Rings, Modules, Fields

4 Polynomial Algebras

Now let’s study polynomial algebras \(R[X]\). We can ask questions like when is it Noetherian or a UFR?

Lemma 4.1. If \(R[X]\) is nontrivial Noetherian, then \(R\) is too, and \(X\) is finite.

Proof. The ideals of \(R\) are a sublattice of those of \(R[X]\), so \(R\) is Noetherian. \(R\) is nontrivial since \(R[X]\) is, so if \(X\) is infinite, then let \(x_i\) be an infinite sequence in \(X\), so that \(\aa _i = (x_1,\dots ,x_i)\) is an infinite ascending chain of ideals. □

The converse is called the Hilbert Basis Theorem.

Proposition 4.2 (Hilbert). If \(R\) is a Noetherian ring and \(X\) a finite set, \(R[X]\) is Noetherian.

Proof. By induction it suffices to show \(R[x]\) is Noetherian. Now given an ideal \(I\) in \(R[x]\), consider the ideal of leading terms \(I_n\) of elements of \(I\) of \(n^{th}\) degree polynomials. Then \(I_n\) is an ascending chain of ideals, hence stabilizes. Up until \(n\) is the stabilization point, choosing finitely many polynomials generating \(I_n\) will generate \(I\). □

As for the second question, it is note true generally that \(R[x]\) is a UFR if \(R\) is. For example, in \(\ZZ /6\ZZ [x]\), \((x-1)x = (x+2)(x+3)\), but \(\ZZ /6\ZZ \) is a UFR. If we restrict to domains however, it is true. The key to proving this is that \(\Frac (R)[x]\) is a UFD, and this

To prove this, we can use Gauss’s Lemma. Note that given a nonzero polynomial in \(R[x]\), we can first remove all common factors in the coefficients so that the gcd of the coefficients is \(1\), which is called a primitive polynomial. Then we have the following:

Lemma 4.3 (Gauss). The product of primitive polynomials is primitive in \(R[x]\), where \(R\) is a UFD. Moreover, if \(0 \neq f = gh\) in \(\Frac (R)[x]\), then \(f = (cg)(c^{-1}h)\) in \(R[x]\), where either factor can be assumed to be primitive.

Proof. For the first statement, reduce mod every prime element \(p\), and note that the product of nonzero polynomials is nonzero. For the second, assume \(f\) is primitive by taking out common factors. Then, choose \(c,d \in R\) so that \(cg,dh\) are primitive. Then \((cg)(dh) = uf\) for some unit \(f\), so \(du^{-1} = c^{-1}\). □

Then we can prove:

Proposition 4.4. \(R[X]\) is a UFD iff \(R\) is.

Proof. Certainly \(R \subset R[X]\) shows that if \(R[X]\) is a UFD, \(R\) is, since factorization in \(R\) stay in \(R\). For the converse, since every element has only finitely many indeterminates, it suffices to prove this when \(X\) is finite, but then by induction it suffices to show that \(R[x]\) is a UFD. First note that \(p \in R\) remains prime in \(R[x]\). Moreover, we can factor nonzero elements of \(R[x]\) into irreducibles by first taking out common factors, and then using Gauss’s Lemma. Now primitive irreducibles \(p\) are prime since if \(p|ab\), WLOG, \(p|a\) in \(\Frac (R)[x]\) as it is a UFD. Then \(px = a\), but by Gauss’s lemma \(x \in R[x]\). By Lemma 2.14 we’re done. □