B I am 47 years old and I trying to learn Mathematics.

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fresh_42

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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: !
 
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.
 

fresh_42

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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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Infrared

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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.
 

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