I'm assuming "the ideal statement" means "If ##\{\pi{x}+B{y}: x,y\in{R}\}=\{gr : r\in{R}\}## then ##gcd(\pi,B)=g##.
This is equivalent to "If d is a common divisor of a and b, and every common divisor of a and b divides d, then d is called a greatest common divisor of a and b."
Because if...
But how do I say ##1## is the greatest common divisor then if ##1## and any unit ##u## generate the same ideal? How can I say ##1## is greater than ##u##?
Also, why do I even need the fact that if ##gcd(\pi,B)=g##, then ##g## is the generator of ##(\pi,B)##? Is it not enough just to show why...
The ideal ##(\pi,B)=\{\pi{x}+By : x,y\in{R}\}## Since ##I## is a PID, we know ##(\pi,B)## has a generator ##g## and you claim (well, not claim-- know) that ##gcd(\pi,B)=g##.
So we want a ##g## such that ##\{gr: r\in{R}\}=\{\pi{x}+B{y} : x,y\in{R}\}##.
Additionally, ##g|\pi## and ##g|B##...
Alright how is this...
Assume BWOC that ##gcd(\pi,B)=d\ne{1}##
Then by the theorem I mentioned in post #4, there do not exist ##x,y\in{R}## such that ##1=x{\pi}+y{B}##.
Since ##\pi## is irreducible, ##d|\pi## implies ##d## is either a unit (has multiplicative inverse), or is a multiple of...
One problem is in my book, the proof of this statement for PID's...
... uses the theorem that this post is about.
Another problem is...
...I don't know if this is necessarily true in general for PID's. Because
##k|\pi## implies ##kt=\pi## for some ##t\in{R}##, and ##\pi|k## implies...
By a theorem in my book, ##gcd(\pi,B) ≠ 1## implies there do not exist ##\sigma,\tau\in{R}## such that ##1=\sigma(\pi)+\tau(B)##.
Assume BWOC that ##gcd(\pi,B) = d## for some ##d\ne{1}##. Well then ##d|\pi## and ##d|B##, so ##\exists{k,t}\in{R}## s.t. ##dt=\pi## and ##dk=B##. ##dt=\pi## implies...
If S is a subring of Z then it either contains a nonzero element in which case it contains all multiples of that element and is of infinite size, or it is just {0} in which case it is size 1.
Is that right?
I was also wondering if you could take a look at my question about PID's? :shy: [edit...
Any subring S of the integers must contain the identity (the integer 1 in this case), but by existence of additive inverses (which S being a subring gives us) if 1 is in there then -1 is in there, and that means S=##\mathbb{Z}##.
Is this right? Also, out of curiousity, do some books not...
Oh and I forgot to add, ##R## being a PID (principle ideal domain) means ##R## is a domain and every ideal in ##R## is a principle ideal.
##R## being a domain means ##R## is a commutative ring (rings have an identity in my book, so even integers is not considered a ring) that has the...
By this do you mean if I insist that subrings of a ring R must have the same identity element as R?
And I agree that proper ideals are not subrings because they don't contain the identity.
2 elements? Being the equivalence classes of even and odd integers? [0] and [1]...
If that's correct...
Thanks, I took out the part about commutative rings having multiplicative inverses. I believe the rest of the definitions are correct?
And do you think the right way to go about this problem is to use an argument about the number of elements in ##\frac{R}{I}## when ##I## is a proper ideal...
Homework Statement
If ##I## is a proper Ideal in a commutative ring ##R##, then ##R## has a subring isomorphic to ##\frac{R}{I}##.
Book says false...
Homework Equations
##I## being a proper ideal in ##R## means ##I## is a proper subset of ##R## where
(i) if ##a,b\in{R}## then ##a+b\in{R}##...
Homework Statement
Let ##R## be a PID and let ##\pi\in{R}## be an irreducible element. If ##B\in{R}## and ##\pi\not{|}B##, prove ##\pi## and ##B## are relatively prime.
Homework Equations
##\pi## being irreducible means for any ##a,b\in{R}## such that ##ab=\pi##, one of #a# and #b# must be...
A homomorphism could me mapping every polynomial to its constant term (term without an x).
Say ##P_{1}(x)=c_{n}x^n+...+c_{1}x+c_{0}## and
##P_{2}(x)=k_{m}x^n+...+k_{1}x+k_{0}##
So ##f(P_{1}(x)P_{2}(x))=c_{0}k_{0}=f(P_{1}(x)f(P_{2}(x))##
Also...
Homework Statement
True or False?
Let R and S be two isomorphic commutative rings (S=/={0}). Then any ring homomorphism from R to S is an isomorphism.
Homework Equations
R being a commutative ring means it's an abelian group under addition, and has the following additional properties...
Thanks for the response.
If we could show Frac(R) is countable, we could conclude Frac(R)=/=reals because the reals are uncountable.
How can we show Frac(R) is countable? My friend said a way to show countability is to find a bijection between Frac(R) and the natural numbers...
Could...
Homework Statement
If R=\mathbb{Q}[\sqrt{2}], then Frac(R)=\mathbb{R}
Homework Equations
\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2} | a,b\in{\mathbb{Q}}\}
Frac(R) is the fraction field of R is basically \{\frac{a+b\sqrt{2}}{c+d\sqrt{2}} | a,b,c,d\in{\mathbb{Q}}\}.
The Attempt at a Solution
Back...
But we need f(a)^2=f(a) for all real numbers a. If we make f(a)=a, then in order for f(a)^2=f(a), we would need a^2=a which is clearly not true in general...
Can you elaborate some more on what you mean? I think I missed it...
Homework Statement
Show there exists a function f: \mathbb{R} \rightarrow \mathbb{R} s.t. f^2=f but f\neq{0,1}.
Homework Equations
Here f^2=f means for arbitrary a\in{\mathbb{R}}, f(a)^2=f(a)
The Attempt at a Solution
I came up with the function f(a)= \begin{cases}
0, & \text{if }a\text{>...
Homework Statement
Prove that every real number x in [0,1] has a decimal expansion.
Homework Equations
Let x\in{[0,1]}, then the decimal expansion for x is an infinite sequence (k_{i})^{\infty}_{i=1} such that for all i, k_i is an integer between 0 and 9 and such that...
Assume BWOC that g^k doesn't generate H. Then there is an element g^t in H such that k doesn't divide t. But that means t=kq+r where r<k. Which means g^r is in the H but that contradicts k being the smallest power of g in H.
Thanks!
Homework Statement
Are all subgroups of a cyclic group cyclic themselves?
Homework Equations
G being cyclic means there exists an element g in G such that <g>=G, meaning we can obtain the whole group G by raising g to powers.
The Attempt at a Solution
Let's look at an arbitrary...
I should have said: "If an element has 2 square roots then two elements in G share a square root."
The reason I feel that's true is because since G is finite, there are only |G| possible options for an element's square root. And there are also |G| elements each of which has a square root. So...
I don't get what you mean by this. Do you mean prove there is no element y such that y is the square root of x1 and x2, x1=/=x2? Or do you mean prove that for any x, the "square root of x" is well-defined?
I don't see what's wrong about what I did? I assumed BWOC that there existed an element...
Oh whoops! I was doing e^{2\pi i+pd} rather than e^{2\pi i(pd)}.
So what euler's theorem actually says is:
e^{2\pi i(pd)}=cos(2(pd)\pi)+isin(2(pd)\pi) and if we let d be the denominator of p, pd will be an integer so cos(2(pd)\pi)+sin(2(pd)\pi)=1+0=1. so letting d be the denominator of p shows...
Every element in G has a square root. If an element has 2 square roots then two elements in G have the same square root (by the pidgeonhole principle). That means there exists a y in G s.t. y^2=x1 and y^2=x2 where x1=/=x2. This is a contradiction because it would suggest y^2=/=y^2.
Is this...
So ##e^{2 \pi i qd}##=##e^{2 \pi qd}## by euler's theorem, but I don't see how that can be the identity for any natural number d? Isn't the identity of this group e^0=1?
I was also wondering if you (or someone) could help answer my questions in the previous posts of this thread about where...
Homework Statement
Let G be a finite group in which every element has a square root. That is, for each x in G, there exists a y in G such that y^2=x. Prove every element in G has a unique square root.
Homework Equations
G being a group means it is a set with operation * satisfying...