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The following is the main question I have encountered.

Let $R$ be a commutative ring and let $P$ be a prime ideal of $R$. Then $\frac{R}{P}$ is an integral domain.($R$ may not have the unity). If the title can be proved I will be done.

Example: $R$ = $2Z$ and $P$ = $6Z$ , $4+6Z$ is the unity in $\frac{2z}{6z}$

I can prove it when $\frac{R}{P}$ is finite. But I can not see the infinite case.

Can anyone help me by giving a hint.

Xam
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cmi
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3 Answers3

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Suppose $(a + P)(b + P) = 0 + P$ in $R/P$. Since $(a + P)(b + P) = ab + P$ we get that $ab \in P$. Now use the definition of a prime ideal.

The question used to be: "Let $R$ be a commutative ring and let $P$ be a prime ideal of $R$. Then $R/P$ is an integral domain." The OP changed the question, and now asks for a proof that $R/P$ has a unity even if $R$ does not. Note to OP: if one of the answers addresses the original question then you should accept it and post a new question rather than changing the question.

Edit: The below is incorrect since the ideal $P$ is not prime as asserted - hat tip @RobertLewis.

To answer the new question, $R/P$ need not have a unity if $R$ does not. Let $R = C_0(\mathbb{R})$, the ring of continuous functions on the real line which vanish at infinity. Let $P$ be the prime ideal in $R$ consisting of functions which vanish on the ray $[0, \infty)$. Consider the map $\phi \colon R \to C_0([0,\infty))$ which sends a continuous function on $\mathbb{R}$ to its restriction to the ray. It is clear that this is a ring homomorphism with kernel $P$, and it is surjective by the Tietze extension theorem. Thus $R/P \cong C_0([0, \infty))$, and this ring does not have a unity: the unity would have to take the value $1$ at every point, but the constant function $1$ does not vanish at infinity.

Paul Siegel
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  • I could not understand – cmi Jan 08 '18 at 03:56
  • @cmi I have updated my answer to address the new question. – Paul Siegel Jan 08 '18 at 07:04
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    I'm having a hard time seeing how $P$ is a prime ideal. Can't I pick $f, g \in R$ such that $fg = 0$ on $[0, \infty)$ but neither $f$ \nor $g$ do? Can't we take $f$ such that $f(1) = 1$ but $f(x) = 0$ for $x \notin (1 - \epsilon, 1 +\epsilon)$ for some positive $\epsilon < 1/4$, and then translate $f$ say, $10$ units to the right to get $g$? – Robert Lewis Jan 08 '18 at 07:16
  • How $P$ belongs to $R$? perhaps I did not understand the Ring $R$ properly..will $e^{-x}$ belong to $R$?@PaulSiegel – cmi Jan 08 '18 at 07:20
  • And I am quite sure that the statement is correct and thank you for your effort but ur earlier answer was wrong as u assumed that $R$ has unity.@PaulSiegel – cmi Jan 08 '18 at 07:26
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    Good point @RobertLewis - that kills the argument. – Paul Siegel Jan 08 '18 at 19:16
  • @PaulSiegel: Thanks for your response. I'm still looking into this. Cheers! – Robert Lewis Jan 08 '18 at 19:22
  • @cmi: The original argument - that $R/P$ has no zero divisors if $P$ is prime - is true weather or not $R$ has unity. I'm still not sure about your new claim, that $R/P$ has unity even if $R$ does not - what evidence do you have of this other than the finite case? – Paul Siegel Jan 08 '18 at 19:26
  • Yes My statement was wrong....Counter example - R=2Z and P=(0)..@PaulSiegel – cmi Jan 08 '18 at 19:29
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The statement you want to prove is false. It's essentially equivalent to: does a commutative ring with no nontrivial zero divisors have $1$? $2\Bbb{Z}$ is an easy counterexample.

Why are these equivalent? Well an ideal $P$ is prime iff $R/P$ has no nontrivial zero divisors, and if $R$ is a commutative ring without 1, with no nontrivial zero divisors, the $P=0$ is prime, and $R/P\cong R$ has no 1.

Then it is natural to ask the question, when does a commutative ring with no nontrivial zero divisors have 1?

Proposition: A nontrivial commutative ring with no nontrivial zero divisors has a 1 if and only if there are $a,b\ne 0\in R$ such that $ab=b$. Then $a$ is the unit.

Proof: If $R$ has 1, then $a=1$ works.

On the other hand, if $ab=b$ with $a\ne 0$, $b\ne 0$, then for any $c\ne 0$, $abc=bc$. Cancelling $b$, since $b\ne 0$, we thus have for all $c\ne 0$, $ac=c$. Thus if it's true for any $b\ne 0$ that $ab=b$, then it's true for all $b\ne 0$. Since it's always true that $a0=0$, we thus have that $a$ is the unit of $R$.

jgon
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OK, integral domains are characterized by the property that

$ab = 0 \Longleftrightarrow a = 0 \; \text{or} \; b = 0; \tag 1$

so if $P$ is a prime ideal in $R$, we look at

$ab + P = (a + P)(b + P) \in R/P; \tag 2$

if

$ab + P = 0 \in R/P, \tag 3$

then

$ab \in P; \tag 4$

thus, since $P$ is a prime ideal,

$a \in P \; \text{or} \; b \in P; \tag 5$

but

$a \in P \Longleftrightarrow a + P = 0 \in R/P, \tag 6$

with a similar assertion binding for $b$. It follows that either

$a + P = 0 \in R/P \; \text{or} \; b + P = 0 \in R/P, \tag 7$

and thus we have shown that

$ab + P = 0 \in R/P \Longleftrightarrow a + P = 0 \in R/P \; \text{or} \; b + P = 0 \in R/P, \tag 8$

which is precisely the criterion for $R/P$ to be an integral domain.

Nota Bene: I guess I would have to say this is a pretty strong hint! End of Note.

Robert Lewis
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