Meaning of mapping R[X]->Maps[R,R]

[spacetimedude]

This is going to be a really silly question, but here it goes.
In a ring theory lecture, I was given a definition to a polynomial $P \in R[X]$ evaluated at the element $\lambda\in R$. I understand the evaluation bit as it is trivial to substitute a lambda into X.
At the end of the definition, it was shown that this process was essentially the mapping $$R[X] \rightarrow Maps(R,R).$$
From my understanding, $Maps(R,R)$ is in itself a linear mapping from R to R, so does $R[X] \rightarrow Maps(R,R)$ mean $R[X]\rightarrow R\rightarrow R?$
Could someone clarify what $R[X] \rightarrow Maps(R,R)$ means?
Thanks in advance.

 

[Rachel duPre]

Dear spacetime

please forgive my language for I am French so try English but often mess up :)

Can I ask perhaps what precise branch of science you study (as ring theory is taught in many branches)?
My reason for asking is the maps function means different things depending on which of the following you are studying:
a) category theory
b) logic
c) graph theory
d) computer sciences

R a real number maps ie changes or morphs in graph theory to (x a real number(R), y a real number(R)) thus results in a shape with no overlapping surfaces

This formula is often used when mapping coastlines in GIS as an example or creating shapes as graphical representation

As often you will see R goes to Maps (x, xy) were both again are R ie real numbers to create snake like figures again used in GIS

But maps can mean morph, an existing function repeated or the combination of a function and a list

A ring is a non overlapping shape of a unique symmetry which requires x infinity = x infinity + y infinity where X and Y are R ie integers
A circle technically does not exist in nature and is a mathematical concept or theory since a circle has never been observed or repeated.

Circular approximations are often drawn but they are imprecise by the width of their circumference and under the microscope are shown to be imprecise given that even when plotting x infinity ... we still get an imprecise circle. Spheres are also imprecise and yet to be observed. Again ring theory helps us approximate a sphere. Sorry I digress and you will know this already, forgive my digression. Key is ring theory does allow us to create circles and spheres visible to the human eye and perceived as perfect circles also accurate enough within tolerance levels to allow us to test the properties of a circle

If you ever get a chance to drive an F1 racing car post race then you will feel the impact of the wheel distortion caused by racing such tyres at intense heat and high speed with frequent late braking. The tyres start almost circular and distort massively. Performance deteriorates as a consequence. Yet to the human eye it is still a perfect wheel.

Please ignore this if not part of your studies as it is a digression

If I have been unclear please ask

Hope that helps

As salaam alaykum

Rachel

 

[mathwonk]

each polynomial defines a map from R to R by substitution, i.e. each polynomial P defines a map P:R-->R. Hence you have defined a function from k[X] to maps from R to R, i.e. function k[X]-->Maps(R,R). I guess I don't know how to make it any more clear. Just ask yourself what it means to define a function k[X]-->Maps(R,R). I.e. for this to make sense, you need to view each polynomial P as a map from R to R. But that is exactly what the substitution process does. Gee I'm going around and around.

 

[Stephen Tashi]

, it was shown that this process was essentially the mapping $$R[X] \rightarrow Maps(R,R).$$

From my understanding, $Maps(R,R)$ is in itself a linear mapping from R to R,

"$Maps(R,R)$" probably denotes the set of mappings from $\mathbb{R}$ to $\mathbb{R}$.There is a function $s$ ("this process") which associates a given polynomial $p(X)$ in one "indeterminate" with the real valued function of one real variable $x$ given by the function $p(x)$.

The domain of $s$ is the set $R[X]$ of polynomials in one "indeterminante".

The co-domain of $s$ is set of polynomial functions in one real variable. The set of polynomial functions is a subset of the set $Maps(\mathbb{R},\mathbb{R})$ of all mappings of $\mathbb{R}$ into itself. So we can say that $s$ is a function from $R[X]$ "into" $Maps(\mathbb{R}, \mathbb{R})$.

A definition of $s$ is: Map the polynomial $p(X)$ in one indeterminante to the real valued function of one real variable that is formed by replacing the symbol for the indeterminante "$X$" in the expression for $p(X)$ with a symbol representing a variable (such as "$x$" or "$\lambda$").

There is distinction between a polynomial "in one indeterminante" versus a a polynomial "in one variable". Even people who are vary familiar with abstract algebra sometimes fail to make this distinction when discussing problems, For example, in the ring of polynomials in one indeterminante with coefficients in the real numbers, the statement $X^2 = 4X$ is simply a false statement because the elements $X^2$ and $4X$ are two distinct elements of the ring. By contrast, the interpretation of $x^2 = 4x$ as claim about real valued functions is an equation that may produce a true or false statement depending on what value of $x$ is used.

 

[Rachel duPre]

Dear mathmonk

We are in the same place or in the same confusion
Maps basically means to change or morph
In this case the key point is R is a real number which includes polynomials AND the result is a non overlapping shape eg an oval or hexagon or pentagram


Ciao

Rachel

 

[fresh_42]

In this case the key point is R is a real number

This is not correct.

"$Maps(R,R)$" probably denotes the set of mappings from $\mathbb{R}$ to $\mathbb{R}$.

I assume it's about an arbitrary ring rather than the real numbers (cp. the mentioning of ring theoy). However, this doesn't change the argument very much. Just wanted to avoid misunderstandings.

$Maps(R,R)$ is in itself a linear mapping from $R$ to $R$.

This doesn't make sense, since $Maps(R,R)$ is a set and not a linear mapping. And its elements aren't linear mappings either, because $f(x)(r+s)=f(r+s) {\neq}_{i.g.} f(r) + f(s) = f(x)(r) + f(x)(s)$, e.g. $f(x) = x^2$.

But the mapping $\varphi \, : \, f(x) \mapsto_\varphi (a \mapsto f(x)(a)=f(a))$ is $R-$linear, if $Map(R,R)$ is considerd as an $R$-module, since $\varphi(\alpha f(x) + \beta g(x))(r) = \alpha f(r) + \beta g(r) = \alpha \varphi(f(x))(r)+ \beta \varphi(g(x))(r)$. It is also multiplicative with the definition $(f(x) \cdot g(x))(r) := f(r) \cdot g(r)$, i.e. a ring homomorphism. Note that this is a different definition than $(f \circ g)(x)=f(g(x))$ which is also often used as a multiplication of functions.

 

[mathwonk]

I think somethign Stephen said helps. I.e. there are many maps from R to R, but not all of them are defined by pol;ynomials. So one can ask whicha re the polynomial maps? and the answer is they are the ones in the image of the function R[X]-->Maps(R,R). This function is not always injective by the way, since R[X} always contains infinitely many elements, but Maps (R,R) does not if R is finite, so a polynomial function can come from more than one polynomial.

 

[spacetimedude]

This is not correct.

I assume it's about an arbitrary ring rather than the real numbers (cp. the mentioning of ring theoy). However, this doesn't change the argument very much. Just wanted to avoid misunderstandings.

This doesn't make sense, since $Maps(R,R)$ is a set and not a linear mapping. And its elements aren't linear mappings either, because $f(x)(r+s)=f(r+s) {\neq}_{i.g.} f(r) + f(s) = f(x)(r) + f(x)(s)$, e.g. $f(x) = x^2$.

But the mapping $\varphi \, : \, f(x) \mapsto_\varphi (a \mapsto f(x)(a)=f(a))$ is $R-$linear, if $Map(R,R)$ is considerd as an $R$-module, since $\varphi(\alpha f(x) + \beta g(x))(r) = \alpha f(r) + \beta g(r) = \alpha \varphi(f(x))(r)+ \beta \varphi(g(x))(r)$. It is also multiplicative with the definition $(f(x) \cdot g(x))(r) := f(r) \cdot g(r)$, i.e. a ring homomorphism. Note that this is a different definition than $(f \circ g)(x)=f(g(x))$ which is also often used as a multiplication of functions.

Thank you. Could you explain to me what Maps(R,R) is? It was not clarified in the lecture and I thought it meant mapping from R to R. This is from my first course in abstract algebra and I haven't covered things like modules so it was a bit difficult to follow what you have stated in the third statement.

Also, I have seen the mapping $M: Hom_F(F^m,F^n)\tilde{\rightarrow} Mat(n\times m;F),$ where F is a field in another section regarding linear mapping, and it would be great if you could tell me what $Hom(F^m,F^n)$ means as well. I assumed it was a mapping from $F^m$ to $F^n$ that was a group homomorphism.

Simple examples will be greatly appreciated.

 

[fresh_42]

Thank you. Could you explain to me what $Maps(R,R)$ is? It was not clarified in the lecture and I thought it meant mapping from $R$ to $R$. This is from my first course in abstract algebra and I haven't covered things like modules so it was a bit difficult to follow what you have stated in the third statement.

Almost. Very likely, even if in an unusual notation, $Maps(R,R) = \{m : R \rightarrow R\}$ is the set of all mappings from $R$ to $R$, whether they are linear in any sense or not.

$R[X] \rightarrow Maps(R,R)$

is a map, which I denoted by $\varphi : R[X] \rightarrow Maps(R,R)$, from the ring of polynominals over (= with coefficients in) $R$ into the set of mappings from $R$ to $R$. It is defined by $\varphi(f(x)) \, : \, R \longrightarrow R$ for all polynomials $f(x) \in R[x]$. Thus $\varphi (f(x))$ is a map from $R$ to $R$. But which one? Well, it is defined by $\varphi (f(x))(r)=f(r)$ for all $r \in R$. So far this is the situation behind:

$R[X] \rightarrow Maps(R,R)$

Now we can ask quite a few things here.

1. $R$ is a ring and so is $R[x]$. Does $Map(R,R)$ also have a ring structure?
2. We can add polynomials by $(f+g)(x)=f(x)+g(x)$. What happens to $\varphi (f(x)+g(x))$?
3. We can multiply polynomials by ring elements (from $R$) by $(\alpha \cdot f)(x)= \alpha \cdot f(x)$. What happens to $\varphi (\alpha f(x))$.
4. We can multiply polynomials by $(f \cdot g)(x)=f(x) \cdot g(x)$. What happens to $\varphi (f(x) \cdot g(x))$?
   
5. A module is basically the same thing as a vector space, only that we don't have a field where the scalars are from, but a ring instead. So $R[x]$ is a $R-$module means quasi $R[x]$ is a "vector space" with coefficients in the ring $R$, given by the addition under point 2 and the scalar multiplication under point 3. This defines $R-$linearity. Now the question is, whether $Map(R,R)$ can also be viewed as a $R-$module?
6. Does 4 define a ring homomorphism?
7. Does 5 (resp. 2. and 3.) define an $R-$module homomorphism (= $R-$linear mapping)?

I leave it to you to answer these questions. It is a straight forward calculation and merely requires to write down the conditions needed.

Also, I have seen the mapping $M: Hom_F(F^m,F^n)\tilde{\rightarrow} Mat(n\times m;F),$ where F is a field in another section regarding linear mapping, and it would be great if you could tell me what $Hom(F^m,F^n)$ means as well.

This is a bit different, since the sets of mappings here are now given a linear structure by definition. $Hom_F(F^m,F^n)$ denotes the set of all $F-$linear functions from the vector space $F^m$ to the vector space $F^n$, which is itself a vector space (again by similar definitions as under 2. and 3.). As you probably know, all linear functions $f$ from $F^m$ to $F^n$ can be represented by a matrix, if we chose a basis in both. And there is (given both basis) only one matrix that represents $f$. So all matrices $Mat(n\times m;F)$ and all $F-$linear functions $Hom_F(F^m,F^n)$ are isomorphic vector spaces: bijective (one function $\stackrel{1:1}{\leftrightarrows}$ one matrix) and $M(f+g)=M(f)+M(g)$ and $M(\alpha f)=\alpha M(f)$ for all $\alpha \in F\, , \, f,g \in Hom_F(F^m,F^n)$, i.e. linear functions $f,g : F^m \rightarrow F^n$ and their matrices $M(f)$ and $M(g)$.

I assumed it was a mapping from $F^m$ to $F^n$ that was a group homomorphism.

If you consider the additive group of the vector space of linear mappings, and the additive group of matrices: yes, in this sense $M$ is a group homomorphism. But it's even more, as they also respect multiples ($\alpha \cdot f$ for $\alpha \in F$, $f:F^m \rightarrow F^n$ linear), which we usually call linear or more precisely $F-$linear, if both properties hold.

The same is true, in case you didn't mean the isomorphism $M$ here but only $Hom_F(F^m,F^n)$. These are all linear mappings from the vector space $F^m$ to the vector space $F^n$. As linear mapping, such a function is of course a group homomorphism of the additive group of the vector spaces. But as it's even more, namely $F-$linear, one calls it linear instead of group homomorphism, which would disregard the multiples.

 

[WWGD]

Almost. Very likely, even if in an unusual notation, $Maps(R,R) = \{m : R \rightarrow R\}$ is the set of all mappings from $R$ to $R$, whether they are linear in any sense or not.

is a map, which I denoted by $\varphi : R[X] \rightarrow Maps(R,R)$, from the ring of polynominals over (= with coefficients in) $R$ into the set of mappings from $R$ to $R$. It is defined by $\varphi(f(x)) \, : \, R \longrightarrow R$ for all polynomials $f(x) \in R[x]$. Thus $\varphi (f(x))$ is a map from $R$ to $R$. But which one? Well, it is defined by $\varphi (f(x))(r)=f(r)$ for all $r \in R$. So far this is the situation behind:

Now we can ask quite a few things here.

1. $R$ is a ring and so is $R[x]$. Does $Map(R,R)$ also have a ring structure?
2. We can add polynomials by $(f+g)(x)=f(x)+g(x)$. What happens to $\varphi (f(x)+g(x))$?
3. We can multiply polynomials by ring elements (from $R$) by $(\alpha \cdot f)(x)= \alpha \cdot f(x)$. What happens to $\varphi (\alpha f(x))$.
4. We can multiply polynomials by $(f \cdot g)(x)=f(x) \cdot g(x)$. What happens to $\varphi (f(x) \cdot g(x))$?
5. A module is basically the same thing as a vector space, only that we don't have a field where the scalars are from, but a ring instead. So $R[x]$ is a $R-$module means quasi $R[x]$ is a "vector space" with coefficients in the ring $R$, given by the addition under point 2 and the scalar multiplication under point 3. This defines $R-$linearity. Now the question is, whether $Map(R,R)$ can also be viewed as a $R-$module?
6. Does 4 define a ring homomorphism?
7. Does 5 (resp. 2. and 3.) define an $R-$module homomorphism (= $R-$linear mapping)?

I leave it to you to answer these questions. It is a straight forward calculation and merely requires to write down the conditions needed.

This is a bit different, since the sets of mappings here are now given a linear structure by definition. $Hom_F(F^m,F^n)$ denotes the set of all $F-$linear functions from the vector space $F^m$ to the vector space $F^n$, which is itself a vector space (again by similar definitions as under 2. and 3.). As you probably know, all linear functions $f$ from $F^m$ to $F^n$ can be represented by a matrix, if we chose a basis in both. And there is (given both basis) only one matrix that represents $f$. So all matrices $Mat(n\times m;F)$ and all $F-$linear functions $Hom_F(F^m,F^n)$ are isomorphic vector spaces: bijective (one function $\stackrel{1:1}{\leftrightarrows}$ one matrix) and $M(f+g)=M(f)+M(g)$ and $M(\alpha f)=\alpha M(f)$ for all $\alpha \in F\, , \, f,g \in Hom_F(F^m,F^n)$, i.e. linear functions $f,g : F^m \rightarrow F^n$ and their matrices $M(f)$ and $M(g)$.

If you consider the additive group of the vector space of linear mappings, and the additive group of matrices: yes, in this sense $M$ is a group homomorphism. But it's even more, as they also respect multiples ($\alpha \cdot f$ for $\alpha \in F$, $f:F^m \rightarrow F^n$ linear), which we usually call linear or more precisely $F-$linear, if both properties hold.

The same is true, in case you didn't mean the isomorphism $M$ here but only $Hom_F(F^m,F^n)$. These are all linear mappings from the vector space $F^m$ to the vector space $F^n$. As linear mapping, such a function is of course a group homomorphism of the additive group of the vector spaces. But as it's even more, namely $F-$linear, one calls it linear instead of group homomorphism, which would disregard the multiples.

I think we need the basis to be ordered for the correspondence between maps and matrices to be well-defined.