## Tag Archives: ring

### Every semigroup is isomorphic to a multiplicative subsemigroup of some ring

Show that every semigroup is isomorphic to a multiplicative subsemigroup of some ring. Exhibit a semigroup which is not isomorphic to the (entire) multiplicative semigroup of any ring.

Let $S$ be a semigroup and let $R$ be a ring. Let $R[S] = \{ r : S \rightarrow R\ |\ r_s = 0\ \mathrm{for\ all\ but\ finitely\ many}\ s \in S\}$. That is, $R[S]$ is the set of all functions $S \rightarrow R$ which vanish everywhere except for a finite subset of $S$.

Given $f,g \in R[S]$, we define $f+g$ and $f \cdot g$ as follows.

1. $(f+g)(s) = f(s) + g(s)$
2. $(f \cdot g)(s) = \sum_{tu = s} f(t)g(u)$

In the definition of $\cdot$, if the summation is empty, then $(f \cdot g)(s) = 0$ as usual.

We claim that $(R[S], +, \cdot)$ is a ring.

1. ($+$ is associative) For all $s \in S$, we have $((f+g)+h)(s) = (f+g)(s) + h(s)$ $= f(s) + g(s) + h(s)$ $= f(s) + (g+h)(s)$ $= (f+(g+h))(s)$. So $(f+g)+h = f+(g+h)$.
2. (A $+$-identity exists) Define $0 : S \rightarrow R$ by $0(s) = 0_R$ for all $s \in S$. Now $(f+0)(s) = f(s) + 0(s)$ $= f(s) + 0$ $= f(s)$, so that $f+0 = f$. Similarly, $0 + f = f$.
3. (Every element has a $+$-inverse) Given $f$, define $\overline{f} : S \rightarrow R$ by $\overline{f}(s) = -f(s)$. Then $(f + \overline{f})(s) = f(s) + \overline{f}(s)$ $= f(s) - f(s) = 0$ $= 0(s)$. So $f + \overline{f} = 0$. Similarly, $\overline{f} + f = 0$. We will denote the additive inverse of $f$ by $-f$.
4. ($+$ is commutative) We have $(f+g)(s) = f(s) + g(s)$ $= g(s) + f(s)$ $= (g+f)(s)$.
5. ($\cdot$ is associative)
 $((f \cdot g) \cdot h)(s)$ = $\displaystyle\sum_{tu = s} (f \cdot g)(t) h(u)$ = $\displaystyle\sum_{tu = s} \left(\displaystyle\sum_{vw = t} f(v)g(w)\right) h(u)$ = $\displaystyle\sum_{tu = s} \displaystyle\sum_{vw = t} f(v)g(w)h(u)$ = $\displaystyle\sum_{vwu = s} f(v)g(w)h(u)$ = $\displaystyle\sum_{vt = s} \displaystyle\sum_{wu = t} f(v) g(w) h(u)$ = $\displaystyle\sum_{vt = s} f(v) \left( \displaystyle\sum_{wu = t} g(w) h(u) \right)$ = $\displaystyle\sum_{vt = s} f(v) (g \cdot h)(t)$ = $(f \cdot (g \cdot h))(s)$

So $f \cdot (g \cdot h) = (f \cdot g) \cdot h$.

6. ($\cdot$ distributes over $+$ from the left)
 $(f \cdot (g + h))(s)$ = $\displaystyle\sum_{tu = s} f(t) (g+h)(u)$ = $\displaystyle\sum_{tu = s} f(t)(g(u) + h(u))$ = $\displaystyle\sum_{tu = s} \left( f(t) g(u) + f(t) h(u) \right)$ = $\displaystyle\sum_{tu = s} f(t) g(u) + \displaystyle\sum_{tu = s} f(t)h(u)$ = $(f \cdot g)(s) + (f + h)(s)$ = $((f \cdot g) + (f \cdot h))(s)$

So $f \cdot (g + h) = (f \cdot g) + (f \cdot h)$

7. ($\cdot$ distributes over $+$ from the right) Similar to the proof for left distributivity.
8. (If $S$ has a left identity, then $R[S]$ has a left identity) If $e \in S$ is a left identity, define $\overline{e} : S \rightarrow R$ by $\overline{e}(e) = 1$ and $\overline{e}(s) = 0$ if $s \neq e$. Now $(\overline{e} \cdot f)(s) = \sum_{tu = s} \overline{e}(t) f(u)$. If $t \neq e$, then $\overline{e}(t) = 0$, and if $t = e$, then $u = s$. (The pair $(e,s)$ is certainly in the index set of the summation.) So this is equal to $f(s)$. Hence $\overline{e} \cdot f = f$.
9. (If $S$ has a right identity, then $R[S]$ has a right identity) Similar to the proof for left identity.

So $R[S]$ is a ring. Rings of this type are called semigroup rings, and their elements are typically written as ‘polynomials’ in the elements of $S$ with coefficients from $R$. Addition and multiplication are then carried out as usual for polynomials – multiply the semigroup elements (preserving order) with their coefficients, and then collect like terms.

Given $s \in S$, we define $\varphi_s : S \rightarrow R$ by $\varphi_s(t) = 1$ if $t = s$ and 0 otherwise. We claim that the map $\Phi : S \rightarrow R[S]$ given by $s \mapsto \varphi_s$ is a semigroup homomorphism. Indeed, $\Phi(st)(a) = \varphi_{st}(a) = 1$ if $a = st$ and $0$ otherwise. Now $(\varphi_s \cdot \varphi_t)(a) = \sum_{uv = a} \varphi_s(u) \varphi_t(v)$. If $u \neq s$, then $\varphi_s(u) = 0$, and if $v \neq t$, then $\varphi_t(v) = 0$. So $(\varphi_s \cdot \varphi_t)(a) = 1$ if $a = st$ and $0$ otherwise. Hence $\Phi(st) = \Phi(s) \cdot \Phi(t)$.

So every semigroup can be embedded in the multiplicative semigroup of a ring. However, not every semigroup is isomorphic to the full multiplicative semigroup of some ring, as we argue. Note that every multiplicative semigroup of a ring has a zero. So any semigroup without a zero cannot be the multiplicative semigroup of a ring.

### Finitely generated modules over R are precisely the module-homomorphic images of Rⁿ

Let $R$ be a ring. Prove that $M$ is a finitely generated module over $R$ if and only if $M$ is a module-homomorphic image of $R^n$.

Suppose first that $M$ is a finitely generated $R$-module – say by $A = \{a_1,\ldots,a_n\}$. Let $e_i$ denote the $i$th standard basis vector in $R^n$, and define $\varphi : R^n \rightarrow M$ by $e_i \mapsto a_i$, and extend linearly. Certainly then $\varphi$ is a surjective module homomorphism.

Conversely, suppose $\varphi : R^n \rightarrow M$ is a surjective module homomorphism, and for each $e_i$ let $a_i = \varphi(e_i)$. Since $\varphi$ is surjective, and since the $e_i$ generate $R^n$, the set $A = \{a_1, \ldots, a_n\}$ generates $M$ over $R$ as desired.

### If R is a Noetherian ring, then Rⁿ is a Noetherian R-module

Let $R$ be a Noetherian ring. Prove that $R^n$, as an $R$-module in the usual way, is also Noetherian.

Recall that a ring is Noetherian if every ideal is finitely generated, and a module is Noetherian if every submodule is finitely generated. (That is, a ring is Noetherian (as a ring) if it is Noetherian (as a module) over itself.)

We will proceed by induction on $n$.

For the base case, $n = 1$, $R^1$ is certainly Noetherian as a module over $R$.

For the inductive step, suppose $R^n$ is Noetherian. Now let $M \subseteq R^{n+1}$ be a submodule; our goal is to show that $M$ must be finitely generated. To that end, let $A = \{r \in R \ |\ (m_i)_0 = r\ \mathrm{for\ some}\ (m_i) \in M \}$. That is, $A$ is the collection (in $R$) of all zeroth coordinates of elements in $M$.

We claim that $A \subseteq R$ is an ideal. If $a,b \in A$, then there exist elements $(m_i)$ and $(n_i)$ in $M$ such that $m_0 = a$ and $n_0 = b$. Now since $M \subseteq R^{n+1}$ is a submodule, we have $(m_i) + r(n_i) \in M$ for all $r \in R$, so that $a+rb \in A$. We clearly have $0 \in A$, so that by the submodule criterion, $A$ is an ideal of $R$.

In particular, since $R$ is Noetherian, the ideal $A$ is finitely generated – say by $\{a_0, a_1, \ldots, a_k\}$. We will let $\alpha_i$ be an element of $M$ whose zeroth coordinate is $a_i$. Now let $m = (m_0,\ldots,m_{n+1}) \in M$. Now $m_0 = \sum c_i a_i$ for some $c_i \in R$, and so $m - \sum c_i m_i = (0,t_1,\ldots,t_{n+1})$ is an element of $M$ whose first coordinate is 0. In particular, we have $M = (m_0,\ldots,m_k) + B$, where $B = M \cap (0 \times R^n)$. (We showed the $(\subseteq)$ direction, and the $(\supseteq)$ inclusion is clear.) Now $M \cap (0 \times R^n)$ is an ideal of $0 \times R^n \cong_R R^n$, which is Noetherian by the induction hypothesis. So $B$ is finitely generated as an $R$-module, and thus $M$ is finitely generated over $R$.

Since $M$ was an arbitrary submodule of $R^{n+1}$, $R^{n+1}$ is Noetherian as a module.

### The quotient of a product is module isomorphic to the product of quotients

Let $R$ be a ring, let $\{A_i\}_{i=1}^n$ be a finite family of (left, unital) $R$-modules, and let $B_i \subseteq A_i$ be a submodule for each $i$. Prove that $(\prod A_i)/(\prod B_i) \cong_R \prod A_i/B_i$.

We did this previously. D&F, why repeat an exercise?

### Representativewise multiplication of cosets of an ideal is well-defined

Let $R$ be a ring and let $I \subseteq R$ be an ideal. Show that if $a_1 - b_1 \in I$ and $a_2 - b_2 \in I$, then $a_1a_2 - b_1b_2 \in I$.

Since $I$ is an ideal, $a_1a_2 - b_1a_2 \in I$ and $b_1a_2 - b_1b_2 \in I$. So $a_1a_2 - b_1a_2 + b_1a_2 - b_1b_2$ $= a_1a_2 - b_1b_2 \in I$ as desired.

### 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 $C \subseteq A$.

### Basic properties of ideal products

Let $R$ be a commutative ring with 1 and let $A,B,C \subseteq R$ be ideals. Prove the following. (1) $A(BC) = (AB)C$, (2) $AB = BA$, (3) If $a \in A$ and $b \in B$ then $ab \in AB$, and (4) $AB \subseteq A$.

Let $a \in A$ and $b \in B$. Certainly then by our equivalent characterization of ideal products, $ab \in AB$.

By this previous exercise, if $z \in A(BC)$ then $z = \sum_i a_i (\sum_j b_{i,j}c_{i,j}) = \sum_i \sum_j a_i b_{i,j} c_{i,j}$. Each $a_ib_{i,j}$ is in $AB$, so that $(a_ib_{i,j})c_{i,j} \in (AB)C$. So $A(BC) \subseteq (AB)C$. The reverse inclusion is similar.

Suppose $\sum a_ib_i \in AB$. Since $A$ is an ideal, each $a_ib_i$ is in $A$. Then $\sum a_ib_i \in A$.

Since $R$ is commutative, $AB = \{\sum_{i=1}^n a_ib_i \ |\ n \in \mathbb{N}, a_i \in A, b_i \in B\}$ $= \{\sum_{i=1}^n b_ia_i \ |\ n \in \mathbb{N}, a_i \in A, b_i \in B\} = BA$.

### An equivalent characterization of ideal products

In TAN, we defined the product of (finitely generated) ideals $I = (A)$ and $J = (B)$ to be $I \star J = (ab \ |\ a \in A, b \in B)$. We can also define an ideal product $IJ = \{\sum_T x_iy_i \ |\ T\ \mathrm{finite}, x_i \in I, y_i \in J\}$. Prove that $IJ = I \star J$.

We did this previously.

### The ideal product is well-defined

In TAN, the product of two finitely generated ideals $(A)$ and $(B)$ was defined to be $(ab \ |\ a \in A, b \in B)$. Argue that this is a well-defined operator on the set of ideals of a fixed ring $R$.

More broadly, $IJ = (xy \ |\ x \in I, y \in J)$. If $I = (A)$ and $J = (B)$, then each $x$ is an $R$-linear combination of the $a_i \in A$, and each $y$ a combination of the $b_i \in B$. Since $R$ is a commutative ring, then, each $xy$ is an $R$-linear combination of $a_ib_j$ where $a_i \in A$ and $b_j \in B$; that is, $IJ \subseteq (ab \ |\ a \in A, b \in B)$. The reverse inclusion follows from this previous exercise.

A more modern treatment would derive the generating set characterization of $IJ$ from the more general definition.

### In a ring with 1, an ideal which contains 1 is the entire ring

Let $R$ be a ring with 1 and let $I \subseteq R$ be an ideal. Show that if $1 \in I$, then $I = R$.

Recall that if $a \in R$ and $b \in I$, then $ab \in I$. Letting $a$ be arbitrary and $b = 1$, then, we have $R \subseteq I$. Hence $R = I$.