2.6 Modules, Vector spaces, Algebras

note

In the forthcoming sections, we will omit the proofs of certain self-evident truths. For example, within the context of modules, it's understood that multiplying by $(−1)$ results in the negation of an element, i.e., $(−1)⋅x=−x$. These facts adhere to the same reasoning principles as those detailed in earlier discussions and can be intuitively grasped by readers who have thoroughly engaged with the material presented in previous sections.

Modules

def: Module

$$\begin{align*} &\sphericalangle \\ &M \in \mathcal G^{\mathcal A}_+ \\ &R \in \mathcal R^1\\ &\cdot: R\times M \to M \\ &\forall a, b \in R, x,y \in M:\\ \text{Associativity:} \,\,&(ab)\cdot x=a\cdot (b \cdot x) \\ \text{Distributivity:} \,\, &a\cdot(x+y) = a\cdot x + a\cdot y \\ &(a+b)\cdot x = a\cdot x + b\cdot x \\ \text{Identity:} \,\,&1 \cdot x = x \\ \hline \\ &M \in \mathcal M_R \,\,(M \text{ is } R\text{-module}) \\ \end{align*}$$

note

Operation $\cdot$ on modules is called scalar multiplication and elements from $R$ - scalars

def: Faithful module

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^1\\ &M \in \mathcal M_R \\ &(\exists r \in R: \forall m \in M: rm=0) \implies r = 0 \\ \hline \\ &M \in \mathcal M^F_R \,\,(M \text{ is a faithful } R\text{-module}) \\ \end{align*}$$

Proposition 2.6.1: Ring is a module

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^1 \\ \hline \\ &R \in \mathcal M_R \end{align*}$$

Proof

Obvious by using ring axioms in module definition

Example: Ideal as a module

$R \in \mathcal R^1, I \lhd_R R \implies I \in \mathcal M_R$

Example: Abelian group

$M \in \mathcal G^{\mathcal A}_+ \implies M \in \mathcal M_\Z$, where $nx:=\underbrace{x+x+\ldots+x}_{n \text{ times}}$.

Example: Polynomials

$R \in \mathcal R^1 \implies R[x] \in \mathcal M_R$

Submodules and factor modules

def: Submodule

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^1 \\ &M \in \mathcal M_R \\ &N \subseteq_G M \\ &\forall r \in R, m \in N: rm \in N & \\ \hline \\ &N \subseteq_M M \end{align*}$$

Proposition 2.6.2 Submodule criterion

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^\mathcal{C} \\ &M \in \mathcal M_R \\ &N \subseteq M \\ \hline \\ &N \subseteq_M M \iff N \ne \empty, \forall r \in R, x, y \in N: x+ry \in N \end{align*}$$

Proof

$\implies$

Obvious

$\impliedby$

$$r = -1 \implies N \ne \empty, \forall x, y \in N: x-y \in N \overset{(2.2.3)}\implies N \subseteq_G M \\ x = 0 \implies \forall r \in R, y \in N: ry \in N$$

$\square$

def: Factor module

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^\mathcal{1} \\ &M \in \mathcal M_R \\ &N \subseteq_M M \\ &\forall a \in R, x + N \in M/N: a(x+N):= ax + N \\ \hline \\ &M / N - \text{factor module} \end{align*}$$

Note that if $x_1+N=x_2+N \implies x_1-x_2 \in N \implies ax_1-ax_2 \in N \implies ax_1 + N = ax_2 + N$, so the multiplication is well-defined.

Homomorphisms and isomorphisms

def: Module homomorphism

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^1 \\ &M_1, M_2 \in \mathcal M_R \\ &\phi: M_1 \to M_2 \\ &\phi(x+y)= \phi(x) + \phi(y), x,y \in M_1 \\ &\phi(ax)=a\phi(x), a \in R, x \in M_1 \\ \hline \\ &\phi: M_1 \rightsquigarrow_M M_2, M_1 \overset{\phi}\rightsquigarrow_M M_2,\\ &\phi - \text{module homomorphism} \end{align*}$$

def: Module isomorphism

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^1 \\ &M_1, M_2 \in \mathcal M_R \\ &\phi: M_1 \rightsquigarrow_M M_2\\ &\phi: M_1 \leftrightarrow M_2 \\ \hline \\ &\phi: M_1 \cong M_2, M_1 \overset{\phi}\cong M_2 \\ &\phi - \text{module isomorphism} \\ \end{align*}$$

Proposition 2.6.3: Homomorphisms kernel is submodule

$$\begin{align*} &\sphericalangle \\ & R \in \mathcal R^{1} \\ &M_1, M_2 \in \mathcal M_R \\ &\phi: M_1 \rightsquigarrow_M M_2 \\ \hline \\ &\ker \phi \subseteq_M M_1 \end{align*}$$

The proof is very similar to $(2.3.8)$ so we'll omit that.

Again we'll state four isomorphism theorems for modules without proofs, as proofs are very similar to that of group isomorphism theorems.

Proposition 2.6.4: First isomorphism theorem

$$\begin{align*} &\sphericalangle \\ & R \in \mathcal R^1 \\ &M_1, M_2 \in \mathcal M_R \\ &M_1 \overset{\psi}{\rightsquigarrow}_M M_2 \\ &\phi:M_1 \to M_1/\ker \psi - \text{natural homomorphism} \\ \hline \\ &\exists! \eta: M_1/\ker \psi \overset{\eta}{\cong}_M \psi(M_1), \psi = \eta \phi \end{align*}$$

Theorem 2.6.5: Second isomorphism theorem

$$\begin{align*} &\sphericalangle \\ & R \in \mathcal R^1 \\ &M \in \mathcal M_R \\ &N, K \subseteq_M M \\ \hline \\ &\begin{align*} & N + K \subseteq_M M \tag{a}\\ & N \cap K \subseteq_M M \tag{b}\\ & (N+K)/N \cong_M K / (K \cap N) \hspace{1cm} \tag{c}\\ \end{align*} \end{align*}$$

Theorem 2.6.6: Third isomorphism theorem

$$\begin{align*} &\sphericalangle \\ & R \in \mathcal R^1 \\ & M \in \mathcal M_R \\ &N, K \subseteq_M M \\ &K \subseteq N \\ \hline \\ &M/N \cong \frac{M/K}{N/K} \end{align*}$$

Theorem 2.6.7: Correspondence theorem (fourth isomorphism theorem)

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^1 \\ &M \in \mathcal M_R \\ &N \subseteq_M M \\ &\mathfrak S :=\{S: N\subseteq_M S \subseteq_M M\} \\ &\mathfrak S_N :=\{S_N: S_N \subseteq_M M/N\} \\ &\phi: \mathfrak S \to \mathfrak S_N, S \mapsto S/N \\ \hline \\ &\begin{align*} & \phi - \text{well-defined} \tag{a} \\ & \phi^{-1}(S_N) =\{x \in M: x+N \in S_N\} \tag{b}\\ & \phi: \mathfrak S \leftrightarrow \mathfrak S_N \tag{c}\\ &N \subseteq_R S \subseteq_M M \implies \phi(S) \subseteq_M M / N \hspace{1cm} \tag{d}\\ & S_N \subseteq_M M/N \implies \phi^{-1}(S_N) \subseteq_M M \hspace{1cm} \tag{e}\\ \end{align*} \end{align*}$$

Finitely generated modules

def: Finitely generated module

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^1 \\ &L \in \mathcal M_R \\ &S:= \{s_1, \ldots, s_n\}, S \subseteq L \\ &\lang S \rang_M : S \subseteq \lang S \rang_M \subseteq_M L\\ &\forall N: S \subseteq N \subseteq_M L: \lang S \rang_M \subseteq N \\ \hline \\ &\lang S \rang_{M_R} \equiv \lang S \rang_M \equiv \lang s_1, \ldots, s_n \rang_M - \text{finitely generated module} \end{align*}$$

Proposition 2.6.8: Finitely generated module explicit form

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^1 \\ &M \in \mathcal M_R \\ &S:= \{s_1, \ldots, s_n\}, S \subseteq M \\ \hline \\ &\lang S \rang_M = \{r_1s_1+\ldots+r_ns_n, \forall r_i \in R\} \end{align*}$$

Proof

$$N:=\{r_1s_1+\ldots+r_ks_k, r_i \in R\} \\ N \ne \empty \\ \forall x, y \in N, a \in A: x+ry = \\ =r_{1,1}s_1+\ldots+r_{k, 1}s_k + rr_{2, 1}s_1+\ldots+rr_{2, k}s_k = \\ =r_{3, 1}s_1+\ldots+r_{3, k}s_k \in N \\ (2.6.2) \implies N \subseteq_M L$$

Finally it's obvious that $S \subseteq N$ and any submodule containing $S$ must at least contain $N$ ($a_is_i$ must be in this submodule and thus their sums as well).

$\square$

def: Linear independence

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^1 \\ &M \in \mathcal M_R \\ &X :=\{x_1, \ldots, x_n\} \subseteq M \\ &\forall a_1, \ldots, a_n \in R: a_1x_1 + \ldots + a_nx_n = 0 \implies a_1 = \ldots = a_n = 0 \\ \hline \\ &X \in \bot_L(M), X - \text{linearly independent over }M \end{align*}$$

def: Module basis and rank

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^1 \\ &N \in \mathcal M_R \\ &L \subseteq N \\ &L \in \bot_L(N)\\ &\lang L \rang_M = N \\ \hline \\ & L \in \mathfrak B_M(N), L - \text{basis of } N \\ & \text{rank}_R N:=n, n- \text{rank of } N \\ \end{align*}$$

Proposition 2.6.9: Each module element is a unique linear combination of basis elements

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^1 \\ &M \in \mathcal M_R \\ &L := \{x_1, \ldots, x_n\} \,(n= \infty \text{ is possible})\\ &L \in \mathfrak B_M(M) \\ \hline \\ & \forall x \in M: \exists! r_1, \ldots, r_n \in R: x = r_1x_1+ \ldots + r_nx_n \\ \end{align*}$$

Proof

$$\exists r_1, \ldots, r_n, s_1, \ldots, s_n \in R: \\ \exists i: r_i\ne s_i, r_1x_1+ \ldots + r_nx_n = s_1x_1+ \ldots + s_nx_n \implies \\ (r_1 - s_1)x_1+ \ldots + (r_n-s_n)x_n = 0 \overset{L \in \bot_L(M)}\implies r_i=s_i$$

$\square$

Proposition 2.6.10: Nakayama's lemma

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &M \in \mathcal M_A \\ &\exists S \subseteq M, |S| < \infty: \lang S \rang_I=M \\ &I \lhd_R A, I \subseteq \bigcap_{J \in \mathfrak M_I(A)}J \\ &IM = M \\ \hline \\ &M = \{0\} \end{align*}$$

Proof

Assume that $M \ne \{0\}$ and $S=\{s_1, \ldots, s_n\}$ is the minimal set such that $\lang S \rang_I=M$. Since

$$IM = M \implies s_n \in IM \implies \exists m \in M, a_i \in A, i \in I: \\ s_n = im = i(a_1s_1 + \ldots + a_ns_n)= \\ i_1s_1 + \ldots + i_ns_n \implies (1-i_n)s_n=i_1s_1 + \ldots + i_{n-1}s_{n-1}$$

If $1-i_n \notin A^*$ then it's contained in some maximal ideal $\mathfrak m$ (because $\lang 1-i_n \rang_I \ne A$ by $(2.4.6)$). But $i_n \in I \subseteq \bigcap_{m \in \mathfrak M_I(A)}m \subseteq \mathfrak m$. This would lead to $1 \in \mathfrak m$ which is impossible for maximal ideal.

So $1-i_n \in A^*$ and thus

$$s_n = (1-i_n)^{-1}i_1s_1 + \ldots + (1-i_n)^{-1}i_{n-1}s_{n-1}$$

Which is a contradiction to $S$ being minimal. Note that in case $n=1$ this would mean that $s_1=0$ which is also a contradiction.

So we proved $M = \{0\}$

$\square$

Corrolary 2.6.11: Module equality check

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &M, N \in \mathcal M_A \\ &\exists L: |L| < \infty, \lang L \rang_M = M \\ &I \lhd_R A, I \subseteq \bigcap_{J \in \mathfrak M_I(A)}J \\ &N \subseteq_M M \\ &M = IM+N \\ \hline \\ &M = N \end{align*}$$

Proof

Notice that $I(M/N)=IM/N=(IM+N)/N=M/N$. Obviously $M/N$ is finitely generated, so we can apply $(2.6.10)$ to $I, M/N$ and thus $M/N = \{0\}$, so $M=N$.

$\square$

Proposition 2.6.12: Cayley–Hamilton theorem

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &M \in \mathcal M_A \\ &\exists L: |L| < \infty, \lang L \rang_M = M \\ & I \lhd_R A \\ & \phi: M \rightsquigarrow M \\ & \phi(M) \subseteq IM \\ & \phi_a:M \rightsquigarrow M, x \to ax \\ \hline \\ &\exists n \in \N, a_i \in I: \phi^n + \phi_{a_1}\phi^{n-1}+\ldots+\phi_{a_n} = 0 \end{align*}$$

note

$$\phi^n: x \mapsto \underbrace{\phi(\phi(\ldots \phi(x)))}_{n\text{ times}}$$

Proof

Consider $L:=\{l_1, \ldots, l_n\}, \lang L \rang_M = M$.

$$\phi(l_i) \in IM \implies \exists r_i \in I, m_i \in M: \phi(l_i) = r_1m_1+\ldots+r_km_k = \\ r_1\sum_j s_{j1}l_j + \ldots + r_n\sum_j s_{jn}l_j = \sum_j a_{j}l_j, a_j \in I, l_j \in L$$

As a side note - it's obvious that $\phi(l_i)=\sum_j a_{j}l_j$ for some $a_j \in A$, condition $\phi(l_i) \in IM$ guarantees that $a_j \in I$.

So we can write $\phi(l_i) = \sum_j a_{ij}l_j$.

Consider a set of endomorphisms $\text{End } M=\{\phi: M \rightsquigarrow M\}$. $\text{End } M$ is a ring with $(\phi+\psi)(x) := \phi(x)+\psi(x)$ and $(\phi\psi)(x):=\phi(\psi(x)))$.

Then we can consider a module with the following scalar multiplication:

$$\cdot: \text{End } M \times M \to M, \phi \cdot x \mapsto \phi(x)$$

And define matrix $A$:

$$A:=\begin{pmatrix} \phi_{a_{11}} & \dots & \phi_{a_{1m}} \\ \vdots & \ddots & \vdots\\ \phi_{a_{n1}} & \dots & \phi_{a_{nm}} \\ \end{pmatrix}$$

Then we can write:

$$\phi(l_i) = \sum_j a_{ij}l_j \implies \phi I\cdot l=A\cdot l \implies \\ (\phi I - A)\cdot l = 0 \implies \text{adj}(\phi I - A)(\phi I - A) \cdot l = 0 \implies \\ \det(\phi I - A)I \cdot l = 0$$

Next note that $\det(\phi I - A) = \phi^n + \phi_{a_1}\phi^{n-1}+\ldots+\phi_{a_n} = 0$ and when applied to each $l_i$ it results in $0$. So it's $0$ on all elements of $L$ and thus is zero on all module elements.

$\square$

Free modules

def: Direct sum

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R \\ &I \subseteq \N \\ &\forall i \in I: M_i \in \mathcal M_R \\ &M:=\{x=(x_1, x_2, \ldots, x_i, \ldots), x_i \in M_i, |i: x_i \ne 0| < \infty\} \\ &\forall x, y \in M, r \in R: \\ &x+y := (x_1 + y_1, \ldots, x_i+y_i, \ldots) \\ &rx := (rx_1, \ldots, rx_i, \ldots) \\ \hline \\ &M:=\bigoplus_{i \in I}M_i \,\, (M \text{ is a direct sum of }M_i) \end{align*}$$

note

For $M \in \mathcal M_A$ we define $M^n:= \underbrace{M \oplus M \oplus \ldots \oplus M}_{n \text{ times}}$

note

The only difference between the direct sum $\bigoplus_{i \in I}M_i$ and the direct product $\prod_{i \in I}M_i$ is the finite amount of non-zero $x_i$ in direct sums.

note

Sometimes, the concepts of external and internal direct sums can be mixed up. The explanation we provided earlier applies to the external direct sum.

However, there's also the internal direct sum, which refers to the sum of submodules of a module that do not overlap. Thus, internal direct sum is a breakdown of a module into submodules.

Despite the potential for confusion, both types of sums use the $\oplus$ symbol. In our discussions, we will reserve this notation specifically for external direct sums.

def: Free module

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal A \overset{(2.6.1)}{\implies} A \in \mathcal M_A \\ &M \in \mathcal M_A \\ &M \cong_M \bigoplus_{i \in I}A \\ \hline \\ &M - \text{free } A\text{-module} \end{align*}$$

Proposition 2.6.13: Free module has a basis

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R \overset{(2.6.1)}{\implies} R \in \mathcal M_R \\ &M \in \mathcal M_R \\ \hline \\ \exists \phi:&M \cong_M \bigoplus_{i = 1}^nR \iff \exists L \in \mathfrak B_M(M) \end{align*}$$

Proof

$\implies$

$$e_i:=(0, 0, \ldots , \underbrace{1}_i, \ldots, 0) \\ L:=\{\phi^{-1}(e_1), \ldots, \phi^{-1}(e_n)\} \\$$

By definition $\lang L \rang_M \subseteq M$. Let's prove that $\lang L \rang_M \supseteq M$:

$$\forall x\in M: \phi(x)=(r_1, \ldots, r_n) \implies \\ x = \phi^{-1}(\phi(x)) = \phi^{-1}((r_1, \ldots, r_n))=\\ \phi^{-1}(r_1e_1+\ldots+r_ne_n)=r_1\phi^{-1}(e_1)+\ldots + r_n\phi^{-1}(e_n) \implies x \mathfrak \in \lang L \rang_M\\$$

Now let's prove that $\forall r_1, \ldots, r_n \in R: r_1\phi(e_1) + \ldots + r_n\phi(x_n) = 0 \implies r_1 = \ldots = r_n = 0$.

$$r_1\phi(e_1) + \ldots + r_n\phi(e_n) = 0 \overset{\phi^{-1}}\implies \\ r_1e_1 + \ldots + r_ne_n = 0 \implies r_1=\ldots= r_n = 0$$

$\impliedby$

Let $L=\{x_1, \ldots, x_n\}$. Define $\phi: M \to \bigoplus_{i =1}^nR, x = r_1x_1+\dots + r_nx_n \mapsto (r_1, \ldots, r_n)$.

By $(2.6.9)$ $\phi$ is well-defined and injective. It's also surjective since $\lang L\rang_M = M$. So $M \leftrightarrow \bigoplus_{i =1}^nR$. Let's prove that $M \overset{\phi}\rightsquigarrow \bigoplus_{i =1}^nR$

$$\forall x, y \in M, x=r_1x_1+\ldots+r_nx_n, y = s_1x_1+\ldots+s_nx_n: \\ \phi(x + y)=\phi((r_1+s_1)x_1+\ldots + (r_n+s_n)x_n)= \\ (r_1+s_1, \ldots, r_n+s_n)=(r_1, \ldots, r_n)+ (s_1, \ldots, s_n)= \\ \phi(x) + \phi(y)$$ $$\forall r \in R, x \in M: \phi(ax)=\phi(ar_1x_1+\ldots+ar_nx_n)=(ar_1, \ldots, ar_n)=a\phi(x)$$

$\square$

Example: Non-free module

Consider $\Z_n \in M_{\Z}$, where $ax:=\underbrace{x+\ldots+x}_{a \text{ times}}$.

If we have a basis $\{x_1, \ldots, x_k\}$ then $nx_1+\ldots+nx_k = 0$ which contradicts linear independence of basis elements.

Proposition 2.6.14: Finitely generated module is a factor module of a fininte free module

$$\begin{align*} &\sphericalangle \\ &R \in \mathcal R^1 \\ &M \in \mathcal M_R \\ \hline \\ &\exists X \subseteq M, |X| < \infty : \lang X \rang_M = M \iff \\ &\exists n \in \N, N \subseteq_M R^n, \eta: M \cong_M R^n/N \end{align*}$$

Proof

$\implies$

Let $X = \{x_1, \ldots , x_n\}, \lang X \rang_M = M$. Now define mapping $\phi: R^n \to M, (r_1, \ldots, r_n) \mapsto r_1x_1 + \ldots + r_nx_n$. Let's prove that $\phi:R^n \rightsquigarrow_M M$:

$$\forall x, y \in M: x = r_1x_1+\ldots+r_nx_n, y = s_1x_1+\ldots+s_nx_n \\ \phi(x+y)=(r_1+s_1)x_1 + \ldots + (r_n+s_n)x_n = \\ =r_1x_1+s_1x_1 + \ldots + r_nx_n+s_nx_n = \phi(x) +\phi(y) \\ \phi(ax) = ar_1x_1 + \ldots + ar_nx_n = a(r_1x_1 + \ldots + r_nx_n) = a \phi(x)$$

It's clear that $\phi(R^n)=M$. So by $(2.6.4)$ $M \cong R^n / \ker \phi, N:=\ker \phi$.

$\impliedby$

We can define homomorphism $\psi: R^n \to M, \psi = \eta^{-1}\phi$, where $\phi: R^n \rightsquigarrow_M R^n / N$ - natural homomorphism.

$$\phi(R^n)=R^n/N, \eta^{-1}(R^n/N) = M \implies \psi(R^n) = M \\ e_i:=(0, 0, \ldots , \underbrace{1}_i, \ldots, 0) \\ X:=\{ \psi(e_1), \ldots, \psi(e_n) \} \\ \lang e_1, \ldots, e_n \rang_M = R^n \\ \lang X \rang_M = \lang \psi(e_1), \ldots, \psi(e_n) \rang_M = \psi(\lang e_1, \ldots, e_n \rang_M) = \psi(R^n)=M$$

$\square$

Vector spaces

def: Vector space

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \implies F \in \mathcal R^{\mathcal C} \\ &V \in \mathcal M_F \\ \hline \\ &V - \text{ vector space} \end{align*}$$

The key distinction between the definitions of modules and vector spaces is that the underlying ring for a vector space is a field. Although vector spaces share conceptual similarities with modules, they use different terminology. Below, we provide a comparison of equivalent terms used in each context:

$$\def\arraystretch{1.5} \begin{array}{c|c} \text{Module} & \text{Vector space} \\ \hline M \text{ is an } A\text{-module} & M \text{ is a vector space over } A\\ \hdashline m \text{ is an element of } M & m \text{ is a vector of } M\\ \hdashline a \text{ is a ring element } & a \text{ is a scalar }\\ \hdashline N \text{ is submodule of } M & N \text{ is subspace of } M\\ \hdashline M/N \text{ is a factor module} & M/N \text{ is a factor space} \\ \hdashline M \text{ is a free module of rank } k & M \text{ is a vector space of dimension } k \\ \hdashline \text{rank}_AM=k & \text{dim}_AM=k\\ \hdashline N \text{ generates } M & N \text{ spans } M \\ \hdashline \phi - A\text{-module homomorphism} & \phi - \text{linear transformation}\\ \hdashline \end{array}$$

For notation purposes, we will use the same symbols for both vector spaces and modules

Algebras

def: Algebra

$$\begin{align*} &\sphericalangle \\ & A, L \in \mathcal R^\mathcal{C} \\ &\phi: A \rightsquigarrow_R L, \phi(1) = 1\\ &\cdot: A \times L \to L, a \cdot b \mapsto \phi(a)\cdot b \implies L \in \mathcal M_A \\ \hline \\ &L \in \mathcal A_{\phi, A} \,\,\, (L - A\text{-algebra}) \end{align*}$$

Example: Commutative ring is a $\Z$-algebra

Consider $L \in \mathcal R^\mathcal{C}$ and $\phi: \Z \to L: n \mapsto \underbrace{1 + \ldots + 1}_{n \text{ times}}$. Then it's easy to check that $L \in \mathcal A_{\phi, \Z}$.

Proposition 2.6.15 Algebra over field contains this field

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F \\ &L \in \mathcal A_{\phi, F} \\ \hline \\ &\exists X \subseteq L: F \overset{\phi}\cong_R X \end{align*}$$

Proof

$$(2.4.20) \implies \ker \phi = \{0\} \overset{(2.3.9)}\implies F \cong \phi(F) \subseteq L$$

def: Algebra homomorphism

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &L_1 \in \mathcal A_{\phi_1, A} \\ &L_2 \in \mathcal A_{\phi_2, A} \\ &\psi: L_1 \rightsquigarrow_R L_2 ,\psi: L_1 \rightsquigarrow_M L_2, \psi(1)=1 \\ \hline \\ &\psi: L_1 \rightsquigarrow_A L_2 - A\text{-algebra homomorphism} \end{align*}$$

def: Finite Algebra

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &L \in \mathcal A_{\phi, A} \\ &\exists X \subseteq L, |X| \lt \infty: \lang X \rang_M = L \\ \hline \\ &L - \text{finite algrebra} \end{align*}$$

def: Finitely generated algebra

$$\begin{align*} &\sphericalangle \\ &A \in \mathcal R^{\mathcal C} \\ &L \in \mathcal A_{\phi, A} \\ &\exists S \subseteq L, S = \{s_1, \ldots, s_n\}: \\ &\forall x \in L: \exists p \in \phi(A)[x_1, \ldots, x_n]: x=p(s_1, \ldots, s_n)\\ \hline \\ &L - \text{finitely generated algrebra} \end{align*}$$

note

Equivalently, if we consider $A[x_1, \ldots, x_n] \in \mathcal A_{id, A}$ ($id$ is identity mapping), then $L$ is a finitely generated algebra if and only if

$$\exists n, \phi: A[x_1, \ldots, x_n] \rightsquigarrow_A L, \phi(A[x_1, \ldots, x_n]) = L$$

def: Finitely generated ring

A ring that is a finetely generated $\Z\text{-algebra}$