I am 47 years old and I trying to learn Mathematics

In summary: Check.For surjectivity, we must show that for any positive rational number ##\dfrac{s}{t}## there is a way to write ##\dfrac{s}{t}=p_na_n+\ldots p_1a_1+p_0a_0## for suitable ##p_0,\ldots,p_n## by our given choice of ##a_n\,.##To do this, we need to show that for any rational number ##s\gt 0## there exists a unique rational number ##t\gt0## such that ##s=t##. This can be done by solving a system of simultaneous equations
  • #1
Marcio reis
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  • #3
Yes, ##\mathbb{Z}[X]## is only a group under addition, and you said multiplicative group of positive rational numbers. As ##0## isn't positive, it can only be the multiplicative group. What I meant was, how do you map a polynomial ##p(X)=a_nX^n+a_{n-1}X^{n-1}+\ldots +a_1X+a_0## into a positive rational number?

Btw: Hello and :welcome: !
 
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  • #4
fresh_42 said:
What isomorphism do you mean?

Z[X] is the additive group and Q is the multiplicative group of all rational numbers without zero.
 
  • #5
fresh_42 said:
Yes, ##\mathbb{Z}[X]## is only a group under addition, and you said multiplicative group of positive rational numbers. As ##0## isn't positive, it can only be the multiplicative group. What I meant was, how do you map a polynomial ##p(X)=a_nX^n+a_{n-1}X^{n-1}+\ldots +a_1X+a_0## into a positive rational number?

Btw: Hello and :welcome: !
Thanks! It's proved somewhere else that Z[X] and the multiplicative group of positive rational numbers are isomorphics. I was wondering why the negative rational numbers must be excluded.
 
  • #6
I don't know whether they are isomorphic without seeing the mapping. If there is one, and I still like to see it, then they are isomorphic. The notation should be more detailed like ##\left(\mathbb{Z},+\right)\cong_\varphi\left(\mathbb{Q}^+,\cdot\right)## and ##\varphi(p(X))## should be defined. I assume it is ##\varphi\, : \,p(X) \longmapsto 10^{p(1)}## or so but I do not see surjectivity. Which polynomial belongs to e.g. ##\dfrac{1}{7}\,?##

Another possibility is, that the author uses "isomorphism into" as being a monomorphism and distinguishes it from "isomorphism onto". This happens occasionally but is very confusing in my opinion.

The correct correspondence is:
monomorphism = injective (bad alternative: isomorphism into)
epimorphism = surjective (alternative: mapping onto)
isomorphism = bijective = injective and surjective (bad alternative: isomorphism onto)

So the answer to your question depends on what you have not given us as information.
 
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  • #7
Note that ##(\mathbb{Q}^{>0},\cdot)## and ##(\mathbb{Q}^{\times},\cdot)## are not isomorphic, since the first group has only one element that squares to the identity (namely ##1##) but the second group has two elements that square to the identity (##\pm 1##). So if ##\mathbb{Z}[X]## is isomorphic to the first group, it cannot be isomorphic to the second group.
 
  • #8
The groups are certainly isomorphic.
Suppose f is a function from z[X] to Q+. We must have f(0) = 1. (identities of the groups)
Suppose f(1) = a0, then f(2) must be a02, f(-1) must be 1/a0, and f(n) must be a0n.
The same can be said about f(0)=1, F(X)=a1, F(2X) =a12 etc, so you get all powers of a1
Now all of those powers must be different numbers, or you'll get strange polynomial identies like a0 = a1 X
What condition must the sequence a0, a1, a2, ... satisfy that all of the powers are different. There is a fairly simple example.
 
  • #9
willem2 said:
The groups are certainly isomorphic.
It doesn't become right by repetition!
Suppose f is a function from z[X] to Q+. We must have f(0) = 1. (identities of the groups)
Suppose f(1) = a0, then f(2) must be a02, f(-1) must be 1/a0, and f(n) must be a0n.
The same can be said about f(0)=1, F(X)=a1, F(2X) =a12 etc, so you get all powers of a1
Now all of those powers must be different numbers, or you'll get strange polynomial identies like a0 = a1 X
You defined ##f\, : \,\mathbb{Z}[x]\longrightarrow \mathbb{Q}^+## by ##f(1)=a_0## which automatically gives us by ##f(z)=f(1+(z-1))=f(1)f(z-1)=a_0f(z-1)## all other values of constant polynomials. Next you defined - I assume - ##f(x^n)=a_n##. In short we have for ##p(x)=p_nx^n+\ldots +p_0## a function ##f(p)=p_na_n+\ldots p_1a_1+p_0a_0##.

Now we have to check the following properties:
  1. Is ##f## well-defined?
  2. Is ##f## a homomorphism?
  3. Is ##f## injective?
  4. Is ##f## surjective?
As polynomials are given by the sequence ##(p_0,\ldots,p_n)## and a sequence ##(a_n)_{n\in \mathbb{N}}## is chosen in advance to set up the function, there will be no two different images for ##f(p)## possible.
1. Check.

Since we defined ##f## by the rule ##f(p-q)=f(p)f(q)^{-1}## we automatically have a homomorphism per construction.
2. Check.

For injectivity, we must show, that ##f(p)=f(q)## implies ##p=q##. So given ##f(p)=p_na_n+\ldots p_1a_1+p_0a_0=q_ma_m+\ldots q_1a_1+q_0a_0=f(q)##, how do you guarantee that ##n=m## and ##p_j=q_j## for all ##j=0,\ldots, n\,?##

For surjectivity, we must show that for any positive rational number ##\dfrac{s}{t}## there is a way to write ##\dfrac{s}{t}=p_na_n+\ldots p_1a_1+p_0a_0## for suitable ##p_0,\ldots,p_n## by our given choice of ##a_n\,.##How do you find those ##p_j\,?##
What condition must the sequence a0, a1, a2, ... satisfy that all of the powers are different. There is a fairly simple example.
My suggestion is to choose ##a_n## to be the inverse of ##n-##th prime number: ##a_0=\dfrac{1}{2},a_1=\dfrac{1}{3},a_2=\dfrac{1}{5},a_3=\dfrac{1}{7},\ldots ##. Can you now check point 3. and 4.?
 
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  • #10
fresh_42 said:
For surjectivity, we must show that for any positive rational number stst\dfrac{s}{t} there is a way to write st=pnan+…p1a1+p0a0st=pnan+…p1a1+p0a0\dfrac{s}{t}=p_na_n+\ldots p_1a_1+p_0a_0 for suitable p0,…,pnp0,…,pnp_0,\ldots,p_n by our given choice of an.an.a_n\,.How do you find those pj?
Well I'd left that as an exercise for the reader.
Every positive rational number can be written in a unique way as a product of powers of prime numbers. (including negative powers). The polynomials corresponding to [itex] \Pi{p_i}^{a_i} [/itex] is [itex] a_0 + a_1 X + a_2 X^2 + \ldots [/itex]
(where the pi are the primes starting with p0 = 2.
 
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  • #11
willem2 said:
Well I'd left that as an exercise for the reader.
Sorry, my fault, I confused you with the OP.
 
  • #12
willem2 said:
Well I'd left that as an exercise for the reader.
Every positive rational number can be written in a unique way as a product of powers of prime numbers. (including negative powers). The polynomials corresponding to [itex] \Pi{p_i}^{a_i} [/itex] is [itex] a_0 + a_1 X + a_2 X^2 + \ldots [/itex]
(where the pi are the primes starting with p0 = 2.
How do you get the negative coefficients?
 
  • #13
WWGD said:
How do you get the negative coefficients?
The ai can be negative. 1/pia corresponds to -a Xi-1
 
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  • #14
WWGD said:
How do you get the negative coefficients?
Ah, yes, mixed up the ##\mathbb Z## and ##\mathbb Q ##, my bad.
 

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