# unique factorization domain

Let R be a unique factorization domain then show that prime element R generates a prime ideal.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

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.

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