I am 47 years old and I trying to learn Mathematics

[Marcio reis]

How did you find PF?: https://www.intmath.com/blog/learn-math/top-10-math-help-sites-5966

Hello everyone.
I know that there is a group isomorphism from Z[X] to the multiplicative group of all positive rational numbers. My question is: Are Z[X] and Q isomorphic groups?

 

[fresh_42]

What isomorphism do you mean?

 

[fresh_42]

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 !

 

[Marcio reis]

What isomorphism do you mean?

Z[X] is the additive group and Q is the multiplicative group of all rational numbers without zero.

 

[Marcio reis]

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 !

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.

 

[fresh_42]

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.

 

[Infrared]

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.

 

[willem2]

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.

 

[fresh_42]

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.?

 

[willem2]

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 $\Pi{p_i}^{a_i}$ is $a_0 + a_1 X + a_2 X^2 + \ldots$
(where the p_i are the primes starting with p₀ = 2.

 

[fresh_42]

Well I'd left that as an exercise for the reader.

Sorry, my fault, I confused you with the OP.

 

[WWGD]

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 $\Pi{p_i}^{a_i}$ is $a_0 + a_1 X + a_2 X^2 + \ldots$
(where the p_i are the primes starting with p₀ = 2.

How do you get the negative coefficients?

 

[willem2]

How do you get the negative coefficients?

The a_i can be negative. 1/p_i^a corresponds to -a X^i-1

 

[WWGD]

How do you get the negative coefficients?

Ah, yes, mixed up the $\mathbb Z$ and $\mathbb Q$, my bad.