## 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.