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

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

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...