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Let $R$ be a commutative ring with identity. Show that $R$ is a field if and only if the only ideals of $R$ are $R$ itself and the zero ideal $(0)$.

I can't figure out where to start other that I need to prove some biconditional statement. Any help?

Santropedro
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Jackson Hart
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4 Answers4

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From some of the comments above you seem a little confused. Since you said you are not familiar with proofs I will try to write this out in a way that you can understand.

You are trying to prove the equivalence of the following statements:

$P:$ A commutative ring $R$ with $1$ is a field.

$Q:$ The only ideals of $R$ are $(0)$ and $(1)$.

Let us look at statement $Q$ closely. Well it is saying that the only ideals of $R$ are the zero ideal (which has only one element, zero) and $(1)$. What is $(1)$? Well by definition of an ideal if you multiply anything in the ideal $(1)$ by anything in $R$, you should get something back in $(1)$ again. But then $1$ is the multiplicative identity of $R$ so multiplying everything in $R$ by $1$ just gives everything in $R$ back again. This means that $(1)$ must be the whole ring $R$.

Now suppose we want to prove $P \implies Q$. Let $I$ be an ideal of a ring $R$. Here "ring" means "commutative ring with a unit". Now here are some things you should know:

(1) $I$ is non-empty

(2) $I$ must at least contain the element $0$ (Why?)

(3) If $I$ has more than one element this means that at least one non-zero element $a$ of the ring must be in $I$. (Why?)

Therefore if $I$ contains only $0$, $I = (0)$. If $I$ does not only contain zero, then by (3) above it contains at least one non-zero element $a$ of $R$. Now recall that we are trying to prove $P \implies Q$. We already know $P$. Therefore this means by definition of a field that $a^{-1}$ exists in $R$.

But then by definition of an ideal $I$, $a^{-1} a = 1$ must be in $I$. Therefore $1 \in I$ so that $I$ must be the whole ring $R$. Hence $I = (1)$. This establishes $P \implies Q$.

Now for the converse:

To show $Q \implies P$ it suffices to show that non-zero every element $a \in R$ contains a multiplicative inverse. So let $a$ be a non-zero element of $R$. The trick now is to consider the principal ideal generated by $a$ (which we denote by $(a)$ ).

Now by assumption of $Q$, since the only ideals of $R$ are $(0)$ and $(1)$, this means that $(a)$ being an ideal of $R$ must be either $(0)$ or $(1)$. Now $(a)$ cannot be $(0)$ for $a \neq 0$. So $(a) = (1)$. But then this means that $1$ is a multiple of $a$, viz. there exists a non-zero $c$ such that

$$ac = 1.$$

However this is precisely saying that $a$ has a multiplicative inverse. Since $a$ was an arbitrary non-zero element of $R$, we are done. Q.E.D.

Does this help you? I can discuss more if you need help.

6

Let $a\in R$, $a\neq 0$. Then the smallest ideal of $R$ that contains $a$ is $Ra=\{ra\mid r\in R\}$.

(Verify that the set $\{ra\mid r\in R\}$ is an ideal of $R$, and that it contains $a$; it's not hard. Then think about why any ideal that contains $I$ must contain this set.).

If $R$ is a field, then what is $Ra$ for any $a\neq 0$? If $I$ is any ideal of $R$, and $a\in I$ with $a\neq 0$, what does that tell you?

Conversely, if the only ideals of $R$ are $(0)$ and $R$, what is $Ra$? What does that tell you about $a$?

Arturo Magidin
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  • Since the asker had no idea where to start, I suspect they may not know that $Ra$ is an ideal. Drew sam, can you prove that? –  Jan 21 '12 at 23:05
  • Ummm, for the first part, I am not sure what you mean. You just multiply the field by that value? Is it just the identity? – Jackson Hart Jan 21 '12 at 23:07
  • And for the second part, is a just e? – Jackson Hart Jan 21 '12 at 23:08
  • @Drew, I don't know what you mean by "e". Arturo has written in his answer what he understands by $Ra$ and by $a$. Just read it carefully. –  Jan 21 '12 at 23:12
  • @drew sam: I explicitly said what "$Ra$" signifies. It is the set of all elements of the form $ra$, with $r\in R$. No, it's not just the identiy, it's *all products of elements of $R$ by $a$"*. And I don't know what $e$ is; do you mean $1_R$? No, it doesn't tell you that $a$ is equal to $1_R$, but it does tell you that you can *get* $1_R$ by multiplying $a$ by something. What does *that* tell you? – Arturo Magidin Jan 21 '12 at 23:13
  • I am thinking, I am sorry, this problem is giving me a lot of trouble and am not that familiar with proofs – Jackson Hart Jan 21 '12 at 23:18
  • @drew sam: Do you know that for a ring with $1$, an ideal is the whole ring if and only if it contains $1$? – Arturo Magidin Jan 21 '12 at 23:19
  • right, i guess that makes sense – Jackson Hart Jan 21 '12 at 23:20
  • does that mean a in in R? – Jackson Hart Jan 21 '12 at 23:21
  • And so, if $R$ is a field, and $I$ is an ideal, and it contains some $a\neq 0$, what do you know about nonzero elements in fields? And what do you know about multiplying elements in an ideal by elements of the ring? And what does that tell you about $I$? Conversely, if for any $a\in R$, $a\neq 0$, you know that $a\in Ra$, and that $Ra$ is an ideal, then $Ra\neq (0)$. So that means that $Ra$ is *what*? And what does that tell you about $Ra$? And about $a$? – Arturo Magidin Jan 21 '12 at 23:22
  • @drewsam: $a$ is in $R$ by hypothesis. Instead of just throwing out statements at random, try to think things through. – Arturo Magidin Jan 21 '12 at 23:23
  • I know that nonzero elements in a field have a multiplicative inverse – Jackson Hart Jan 21 '12 at 23:26
  • And ideals absorb elements of rings under multiplication – Jackson Hart Jan 21 '12 at 23:26
  • @drew: So, if $R$ is a field, and $a\neq 0$, then $a$ has a multiplicative inverse. What happens when you multiply an element by its multiplicative inverse? What can you say about the result of multiplying *any* element of an ideal $I$ by *any* element of the ring $R$? So, if $I$ is an ideal in $R$, and $a\in I$ is not zero, what can you say about $I$? What does that mean? Don't stop at the first question in my sequence of questions, try to go through all of them. – Arturo Magidin Jan 21 '12 at 23:27
  • Multiplying by the inverse yields the identity. So the ideals absorbs the elements? This means that I absorbs all elements, right? – Jackson Hart Jan 21 '12 at 23:31
  • @drew sam: You're just not putting things together, and are hoping I will put them together for you. It's not my homework, it's yours. Here is what you know (for one direction): If $R$ is a field, then every nonzero element has a multiplicative inverse. If $I$ is an ideal, and it's not the zero ideal, then you need to show that $I=R$. If $I$ not the zero ideal, then it contains a nonzero element. It also contains the product of any element in the ring by an element of the ideal. And if it contains the identity, then it contains everything. If it contains everything, then it *is* all of $R$. – Arturo Magidin Jan 21 '12 at 23:34
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    I am sorry, I do not expect you to do it, that's not what I wanted. I have a learning disability to it takes me really long to understand things. I am sorry I will try and work harder – Jackson Hart Jan 21 '12 at 23:36
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    @drew sam: If that's the case, then let me suggest that this website and quick interaction through comments is *not* an appropriate forum for you; you should try to find one-on-one, personal help at your institution instead, or try to think through things before posting quick follow-up questions as you have done here. – Arturo Magidin Jan 21 '12 at 23:37
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It is given that the $0$-ring $\{0\}$ and the whole ring $R$ are ideals of any ring $R$. (Here I assume $R = \mathbb{Q}$, the rationals)

Suppose by contradiction that there is another ideal $N$ of $Q$ that is not $\{0\}$ or all of $Q$.

$N \neq \{0\}$ implies there is a nonzero rational number $a$ contained in $N$. Since $\frac{1}{a}$ is contained in $\mathbb{Q}$, the ideal $N$ must contain $(\frac{1}{a})\cdot a = 1$.

$N \neq Q$ implies that there is a rational number $b$ which is not contained in $N$. But $N$ contains $1$, so $N$ must also contain $b\cdot 1 = b$ by the definition of an ideal, contradicting the assumption that $N$ is not all of $\mathbb{Q}$.

Therefore there are no proper, non-trivial ideals of $\mathbb{Q}$.

Julian Kuelshammer
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  • For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latex). – Julian Kuelshammer Nov 02 '12 at 19:57
  • Also this answers only one direction of the question (only for $\mathbb{Q}$, but that generalises easily). – Julian Kuelshammer Nov 02 '12 at 20:00
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Hint $\ $ field $\rm R\! \iff 0\:|\:r\ \ or\ \ r\:|\:1\! \iff (0)\supset (r)\ \ or\ \ (r)\supset (1)\!\iff\! (0)\supset I\ \ or\ \ I\supset (1)$

Alternatively it is the special case $\rm\: I = 0\:$ of: $\, $ field $\rm\: R/I \iff I\:$ maximal $ $ (proof).

Bill Dubuque
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