unique factorization domain

Proof:
Let us consider an arbitrary prime element r in R and the ideal I = <r> generated by r. We want to show that I is an prime ideal. We think about two arbitrary elements a and b in R with ab in I.
Since R is an unique factorization domain (UFD), then we have unique factorizations for both a and b
a = p_1 * p_2 * ... * p_k
b = q_1 * q_2 * ... * q_t
where p_1, ..., p_k are prime factors of a, q_1, ..., q_t are prime factors of b
Then ab = p1 * p_2 * ... * p_k * q_1 * q_2 * ... * q_t
Since ab is in I = <r>, then we can find some x in R such that ab = xr
Then xr also has unique factorization.
Assume x = w_1 * w_2 * ... * w_n, where w_1, ..., w_n are prime factors of x
Then we have
xr = w_1 * w_2 * ... * w_n * r is the unique factorization of xr = ab because r itself is a prime element.
Thus we have
p1 * p2 * ... * p_k * q_1 * q_2 * ... * q_t = w_1 * w_2 * ... * w_n * r
By uniqueness, r should be in the set { p_1, p_2, ..., p_k, q_1, q_2, ..., q_t}
We have two cases.
case 1: r = p_i for some 1<= i <= k, then a is a mulitple of r and thus a belongs to I = <r>
case 2: r = q_j for some 1<= j <= t, then b is a mulitple of r and thus b belongs to I = <r>
Therefore, ab is in I = <r> induces that either a or b is in I = <r>
Hence I = <r> is a prime ideal.
Done.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!