Let be a commutative ring with 1. Prove that a polynomial ring in more than one variable over is not a principal ideal domain.
[Short, Nice Proof]
If is a PID, then by Corollary 8 on page 281 of D&F, is a field- a contradiction.
[Long, Complicated Proof]
We begin with a lemma.
Lemma: Let be a commutative ring with 1. If is a principal ideal domain, then every element of is either 0, a unit, or a zero divisor. Proof: Let be nonzero. The ideal is principal. Since for some , the constant coefficient of is nonzero. Now for some . Let , , and be the constant terms of , , and , respectively. If , then is a zero divisor since . Thus is a zero divisor. If , then letting and be the linear coefficients of and , respectively, we have , so that . In particular, is a unit. Thus we in fact have for some . Now for some , so that . Note that has no constant term, so that comparing coefficients, the constant term of is zero. Write . Then , so that . Note that is not a zero divisor, so that . That is, we have . Using a lemma proved in this previous exercise, we have , so that , and thus is a unit.
Now let be a commutative ring with 1 and consider . If is a principal ideal domain, then every element of is either zero, a unit, or a zero divisor. However, is not zero and not a zero divisor. Moreover, is not a unit since has constant term 0 for all . Every polynomial ring in two or more variables is isomorphic to for some , and thus the result is proved.
This is a handy lemma that we ended up not using here, but don’t want to throw out.
Lemma: Let be a commutative ring with 1, and suppose is irreducible and not a unit. Then is not a principal ideal. Proof: Suppose to the contrary that . Since for some and is irreducible, there are two cases. If is a unit, then in fact for some . Comparing degrees, we see that , so that . This is a contradiction since is not a unit. If is a unit, then . In particular, for some . Comparing constant coefficients, we have that is a unit, a contradiction. Thus is not principal. .