Evaluation Homomorphism for Polynomial Rings Part 3

In this video we discuss the evaluation homomorphism applied to polynomial rings.

[turokg]

when did we talk bout images of homomorphisms being subrings? i missed it?

[debendragurung]

Coming back here from Field Extension video. I have totally misunderstood certain things.
On Evaluation Homomorphism - I thought it was a mapping for certain polynomial p(x)∈R[x]. evaluation of p(x) to be
p(f(x)): R↦R and proceeding onto the structure. For a and b∈R, just found out p(a)+p(b) ≠ p(a+b).
p(a)p(b) ≠p(ab).
Also p(0) not always equal to 0 and p(1) not always equal to 1.

While infact what Evaluation Homomorphism is about is a mapping
φ_a: R[x]↦R.
Such that if p(x), q(x), r(x) ∈R[x], and a∈R
p(a)+q(a)= (p+q)(a)
p(a)q(a)=pq(a)
If p(x) = 0, then φ:0↦0
If p(x) = 1, then φ:1↦1
And any constant polynomial is mapped onto itself there's no need to denote φ_a for it.