Question

Problem 2. Suppose that R is a commutative ring, and that

R[X] := { (a_0, a_1, a_2, ...)^T | a_i is in R, a_i not equal to 0 for only finitely many i}

is the set of polynomials over R, where we have named one particular element

X := (0, 1, 0, 0, . . .)T .

Show that R[X] forms a commutative ring with a suitably-chosen addition and
multiplication on R[X]. This will involve specifying a “zero” element of R[X]
that is the identity element with respect to addition, and a “unit” element of
R[X] that is the identity element with respect to multiplication. For example,
the usual addition we use for vectors in Rn should extend nicely even though
the entries here are only in a ring and not a field (and there are infinitely many
of them).
Your operations should act like how polynomial addition and multiplication
normally act. That is: to each element p = (a_0, a_1, a_2, . . . , a_k, 0, 0, . . . ,)^T is in R[X],

we can associate a polynomial function

p hat = x mapped to a_0 + a_1*x + a_2*x^2 + · · · + a_k*x^k.

Show that the mapping p mapped to p hat is a ring homomorphism from R[X] to the
“ring of polynomial functions” (you don’t need to show that the latter is a ring)
which has the following two operations defined for two functions f, g : R implies R:

f + g := x mapped to f (x) + g (x)
fg := x mapped to f (x) g (x) .

Really all you want to do here is think of two polynomials like
p = (a_0, a_1, a_2)^T is equivalent to p hat = x mapped to a_0 + a_1*x + a_2*x^2
and
q = (b_0, b_1, b_2)^T is equivalent to q hat = x mapped to b_0 + b_1*x + b_2*x^2

and then figure out what the the coordinates of p·q should be in R[X] by looking
at the coefficients of the powers in p hat · q hat. When you have it right, (p · q) hat = p hat · q hat
should hold.

Answer 1