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 to0 for only finitely many i}

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

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

Show that R[X] forms a commutative ring with a suitably-chosenaddition 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 nicelyeven though
the entries here are only in a ring and not a field (and there areinfinitely many
of them).
Your operations should act like how polynomial addition andmultiplication
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 homomorphismfrom R[X] to the
“ring of polynomial functions” (you don’t need to show that thelatter 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 polynomialslike
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 inR[X] by looking
at the coefficients of the powers in p hat · q hat. When you haveit right, (p · q) hat = p hat · q hat
should hold.