Tag Archives: polynomial ring

A module which has rank 1 but is not free

Let R = \mathbb{Z}[x] and let M = (2,x) be considered as an R-submodule of R. Show that \{2,x\} is not a basis of M. Show that M has rank 1, but is not free with free rank 1.


Note that 2 \cdot x + (-x) \cdot 2 = 0, so that x and 2 are not R-linearly independent. Thus \{2,x\} has no hope of being a basis of M.

Now suppose \alpha, \beta \in M are nonzero. Similarly, \beta \cdot \alpha + (-\alpha) \cdot \beta = 0, so that \{\alpha, \beta\} is not R-linearly independent. In particular, the rank of M as an R-module is at most 1. On the other hand, since R is an integral domain, every singleton set is linearly independent. So the rank of M over R is exactly 1.

Suppose now that M is free with free rank 1. That is, M = R\alpha for some \alpha. Now 2 = r\alpha for some r \in R, so that \alpha has degree 0. That is, \alpha \in \mathbb{Z}, and we have \alpha \in \{\pm 1, \pm 2\}. If \alpha = \pm 1, then M = R. This is a contradiction since (for example) 1 \notin (2,x). So, without loss of generality, \alpha = 2. But then M = 2\mathbb{Z}[x], and so x \notin M, a contradiction. Thus M is not free of free rank 1 as an R-module.

Exhibit a nontrivial alternating bilinear map from a tensor square

Let R = \mathbb{Z}[x,y] and I = (x,y).

  1. Prove that if ax + by = a^\prime x + b^\prime y in R, then there exists h \in R such that a^\prime = a + yh and b^\prime = b - xh.
  2. Prove that the map \varphi(ax+by,cx+dy) = ad-bc \mod (x,y) is a well-defined, alternating, bilinear map I \times I \rightarrow R/I.

Note that (a-a^\prime)x = (b^\prime - b)y. Since x and y are irreducible (hence prime) in the UFD R, we have that y divides a - a^\prime and x divides b^\prime - b. Say a - a^\prime = yh and b^\prime - b = xh^\prime. Since R is a domain, we see that h = h^\prime, and thus a^\prime = a - yh and b^\prime = b + xh.

Suppose now that a_1x + b_1y = a_2x + b_2y and c_1x + d_1y = c_2x + d_2y. Then there exist h,g \in R such that a_2 = a_1 - yh, b_2 = b_1 + xh, c_2 = c_1 - yg, and d_2 = d_1 + xg. Evidently, a_2d_2 - b_2c_2 \equiv a_1d_1 - b_1c_1 \mod (x,y). Thus \varphi is well defined. Note that \varphi((a_1+a_2)x + (b_1+b_2)y, cx+dy) = (a_1+a_2)d - (b_1+b_2)c = (a_1d - b_1c) + (a_2d - b_2c) = \varphi(a_1x+b_1y,cx+dy) + \varphi(a_2x+b_2y,c_x+d_y), so that \varphi is linear in the first argument; similarly it is linear in the second argument. Moreover, \varphi((ax+by)r,cx+dy) = \varphi(arx+bry,cx+dy) = ard-brc = \varphi(ax+by,r(cx+dy)). Thus \varphi is R-bilinear. Since \varphi(ax+by,ax+by) = ab-ab = 0, \varphi is alternating. Finally, note that \varphi(1,0) = 1. Thus there exists a nontrivial alternating R-bilinear map on I \times I.

Polynomial rings over commutative rings are flat modules over their coefficient rings

Let R be a commutative ring. Show that as a left unital R-module, R[x] is flat.


We claim that R[x] is in fact a free R-module with basis B = \{x^t \ |\ t \in \mathbb{N} \}, using the convention that x^0 = 1. Certainly every element of R[x] is an R-linear combination of elements of B, and if \sum r_ix^i = 0 then r_i = 0 for all i.

Free modules are flat, so R[x] is flat.

Characterize the ideals in F[x] principally generated by monic irreducibles

Let F be a field, let E be an extension of F, and let \alpha \in E be algebraic over F with minimal polynomial p(x). Prove that the ideal (p(x)) in F[x] is equal to \{ q(x) \in F[x] \ |\ q(\alpha) = 0 \}.


If t(x) \in (p(x)), then t(x) = p(x)s(x) for some s(x) \in F[x]. Thus t(\alpha) = p(\alpha)s(\alpha) = 0.

Conversely, if t(\alpha) = 0, then the minimal polynomial p(x) divides t(x). So t(x) \in (p(x)).

Establish an isomorphism between two R-algebras

Let S be a commutative ring with 1 and let R \subseteq S be a subring containing 1. Let I \subseteq R[x_1,\ldots,x_n] be an ideal. Prove that as S-algebras, S \otimes_R R[x_1,\ldots,x_n]/I and S[x_1,\ldots,x_n]/IS[x_1,\ldots,x_n] are isomorphic.


Define \varphi : S \times R[x_1,\ldots,x_n] \rightarrow S[x_1,\ldots,x_n] by \varphi(sr) = sr. This mapping is clearly R-bilinear, and so induces an abelian group homomorphism \Phi : S \otimes_R R[x_1,\ldots,x_n] \rightarrow S[x_1,\ldots,x_n]. Since S is commutative, \Phi is also a ring homomorphism. Moreover, \Phi(1 \otimes 1) = 1 and \Phi(a(s \otimes r)) = \Phi(as \otimes r) = asr = a \Phi(s \otimes r), so that \Phi is an S-algebra homomorphism. Now every simple tensor (hence every element) of S \otimes_R R[x_1,\ldots,x_n] can be written in the form \sum s_i \otimes m_i where the m_i are monic monomials in R[x_1,\ldots,x_n]. Note that \Phi(s \otimes m) = sm for any monomial m and s \in S, so that \Phi is surjective. Moreover, if \Phi(\sum s_i \otimes m_i) = \sum s_im_i = 0 where the m_i are distinct, then s_i = 0 for all i, and thus \sum s_i \otimes m_i = 0. So \mathsf{ker}\ \Phi = 0, and thus \Phi is injective. So \Phi is an S-algebra isomorphism.

Let \pi : R[x_1,\ldots,x_n] \rightarrow R[x_1,\ldots,x_n]/I denote the natural projection, and let \Psi = (1 \otimes \pi) \circ \Phi^{-1}. Then \Psi : S[x_1,\ldots,x_n] \rightarrow S \otimes_R R[x_1,\ldots,x_n]/I is a surjective S-algebra homomorphism. Now let r \in I and s = \sum s_im_i \in S[x_1,\ldots,x_n]. Evidently, \Psi(rs) = \Psi(\sum s_im_ir) = (1 \otimes \pi)(\sum s_i \otimes m_ir) = 0. In particular, IS[x_1, \ldots, x_n] \subseteq \mathsf{ker}\ \Psi. By the (generalized) first isomorphism theorem, the mapping \overline{\Psi} : S[x_1,\ldots,x_n]/IS[x_1,\ldots,x_n] \rightarrow S \otimes_R R[x_1,\ldots,x_n] given by \overline{\Psi}(\overline{s}) = \Psi(s) is well-defined and a surjective S-algebra homomorphism.

Now define \theta : S \times R[x_1,\ldots,x_n]/I \rightarrow S[x_1,\ldots,x_n]/IS[x_1,\ldots,x_n] by \theta(s,r+I) = sr+IS[x_1,\ldots,x_n]. Note that if p-q \in I, then sp-sq \in IS[x_1,\ldots,x_n], so that \theta(s,p+I) = \theta(s,q+I). That is, \theta is well-defined. Clearly \theta is R-bilinear, so that we have an S-algebra homomorphism \Theta : S \otimes_R R[x_1,\ldots,x_n]/I \rightarrow S[x_1,\ldots,x_n]/IS[x_1,\ldots,x_n] such that \Theta(s otimes \overline{r}) = \overline{sr}.

We claim that \overline{\Psi} and \Theta are mutual inverses. To see this, note that for all s \in S and all monic monomial m, we have (\Theta \circ \overline{\Psi})(\overline{sm}) = \Theta(\overline{\Psi}(\overline{sm})) = \Theta(\Psi(sm)) = \Theta((1 \otimes \pi)(\Phi^{-1}(sm))) = \Theta((1 \otimes \pi)(s \otimes m)) = \Theta(s \otimes m+I) = \overline{sm}. Thus \Theta \circ \overline{\Psi} = \mathsf{id}. Similarly, (\overline{\Psi} \circ \Theta)(s \otimes m+I) = \overline{\Psi}(\Theta(s \otimes m+I)) = \overline{\Psi}(\overline{sm}) = \Psi(sm) = (1 \otimes \pi)(\Phi^{-1}(sm)) = (1 \otimes \pi)(s \otimes m) = s \otimes m+I. Thus \overline{\Psi} \circ \Theta = \mathsf{id}.

Hence, as S-algebras, we have S \otimes_R R[x_1,\ldots,x_n]/I \cong S[x_1,\ldots,x_n]/IS[x_1,\ldots,x_n].

Establish an isomorphism between two R-algebras

Let S be a commutative ring with 1 and let R \subseteq S be a unital subring. Prove that as S-algebras, S[x] and S \otimes_R R[x] are isomorphic.


Define \varphi : S \times R[x] \rightarrow S[x] by \varphi(s,r) = sr. This mapping is certainly R-bilinear, and so (since S is commutative) induces an additive group homomorphism \Phi : S \otimes_R R[x] \rightarrow S[x] such that s \otimes r \mapsto sr. Note that \Phi((s_1 \otimes r_1)(s_2 \otimes r_2)) = \Phi(s_1s_2 \otimes r_1r_2) = s_1s_2r_1r_2 = s_1r_1s_2r_2 = \Phi(s_1 \otimes r_1) \Phi(s_2 \otimes r_2), so that \Phi is a ring homomorphism. Moreover, we have \Phi(1 \otimes 1) = 1 and \Phi(a (s \otimes r)) = \Phi(as \otimes r) = asr = a\Phi(s \otimes r), so that \Phi is an S-algebra homomorphism.

Since \Phi(s \otimes x^k) = sx^k, \Phi is surjective. Note that every simple tensor (hence every element) of S \otimes_R R[x] can be written in the form \sum s_i \otimes x^i. If \Phi(\sum s_i \otimes x^i) = \sum s_ix^i = 0, then we have s_i = 0 for all i, and thus \sum s_i \otimes x^i = 0. In particular, \mathsf{ker}\ \Phi = 0, so that \Phi is injective.

Thus S[x] and S \otimes_R R[x] are isomorphic as S-algebras.

Relatively prime polynomials over a field have no common roots

Let F be a field and let p,q \in F[x] be relatively prime. Prove that p and q can have no roots in common.


We noted previously that F[x] is a Bezout domain. In particular, if p and q are relatively prime then there exist a,b \in F[x] such that ap + bq = 1. If \zeta is a root of both p and q, then we have 1 = 0, a contradiction.

In a polynomial ring over a field, greatest common divisors exist and are unique up to units and Bezout’s identity holds

Let F be a field and let a(x),b(x) \in F[x]. Then a greatest common divisor d(x) of a(x) and b(x) exists and is unique up to multiplication by a unit. Moreover, there exist s(x) and t(x) such that d = as+bt.


This follows because F[x] is a Euclidean domain.

Divisors of irreducible polynomials over a field

Let F be a field and suppose p(x) \in F[x] is irreducible. If q(x)|p(x), what can we say about q(x)?


Note that the units in F[x] are precisely the nonzero constant polynomials. If q(x)|p(x), then we have p(x) = q(x)t(x) for some t(x) \in F[x]. Since p(x) is irreducible, either q(x) \in F^\times is a unit, or q is a constant multiple of p(x).

Hilbert’s Basis Theorem for polynomial rings: Every ideal in a polynomial ring over a field is finitely generated

Let F be a field and let R = F[x_1, \ldots, x_t]. Let I \subseteq R be a nonzero ideal. Fix a monomial order. Prove that I has a Gröbner basis with respect to the monomial order and conclude that I is finitely generated.


Let I \subseteq R be a nonzero ideal, and consider the monomial ideal LT(I). By Dickson’s lemma, LT(I) is finitely generated by monomials; say LT(I) = (a_1, \ldots, a_k). Without loss of generality we may assume that no a_i divides another. Recall that LT(I) = (LT(g_i) \ |\ g_i \in I). In particular, we have that for each i \in [1,k], there exist g_i \in I and j \in [1,k] such that a_j|LT(g_i)|a_i. In fact, a_j = a_i, so that a_i and LT(g_i) are associates. That is, we have LT(I) = (LT(g_1), \ldots, LT(g_k)) for some g_i \in I. By Proposition 24 on page 322 of D&F, G = \{g_1, \ldots, g_k\} is a Gröbner basis for I. Then I = (G) is finitely generated.