associates in integral domains
lite.o...@gmail.com
for some unit u)
a is element of R.
1) Let a be irreducible. Then a~b if and only if b is irreducible.
2) Let a be prime. Then a~b if and only if b is prime.
3) Let a be unit. Then a~b if and only if b is a unit.
Which of these is incorrect (which direction of the 3 biconditionals)?
Not asking for proof, just wondering if the above 3 are correct.
quasi
wrote:
They are all correct, and the proofs are instant -- try them.
quasi
quasi
Oops.
I read the statements too quickly.
The forward implications are all correct, that is:
Let a be irreducible. Then a ~ b => b is irreducible.
Let a be prime. Then a ~ b => b is prime.
Let a be a unit. Then a ~ b => b is a unit.
and, as previously noted, the proofs are instant.
For the reverse implications,
Let a be irreducible. Then b irreducible => a ~ b.
Let a be prime. Then b prime => a ~ b
Let a be a unit. Then b a unit => a ~ b.
the first two are silly (false in almost any ring -- try Z for
example), and the third one is trivially true.
quasi
lite.o...@gmail.com
Now, tell me if this is true or not (or how to modify it to make it
true).
If D is a commutative ring and D[t] is a P.I.D., then D is a field.
What is your favourite reference for algebra?
quasi
wrote:
>Yes, thank you quasi.
>
>Now, tell me if this is true or not (or how to modify it to make it
>true).
>
>If D is a commutative ring and D[t] is a P.I.D., then D is a field.
Yes, it's true.
Suppose D[t] is a PID, and d in D is a nonzero nonunit element of
Consider the ideal
I = (d,t).
By hypothesis, I is principal, so we have I = (h), for some h in D[t].
Then d,t are in (h).
You take it from there.
>What is your favourite reference for algebra?
There are lots of great algebra textbooks, but none that I would
regard as a single, all-in-one reference.
quasi
Arturo Magidin
Only one of the biconditionals is true; in the other two cases, only
one of the two implications holds, the other one does not.
Here's a hint: try it in a familiar setting, say, the
integers. Extrapolate.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
Muhammad
> In article <d5c77686-b2fa-4f4f-a022-15b99bec4...@v72g2000hsv.googlegroups.com>,
Am I missing something here? From what I know for an integral domain R
(commutative ring without zeo divisors and with 1 =! 0), a~b iff
aR = bR. A nonzero element p of R is a prime if (and only if) pR is a
prime ideal. (So -3 is a prime in Z the ring of integers.)Also an
element a is a nonunit if and only if aR =! R. Finally a nonunit is
irreducible if and only if a is not expressible as a product of two or
more nonunits. (A tongue in the cheek definition of an irreducible
element is: a is irreducible if a = xy implies a divides x or a
divides y.)
Muhammad
Bill Dubuque
>mag...@math.berkeley.edu (Arturo Magidin) wrote:>> (i.e, a = bu for some unit u). a is element of R.
>>>
>>> 1) Let a be irreducible. Then a~b <=> b is irreducible.
>>> 2) Let a be prime. Then a~b <=> b is prime.
>>> 3) Let a be unit. Then a~b <=> b is a unit.
>>>
>>> Which of these is incorrect (which direction of the 3 biconditionals)?
>>> Not asking for proof, just wondering if the above 3 are correct.
>>
>> Only one of the biconditionals is true; in the other two cases, only
>> one of the two implications holds, the other one does not. Here's a
>> hint: try it in a familiar setting, say, the integers. Extrapolate.
>
Beware that various notions of "associate" equivalent in domains
do not generally remain equivalent in rings, e.g. see [1], [2].
> A non0 element p of R is a prime if (and only if) pR is a prime ideal.
> An element a is a nonunit if and only if aR =! R. Finally a nonunit is
> irreducible iff its not expressible as a product of two or more nonunits.
Arturo's reply is correct, so it's not clear what your point is.
> (A tongue in the cheek definition of an irreducible element is:
> a is irreducible if a = xy implies a|x or a|y.)
Why tongue-in-cheek? It's useful pedagogically in order to make clear
the implication prime => irreducible, e.g. from one of my 2005 posts:
for nonunit p
p irreducible iff p = ab => p|a or p|b
p prime iff p | ab => p|a or p|b
Thus p prime => p irreducible
follows immediately from p = ab => p|ab
--Bill Dubuque
[1] http://google.com/groups?threadm=y8zznru8v1d.fsf%40nestle.ai.mit.edu
[1] When are Associates Unit Multiples?
D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles
Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.
http://math.la.asu.edu/~rmmc/rmj/vol34-3/andepage1.pdf
http://projecteuclid.org/handle/euclid.rmjm/1181069828
Bill Dubuque
>
> tell me if this is true or not (or how to modify it to make it true).
>
The same question was asked here last week, see my simple hint there:
http://google.com/groups?threadm=y8zljx0c7gp.fsf%40nestle.csail.mit.edu
--Bill Dubuque
Muhammad
> Muhammad <mzafrul...@usa.net> wrote:
> >magi...@math.berkeley.edu (Arturo Magidin) wrote:
> >><lite.on.b...@gmail.com> wrote:
>
> >>> R is integral domain. let a~b mean a,b are associates
> >> (i.e, a = bu for some unit u). a is element of R.
>
> >>> 1) Let a be irreducible. Then a~b <=> b is irreducible.
> >>> 2) Let a be prime. Then a~b <=> b is prime.
> >>> 3) Let a be unit. Then a~b <=> b is a unit.
>
> >>> Which of these is incorrect (which direction of the 3 biconditionals)?
> >>> Not asking for proof, just wondering if the above 3 are correct.
>
> >> Only one of the biconditionals is true; in the other two cases, only
> >> one of the two implications holds, the other one does not. Here's a
> >> hint: try it in a familiar setting, say, the integers. Extrapolate.
>
> > Am I missing something here? From what I know for an integral domain R
> > (commutative ring without zero divisors and 1 =! 0), a~b iff aR = bR.
>
> Beware that various notions of "associate" equivalent in domains
> do not generally remain equivalent in rings, e.g. see [1], [2].
The question was for R an integral domain! So, your warning is
unwarranted.
>
> > A non0 element p of R is a prime if (and only if) pR is a prime ideal.
> > An element a is a nonunit if and only if aR =! R. Finally a nonunit is
> > irreducible iff its not expressible as a product of two or more nonunits.
>
> Arturo's reply is correct, so it's not clear what your point is.
If you claim that, then you should provide counter-examples, using
elements of an integral domain, for the implications that do not hold.
It is a good thing we are talking Mathematics, you know how to correct
me.
>
> > (A tongue in the cheek definition of an irreducible element is:
> > a is irreducible if a = xy implies a|x or a|y.)
>
> Why tongue-in-cheek? It's useful pedagogically in order to make clear
> the implication prime => irreducible, e.g. from one of my 2005 posts:
> for nonunit p
>
> p irreducible iff p = ab => p|a or p|b
>
> p prime iff p | ab => p|a or p|b
>
> Thus p prime => p irreducible
>
> follows immediately from p = ab => p|ab
I have no problem with that, the tongue in the cheek part comes from
this definition, of irreducible, being close to the definition of a
prime. When used without warning it does have an interesting effect.
(I once heard someone say: But that is how you define a prime! Of
course after a while the guy calmed down.)
>
> --Bill Dubuque
>
> [1]http://google.com/groups?threadm=y8zznru8v1d.fsf%40nestle.ai.mit.edu
>
> [1] When are Associates Unit Multiples?
> D.D. Anderson, M. Axtell, S.J. Forman, and Joe Stickles
> Rocky Mountain J. Math. Volume 34, Number 3 (2004), 811-828.http://math.la.asu.edu/~rmmc/rmj/vol34-3/andepage1.pdfhttp://projecteuclid.org/handle/euclid.rmjm/1181069828
The paper that you mention in the reference is a good paper, but as I
pointed out before, the OP had a question about elements of a
particular integral domain where an associate means what the OP said.
Muhammad
Muhammad
> R isintegraldomain. let a~b mean a,b areassociates(i.e, a = bu
> for some unit u)
>
> a is element of R.
> 1) Let a be irreducible. Then a~b if and only if b is irreducible.
> 2) Let a be prime. Then a~b if and only if b is prime.
> 3) Let a be unit. Then a~b if and only if b is a unit.
>
> Which of these is incorrect (which direction of the 3 biconditionals)?
> Not asking for proof, just wondering if the above 3 are correct.
correctness I am including below the proofs.
R is integral domain. let a~b mean a,b are associates (i.e, a = bu
for some unit u)
a is element of R.
1) Let a be irreducible. Then a~b if and only if b is irreducible.
product of two nonunits x,y. I.e. b=xy. But then a =bu =xyu = (x)(yu)
a product of two nonunits. Conversely suppose that b is irreducible
and suppose by way of contradiction that a is not and say a = rs where
r and s are nonunits. Then rs =bu which gives rs(u^{-1})=
r(s(u^{-1}))= b which contradicts the assumption that b is
irreducible.
2) Let a be prime. Then a~b if and only if b is prime.
bR. So a being a prime implies that b is a prime the converse can be
proved in the same manner.
3) Let a be unit. Then a~b if and only if b is a unit.
x is a unit iff xR = R.
Which of these is incorrect (which direction of the 3
biconditionals)?
Not asking for proof, just wondering if the above 3 are correct.
made you ask those questions in the first place.
Muhammad
quasi
<mzafr...@usa.net> wrote:
>On Oct 13, 10:48 pm, lite.on.b...@gmail.com wrote:
>> R is integral domain. let a~b mean a,b are associates (i.e, a = bu
>> for some unit u)
>>
>> a is element of R.
>> 1) Let a be irreducible. Then a~b if and only if b is irreducible.
>> 2) Let a be prime. Then a~b if and only if b is prime.
>> 3) Let a be unit. Then a~b if and only if b is a unit.
>>
>> Which of these is incorrect (which direction of the 3 biconditionals)?
>> Not asking for proof, just wondering if the above 3 are correct.
>
>All are correct.
I think you are misinterpreting the biconditionals.
Each of the statements has the form:
Let A. Then B <=> C.
The forward direction is:
Let A. Then B => C.
The reverse direction is:
Let A. Then C => B.
With that interpretation, the statements
(1) Let a be irreducible. Then a~b if and only if b is irreducible.
(2) Let a be prime. Then a~b if and only if b is prime.
are true in the forward direction, but the reverse direction fails
badly for most rings (for example Z).
quasi
Muhammad
> On Thu, 16 Oct 2008 21:54:10 -0700 (PDT), Muhammad
>
You are right. If b is a prime and a is a prime it does not mean that
a~b, unless in a domain with only one prime. I did overlook that.
Same with irreducible. I think I should retire.
Oh, and I do apologize to every one involved.
Muhammad
Muhammad
> Muhammad <mzafrul...@usa.net> wrote:
> >magi...@math.berkeley.edu (Arturo Magidin) wrote:
>
> - Show quoted text -
Sorry about6 that. I was not thinking straight.
Muhammad
Muhammad
> In article <d5c77686-b2fa-4f4f-a022-15b99bec4...@v72g2000hsv.googlegroups.com>,
>
Arturo, you were right and I was wrong. I must not be thinking
straight these days.
Muhammad
Muhammad
Arturo's answers was right and mine wrong.
Muhammad