Answer: Prove Injectivity of Bijection in F[x] Modulo a Fixed Polynomial P(x)

In summary, we are trying to prove that \phi is a bijection between the set S and the ring F[x]/(p(x)). To prove injectivity, we assume two elements of S have the same image in F[x]/(p(x)) and show that the only way this is possible is if the two elements are equal. To prove surjectivity, we show that every element in the ring F[x]/(p(x)) has a corresponding element in S.
  • #1
VinnyCee
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0

Homework Statement



Let [tex]n\,\in\,\mathbb{N}[/tex]. Let F be a field, and suppose that [tex]p(x)\,\in\,F[x][/tex] is a polynomial of degree (n + 1).

Let S be the set:

[tex]S\,=\,\left\{\left(a_0,\,\ldots,\,a_n\right)\,:\,a_i\,\in\,F\right\}[/tex]

Define [tex]\phi[/tex]: [tex]S\,\rightarrow\,F[x]/\left(p(x)\right)[/tex] via

[tex]\phi\left(\left(a_0,\,\ldots,\,a_n\right)\right)\,=\,\left[a_0\,+\,a_1\,x\,+\,\cdots\,+\,a_n\,x^n\right][/tex]

Prove that [tex]\phi[/tex] is a bijection.

Homework Equations



[tex]f\,:\,B\,\rightarrow\,C[/tex] is injective provided that whenever f(a) = f(b) in C, then a = b in B.

[tex]f\,:\,B\,\rightarrow\,C[/tex] is surjective iff I am f = C.

[tex]f\,:\,B\,\rightarrow\,C[/tex] is bijective provided that f is both injective and surjective.

If [tex]f\,:\,B\,\rightarrow\,C[/tex] is a function, then the image of f is this subset of C:

[tex]Im\,f\,=\,\left{c\,|\,c\,=\,f(b)\,for\,some\,b\,\in\,B\right}\,=\,\left{f(b)\,|\,b\,\in\,B\right}[/tex]

The Attempt at a Solution



Prove Injectivity:

Here, I need to prove that whenever F[a]/(p(a)) = F/(p(b)) in F[x]/(p(x)), then a = b in S.

Now I need expressions for F[a]/(p(a))...

[tex]F[a]\,\equiv\,g(a)\,\left(mod\,p(a)\right)[/tex]

And F/(p(b))...

[tex]F\,\equiv\,h(b)\,\left(mod\,p(b)\right)[/tex]

Now how do I show that a = b in S?
 
Last edited:
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  • #2


VinnyCee said:
[
Here, I need to prove that whenever F[a]/(p(a)) = F/(p(b)) in F[x]/(p(x)), then a = b in S.


What do you mean with F[a]/(p(a))? In fact, what do you mean with F[a], I'm not used to such a notations...
 
  • #3


The brackets denote congruence class...

Definition: Let a and n be integers with n > 0. The congruence class of a modulo n (denoted [a]) is the set of all those integers that are congruent to a modulo n, that is, [tex][a]\,=\,\left\{b\,|\,b\,\in\,\mathbb{Z}\,\,and\,\,b\,\equiv\,a\,\left(mod\,n\right)\right\}[/tex].

The ring P of polynomials with coefficients in R is denoted by R[x].
To prove injectivity, assume that you have

[tex]\phi\left(\left(a_0,\,\ldots,\,a_n\right)\right)\,=\,F_1[x]\,=\,\left[a_0\,+\,a_1\,x\,+\,\cdots\,+\,a_n\,x^n\right][/tex]

[tex]\phi\left(\left(b_0,\,\ldots,\,b_n\right)\right)\,=\,F_2[x]\,=\,\left[b_0\,+\,b_1\,x\,+\,\cdots\,+\,b_n\,x^n\right][/tex]Assume that [tex]F_1[x]\,\equiv\,F_2[x]\,\left(p(x)\right)[/tex].

[tex]\left[a_0\,+\,a_1\,x\,+\,\cdots\,+\,a_n\,x^n\right]\,-\,\left[b_0\,+\,b_1\,x\,+\,\cdots\,+\,b_n\,x^n\right]\,|\,p(x)[/tex]

[tex]\left[a_0\,-\,b_0\right]\,+\,\left[a_1\,-\,b_1\right]\,x\,+\,\cdots\,+\,\left[a_n\,-\,b_n\right]\,x^n\,|\,p(x)[/tex]

But, p(x) is a polynomial of degree n + 1, making the above statement impossible (since a higher degree polynomial cannot divide a lower degree polynomial) unless each of the coefficients on the left side are zero ([tex]a_i\,=\,0\,\forall\,i\,\in\,\mathbb{N})[/tex].

[tex]deg\left(F_1[x]\right)\,=\,deg\left(F_2[x]\right)\,<\,deg\left(p(x)\right)[/tex]

This means the coefficients are equal, and thus we have shown that [tex]\phi[/tex] is injective.

Now, to prove that it is surjective...
 
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  • #4


Your proof is good, but you must beware of your notations, things like

[tex]F_1[x]=[a_0+...+a_nx^n][/tex] or [tex]deg(F_1[x])[/tex]

make no sense. [tex]F_1[x][/tex] is a ring of polynomials, while [tex][a_0+...+a_nx^n][/tex] is an equivalence class of polynomials. You can't say that they're equal...

Likewise, you can't take the degree of a ring of polynomials...
 

1. What is a bijection?

A bijection is a function that has a one-to-one correspondence between its domain and range. This means that each element in the domain is paired with a unique element in the range, and vice versa.

2. How do you prove injectivity of a bijection in F[x] modulo a fixed polynomial P(x)?

To prove injectivity of a bijection in F[x] modulo a fixed polynomial P(x), we need to show that for any two elements a and b in the domain, if f(a) = f(b), then a = b. This can be done by showing that the remainder of f(a) when divided by P(x) is equal to the remainder of f(b) when divided by P(x).

3. What is F[x] modulo a fixed polynomial P(x)?

F[x] modulo a fixed polynomial P(x) is a mathematical structure that represents the set of all polynomials with coefficients in the field F, divided by the fixed polynomial P(x). This is also known as a polynomial ring or a quotient ring.

4. Why is proving injectivity important for a bijection in F[x] modulo a fixed polynomial P(x)?

Proving injectivity is important for a bijection in F[x] modulo a fixed polynomial P(x) because it ensures that the function is one-to-one, meaning that each input has a unique output. This is a necessary condition for a bijection to be invertible.

5. Can a bijection in F[x] modulo a fixed polynomial P(x) be proven to be injective and not surjective?

Yes, it is possible for a bijection in F[x] modulo a fixed polynomial P(x) to be proven to be injective but not surjective. This means that the function has a one-to-one correspondence between its domain and range, but not all elements in the range are mapped to by the function.

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