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

[VinnyCee]

Homework Statement

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

Let S be the set:

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

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

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

Prove that \phi is a bijection.

Homework Equations

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

f\,:\,B\,\rightarrow\,C is surjective iff I am f = C.

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

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

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

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))...

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

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

F<b>\,\equiv\,h(b)\,\left(mod\,p(b)\right)</b>

Now how do I show that a = b in S?

 

[micromass]

[
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...

 

[VinnyCee]

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, [a]\,=\,\left\{b\,|\,b\,\in\,\mathbb{Z}\,\,and\,\,b\,\equiv\,a\,\left(mod\,n\right)\right\}.

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

\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]

\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]Assume that F_1[x]\,\equiv\,F_2[x]\,\left(p(x)\right).

\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)

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

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 (a_i\,=\,0\,\forall\,i\,\in\,\mathbb{N}).

deg\left(F_1[x]\right)\,=\,deg\left(F_2[x]\right)\,&lt;\,deg\left(p(x)\right)

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

Now, to prove that it is surjective...

 

[micromass]

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

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

make no sense. F_1[x] is a ring of polynomials, while [a_0+...+a_nx^n] 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...