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

#### Marcio reis

Related Linear and Abstract Algebra News on Phys.org

#### fresh_42

Mentor
2018 Award
What isomorphism do you mean?

#### fresh_42

Mentor
2018 Award
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 !

• berkeman

#### 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

Mentor
2018 Award
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.

Last edited:

#### 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.