Tag Archives: commutative ring

The tensor algebra of a cyclic module is commutative

Let R be a commutative ring with 1, and let M be an (R,R)-bimodule in the usual way. (I.e. rm = mr.) Prove that if M is cyclic as an R-module, then the tensor algebra \mathcal{T}(M) is commutative.


Note that as a ring, \mathcal{T}(M) is generated by 0- and 1-tensors. Thus to show that \mathcal{T}(M) is commutative, it suffices to show that these generators commute pairwise.

Certainly the 0-tensors commute with each other, since R is commutative. Similarly, 0- and 1-tensors commute pairwise using the “commutativity condition” rm = mr. It remains to be seen that 1-tensors commute with each other. To that end, suppose M = Ra, and let ra,sa \in M. Note that ra \otimes sa = ar \otimes sa = as \otimes ra = sa \otimes ra, and indeed the 1-tensors commute. Thus \mathcal{T}(M) is a commutative R-algebra.

In particular, we have m_1 \otimes m_2 - m_2 \otimes m_1 = 0 for all m_1 and m_2, so that \mathcal{S}(M) \cong_R \mathcal{T}(M).

Interaction between the determinant of a square matrix and solutions of a linear matrix equation over a commutative unital ring

Let R be a commutative ring with 1, let V be an R-module, and let X = [x_1\ \cdots\ x_n]^\mathsf{T} \in V^n. Suppose that for some matrix A \in \mathsf{Mat}_{n \times n}(R) we have AX = 0. Prove that \mathsf{det}(A)x_i = 0 for all x_i.


Let B be the transpose of the matrix of cofactors of A. That is, B = [(-1)^{i+j} \mathsf{det}(A_{j,i})]_{i,j=1}^n. By Theorem 30 in D&F, we have BA = \mathsf{det}(A)I, where I is the identity matrix.

Now if AX = 0, then BAX = B0 = 0, so that \mathsf{det}(A)X = 0. Comparing entries, we have \mathsf{det}(A)x_i = 0 for all i.

Facts about sums of ideals

Let R be a commutative ring with 1, and let A,B,C \subseteq R be ideals. Recall that A+B = \{a+b \ |\ a \in A, b \in B\}. Prove the following.

  1. A+B is an ideal
  2. A,B \subseteq A+B
  3. If A,B \subseteq C then A+B \subseteq C
  4. A+B = (A,B)
  5. If (A,B) = (1), then there exist \alpha \in A and \beta \in B such that \alpha+\beta = 1.
  6. If (A,B) = (1) and BC \subseteq A, then C \subseteq A.

Suppose a_1+b_1,a_2+b_2 \in A+B, and let r \in R. Then (a_1+b_1) + r(a_2+b_2) = (a_1+ra_2) + (b_1+rb_2) \in A+B since A and B are ideals. Moreover, 0 = 0+0 \in A+B. By the submodule criterion, A+B \subseteq R is an ideal.

For all a \in A, a = a+0 \in A+B. So A \subseteq A+B, and similarly B \subseteq A+B.

Suppose A, B \subseteq C for some ideal C. Since C is an ideal, it is closed under sums, so that a+b \in C for all a \in A and b \in B. Thus A+B \subseteq C.

Note that A, B \subseteq (A,B), so that A+B \subseteq (A,B) by the previous point. Now let x \in (A,B); by definition, x = \sum r_ic_i for some r_i \in R and c_i \in A \cup B. Collecting terms in A and B, we have x = \alpha + \beta for some \alpha \in A and \beta \in B. Thus A+B = (A,B).

Suppose (A,B) = (1). By the previous point, A+B = (1), and in particular 1 = \alpha+\beta for some \alpha \in A and \beta \in B.

Suppose (A,B) = (1). By the previous point, there exist \alpha \in A and \beta \in B such that \alpha+\beta = 1. Now let \gamma \in C. We have \gamma = \gamma(\alpha+\beta) = \gamma\alpha + \gamma\beta. Since BC \subseteq A, \gamma\beta \in A, and so \gamma \in A. Thus C \subseteq A.

Over a commutative ring, the tensor product of two flat modules is flat

Let R be a commutative ring and let M and N be R-modules. Show that if M and N are flat over R, then M \otimes_R N is flat over R.


This is a special case of the previous exercise, since both M and N can naturally be considered (R,R)-bimodules.

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.

Over a commutative ring, the tensor product of two free modules is free

Let R be a commutative ring with 1.

  1. Prove that the tensor product of two free R-modules is free.
  2. Prove that the tensor product of two projective R-modules is projective.

Suppose A and B are free R-modules; we can assume (since R is commutative) that A = \bigoplus_I R and B = \bigoplus_J R for some nonempty index sets I and J. By this previous exercise, binary tensor products essentially commute with direct sums. Thus, A \otimes_R B = (\bigoplus_I R) \otimes (\bigoplus_J R) \cong_R \bigoplus_I (R \otimes_R (\bigoplus_J R)) \cong_R \bigoplus_I \bigoplus_J (R \otimes_R R) \cong_R \bigoplus_{I \times J} R. Thus A \otimes_R B is free.

Now suppose A and B are projective. Then there exist M and N so that A \oplus M and B \oplus N are free. By the previous argument, (A \oplus M) \otimes_R (B \oplus N) is free. Now (A \oplus M) \otimes_R (B \oplus N) \cong_R A \otimes_R (B \oplus N) \oplus M \otimes_R (B \oplus N) \cong_R (A \otimes_R B) \oplus (A \otimes_R N) \oplus (M \otimes_R (B \oplus N)) is free, so that A \otimes_R B is projective.

In an (R,R)-bimodule on which elements act the same from the left or right, ring elements commute up to action

Let R be a ring and let M be an (R,R)-bimodule such that r \cdot m = m \cdot r for all m \in M and r \in R. Prove that for all r,s \in R and m \in M, rs \cdot m = sr \dot m. (That is, the assumption that R is commutative in the definition of an R-algebra is a rather natural one.)


Let r,s \in R and m \in M. Then we have rs \cdot m = r \cdot (s \cdot m) = r \cdot (m \cdot s) = (m \cdot s) \cdot r = m \cdot sr = sr \cdot m, as desired.

The tensor product of the quotients of a commutative ring by two ideals is isomorphic to the quotient by their sum

Let R be a commutative ring with ideals I and J. Let R/I and R/J be R-modules (in fact (R,R)-bimodules) in the usual way.

  1. Prove that every element of R/I \otimes_R R/J can be written as a simple tensor of the form (1 + I) \otimes (r + J).
  2. Prove that R/I \otimes_R R/J \cong_R R/(I+J).

We will prove the first result first for simple tensors; the extension to arbitrary sums of tensors follows by tensor distributivity. Let (a+I) \otimes (b+J) be an arbitrary simple tensor in R/I \otimes_R R/J. Now (a+I) \otimes (b+J) = (1+I)a \otimes (b+J) = (1+I) \otimes a(b+J) = (1+I) \otimes (ab+J), as desired.

Now define \varphi : R/I \times R/J \rightarrow R/(I+J) by (a+I, b+J) \mapsto (a+b) + (I+J). First, suppose a_1-a_2 \in I and b_1-b_2 \in J. Then b_1(a_1-a_2) \in I and a_2(b_1-b_2) \in J, so that a_1b_1 - a_2b_2 \in I+J. Thus \varphi is well-defined. It is clear that \varphi is R-balanced, and so induces an R-module homomorphism \Phi : R/I \otimes_R R/J \rightarrow R/(I+J). Since \varphi((1+I) \otimes (r+J)) = r + (I+J), \Phi is surjective. Now suppose (1+I) \otimes (r+J) is in the kernel of \Phi; then r \in I+J. Say r = a+b where a \in I and b \in J. Then (1 + I) \otimes (r + J) = (1+I) \otimes (a+J) + (1+I) \otimes (b+J) = (a+I) \otimes (1+J) + (1+I) \otimes (b+J) = 0+0 = 0, and hence \Phi is an isomorphism.

Over a commutative ring, the tensor product of a module by a free module of finite rank is a direct power

Let R be a commutative ring and let N be a free R-module of finite rank n; say B = \{e_i\}_{i=1}^n is a basis for N.

  1. Let M be any nonzero R-module. (Since R is commutative, M is naturally an (R,R)-bimodule.) Show that M \otimes_R N \cong_R M^n. In particular, every element of M \otimes_R N can be written uniquely in the form \sum_{i=1}^n m_i \otimes e_i.
  2. Show that if \sum m_i \otimes n_i = 0, where the n_i are merely assumed to be R-linearly independent in N, it need not be the case that the m_i are all zero. (That is, it is crucial that the n_i generate N as an R-module.)

Define a mapping \varphi_B : M \times N \rightarrow M^n by \varphi_B(m, \sum r_ie_i) = (r_i \cdot m). (This is well defined since B is a basis for N.) Moreover, \varphi_B(m_1 + m_2, \sum r_ie_i) = (r_i(m_1 + m_2)) = (r_i m_1) + (r_i m_2) = \varphi_b(m_1, \sum r_ie_i) + \varphi_b(m_2, \sum r_ie_i), \varphi(m, (\sum r_i e_i) + (\sum s_i e_i)) = \varphi_B(m, \sum (r_i+s_i)e_i) = ((r_i+s_i)m) = (r_im) + (s_im) = \varphi_B(m, \sum r_ie_i) + \varphi_B(m, \sum s_i e_i), and \varphi_B(m \cdot a, \sum r_i e_i) = (r_i (ma)) = (ar_i m) = \varphi_B(m, \sum ar_ie_i) = \varphi_B(m, a \sum r_ie_i), so that \varphi_B is bilinear. By the universal property of tensor products, \varphi_B induces a unique group homomorphism (indeed, R-module homomorphism) \Phi_B : M \otimes_R N \rightarrow M^n such that \Phi_B(m \otimes \sum r_i e_i) = (r_im).

Now define \Psi_B : M^n \rightarrow M \otimes_R N by (m_i) \mapsto \sum m_i \otimes e_i. Clearly \Psi_B is an R-module homomorphism.

Moreover, note that (\Psi_B \circ \Phi_B)(m \otimes \sum r_i e_i) = \Psi_B(r_im)) = \sum r_im \otimes e_i = \sum m \otimes r_ie_i = m \otimes \sum r_ie_i. Similarly, (\Phi_B \circ \Psi_b)(m_i) = \Phi_B(\sum m_i \otimes e_i) = \sum \Phi_B(m_i \otimes e_i) = (m_i). Thus \Phi_B and \Psi_B are mutual inverses, and we have M \otimes_R N \cong_R M^n.

In particular, \sum m_i \otimes e_i = 0 if and only if m_i = 0 for all i.

Now for the counterexample, let R = \mathbb{Z}, let N = \mathbb{Z}^1 be the free \mathbb{Z}-module of rank 1, and let M = \mathbb{Z}/(2) be a \mathbb{Z}-module in the usual way. Consider the simple tensor \overline{1} \otimes 2 in M \otimes_R N; note that 1 \neq 0, but that \overline{1} \otimes 2 = \overline{1} \otimes 2 \cdot 1 = \overline{1} \cdot 2 \otimes 1 = \overline{2} \otimes 1 = 0 \otimes 1 = 0.

The Euclidean norm on ZZ[sqrt(-5)]

Let N : \mathbb{Z}[\sqrt{\text{-}5}] \rightarrow \mathbb{N} be the Euclidean norm given by N(a+b\sqrt{\text{-}5}) = a^2 + 5b^2.

  1. Show that N(\alpha\beta) = N(\alpha)N(\beta) for all \alpha,\beta.
  2. Show that N(\alpha) = 1 if and only if \alpha is a unit.
  3. Show that N(\alpha) = 0 if and only if \alpha = 0.
  4. Show that if N(\alpha) is prime as a natural number then \alpha is irreducible in \mathbb{Z}[\sqrt{\text{-}5}].
  5. Show that 3 + 2\sqrt{\text{-}5} and 2 + \sqrt{\text{-}5} are prime.

Let \alpha = a_1 + a_2\sqrt{\text{-}5} and \beta = b_1 + b_2\sqrt{\text{-}5}, and note the following.

N(\alpha\beta)  =  N((a_1+a_2\sqrt{\text{-}5})(b_1+b_2\sqrt{\text{-}5}))
 =  N((a_1b_1 - 5a_2b_2) + (a_1b_2+a_2b_1)\sqrt{\text{-}5})
 =  (a_1b_1 - 5a_2b_2)^2 + 5(a_1b_2+a_2b_1)^2
 =  a_1^2b_1^2 + 5a_1^2 b_2^2 + 5a_2^2b_2^2 + 25 a_2^2b_2^2
 =  (a_1^2 + 5a_2^2)(b_1^2 + 5b_2^2)
 =  N(\alpha)N(\beta)

Suppose \alpha is a unit; then there exists \beta such that \alpha\beta = 1. Considering norms, we have N(\alpha)N(\beta) = 1 (using the previous result). So N(\alpha) = 1. Conversely, suppose N(a+b\sqrt{\text{-}5}) = a^2+5b^2 = 1; then b = 0, and we have a = \pm 1. Thus \alpha is a unit.

Certainly N(0) = 0. Now suppose N(a+b\sqrt{\text{-}5}) = a^2+5b^2 = 0; then a = b = 0.

If N(\alpha) is prime and \alpha = \beta\gamma, then N(\beta)N(\gamma) is prime. Without loss of generality, N(\beta) = 1, so that \beta is a unit. Thus \alpha is irreducible.

Note that N(3+2\sqrt{\text{-}5}) = 3^2 + 5\cdot 2^2 = 29 is prime; thus 3+2\sqrt{\text{-}5} is irreducible.

Note that a^2 + 5b^2 = 3 has no solutions in the integers since 3 is not a perfect square. In particular, no element of \mathbb{Z}[\sqrt{\text{-}5}] has norm 3. Now N(2 + \sqrt{\text{-}5}) = 9. If 2 + \sqrt{\text{-}5} = \alpha\beta, then N(\alpha)N(\beta) = 9. since no element of this ring has norm 3, without loss of generality we have N(\alpha) = 1 so that \alpha is a unit. Thus 2 + \sqrt{\text{-}5} is irreducible.