difficult to learn in more complex complex contexts, where additional complexity obfuscates matters. You can find further discussion, including examples by the master algebraist Irving Kaplansky in my post here.

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There is an assumption of commutativity lurking around, namely in the equivalences given. E.g., the trivial ideal is maximal in the ring of $n\times n$ matrices with coefficients in a field, but the ring is not a field if $n\gt 1$.

For non-commutative rings with identity, recall that an ideal $P$ is a prime ideal if and only if for any two ideals $A$ and $B$, if $AB\subseteq P$ then $A\subseteq P$ or $B\subseteq P$. (This is equivalent to the element-wise condition in commutative rings, but for noncommutative rings, the element-wise condition is stronger; such ideals are called "completely prime").

To show that a maximal ideal must be prime, let $\mathfrak{M}$ be a maximal ideal. We show that if $A$ and $B$ are ideals and neither is contained in $\mathfrak{M}$, then their product is not contained in $\mathfrak{M}$.

If $A$ is not contained in $\mathfrak{M}$, then by maximality we have $A+\mathfrak{M}=R$. Similarly, $B+\mathfrak{M}=R$. Therefore, $$R = RR = (A+\mathfrak{M})(B+\mathfrak{M})=AB + A\mathfrak{M} + \mathfrak{M}B + \mathfrak{M}^2 \subseteq AB+\mathfrak{M}\subseteq R.$$ Therefore, $AB+\mathfrak{M}=R$, so $AB$ is not contained in $\mathfrak{M}$, as claimed. Thus, if $\mathfrak{M}$ is maximal, then it is prime.

Of course, this proof works equally well for commutative rings, and does not require the equivalence via structure of the quotients.


For completelyness:

Proposition. Let $R$ be a ring with identity, and let $\mathfrak{P}$ be an ideal.

  1. If for every $a$ and $b$ in $R$, $ab\in \mathfrak{P}\Rightarrow a\in\mathfrak{P}\text{ or }b\in\mathfrak{P}$, then $\mathfrak{P}$ is a prime ideal (if $A,B$ are ideals and $AB\subseteq \mathfrak{P}$, then $A\subseteq \mathfrak{P}$ or $B\subseteq \mathfrak{P}$).

  2. If $R$ is commutative, then the converse of (1) also holds.

Proof.

  1. Let $A$, $B$ be ideals such that $AB\subseteq \mathfrak{P}$; if $A\subseteq\mathfrak{P}$ we are done. Otherwise, let $a\in A$ such that $a\notin \mathfrak{P}$. Then for every $b\in B$ we have $ab\in AB\subseteq\mathfrak{P}$, hence $ab\in\mathfrak{P}$. By our assumption, $b\in\mathfrak{P}$. Thus, $B\subseteq \mathfrak{P}$, as claimed.

  2. Assume $R$ is commutative and that for any ideals $A$, $B$, if $AB\subseteq \mathfrak{P}$, then $A\subseteq \mathfrak{P}$ or $B\subseteq \mathfrak{P}$. Let $a,b\in R$ be such that $ab\in \mathfrak{P}$. Then $(a)(b) = (ab)\subseteq \mathfrak{P}$, so either $(a)\subseteq \mathfrak{P}$ or $(b)\subseteq\mathfrak{P}$. Hence either $a\in \mathfrak{P}$ or $b\in\mathfrak{P}$, as claimed. $\Box$

The assumption of commutativity in 2 is behind the equality $(a)(b)=(ab)$. In a noncommutative ring, $(a)$ consists of all sums of elements of the form $ras$ with $r,s\in R$, and similarly for $(b)$. The product will be generated by all elements of the form $rasbt$ with $r,s,t\in R$, and without commutativity we may be unable to rewrite them as sums of elements of the form $u(ab)v$.

For an example of a noncommutative ring and an ideal $\mathfrak{P}$ that is prime but not completely prime, let $S$ be the ring of $n\times n$ matrices over a field $F$. It's known that if $R$ is a ring with identity, then the ideals of $M_n(R)$, the ring of $n\times n$ matrices over $R$ are exactly of the form $M_n(\mathfrak{J})$, where $\mathfrak{J}$ is an ideal of $R$. So if $R=F$ is a field, then the only ideals of $R$ are the trivial ideal and the whole field, so the only ideals of $M_n(R)$ are the trivial ideal and the whole ideal. Therefore, the zero ideal is prime. However, you can find nonzero zero divisors in $S$, so you can find products of two elements which lie in the zero ideal with neither element being the zero element. So the zero ideal is prime but not completely prime.


For rings without identity things get more annoying. For example, consider the ring $2\mathbb{Z}$ of even integers. An ideal is maximal if and only if it is of the form $2p\mathbb{Z}$ for some prime $p$. In particular, $4\mathbb{Z}$ is maximal; but $4\mathbb{Z}$ is not a prime ideal, since $(2\mathbb{Z})(2\mathbb{Z}) \subseteq 4\mathbb{Z}$, but $2\mathbb{Z}$ is not contained in $4\mathbb{Z}$. Or take a ring with zero multiplication; the ideals correspond exactly to the subgroups of its additive structure, so maximal ideals correspond to maximal subgroups. However, no proper ideal is prime (under either definition).

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Thanks, this looks like a more useful form to do these calculations. –  Adam Nov 29 '11 at 22:33
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Everything's fine in your answer ; there is absolutely no problem in your proof. +1!

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Thanks for the replies, very quick and also I appreciate the edit - how are cosets typed in TeX? e.g. a with a bar on the top? Like the complex conjugate of a? –  Adam Nov 29 '11 at 20:17
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@Adam You have choices like \bar{a} $\bar{a}$ or \overline{a} $\overline{a}$. \overline works better when you want to put a bar over something longer than 1 character e.g. $\overline{abcde}$ vs $\bar{abcde}$. –  yunone Nov 29 '11 at 20:23
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