Solving Polynomials in $\mathbb{Z}_{n}$ | Algebra Course

[Essnov]

I am taking a first course in algebra and this is a problem in my textbook that has me stumped:

$$Fix \ a \in \mathbb{Z}_{n} \ and \ f \in \mathbb{Z}_{n}[x] \ with \ \deg(f) = m. Show \ there \ is \ h_{m-1} \in \mathbb{Z}_{p}[x] \ with \ \deg(h_{m-1}) \leq (m-1) \ so \ f = a_{m}(x - a)^m + h_{m-1}$$

We haven't covered polynomial rings or anything like that.

To be honest I don't understand how it makes sense to add coefficients in Z_n with coefficients in Z_p, although maybe it makes sense if p | n?

If it's true then:
$$f(a) = \sum_{k = 0}^{m} a_{k}a^{k} = h_{m-1}(a)$$
But it's unclear to me how that's possible.

Any hint at all is really appreciated.

 

[lippyka]

It sounds like you are getting confused because of the different polynomials being considered. In this problem, f is a polynomial in Zn[x] with degree m and a is an element of Zn. We want to find a polynomial h_{m-1} in Zp[x] with degree less than or equal to m-1 such that f = a_{m}(x-a)^m + h_{m-1}. To solve this problem, let's start by using the fact that f(a) = \sum_{k=0}^{m} a_{k}a^{k}. We can write this as f(a) = a_{m}a^m + h_{m-1}(a), where h_{m-1}(a) = \sum_{k=0}^{m-1} a_{k}a^{k}. Then, we can see that h_{m-1} is a polynomial in Zp[x] with degree m-1. Now, all we need to do is show that f = a_{m}(x-a)^m + h_{m-1}. To do this, we just need to substitute a into both sides of the equation f(a) = a_{m}a^m + h_{m-1}(a). This gives us f(a) = a_{m}(a-a)^m + h_{m-1}(a). Since a - a = 0, this simplifies to f(a) = a_{m}(0)^m + h_{m-1}(a). Since any number raised to the 0 power is equal to 1, this becomes f(a) = a_{m} + h_{m-1}(a). Since this holds for any a in Zn, it follows that f = a_{m}(x-a)^m + h_{m-1}.