Ring Ideals

Proposition: let R be a ring and $(I_j), j \in J$ be a collection of ideals of R. Then $\bigcap_{i \in J}I_j \subseteq R$ is an ideal.

Proof: $\bigcap_{j \in J}I_j$ is an additive subgroup (by an earlier proposition about the intersection of subgroups being a subgroup). Let $x \in \bigcap_{j \in J}I_j$ then for all $j, x \in I_j$ so $yx \in I_j$ and $xy \in \bigcap_{j \in J}I_j$. Similarly $xy \in I_j$ so $yx \in \bigcap_{j \in J}I_j$.

Let R be a ring, $I \subseteq I$ an ideal and $A \subseteq R$ a subset. We define $(A) = \bigcap_{A \subseteq I \subseteq R} I$. (A) is the smallest ideal that contains A and we call it the ideal generated by A.

Proposition: let R be a unitary ring, $A \subseteq R$, we define $RAR = \{ \sum_{i=0}^n r_ia_is_i : r_i,s_i \in R, a_i \in A\}$. Then (A) = RAR.

Proof: RAR is an ideal: $\sum_{i=0}^n r_ia_is_i + \sum_{j=0}^m u_jb_jv_j = \sum_{k=0}^{n+m}x_kc_ky_k$ where $x_k = r_k, x_k = a_k, y_k = s_k; 0 \leq k < n$ and $x_k = u_{k-n}, c_k = b_{k-n}, y_k = v_{k-n}; n \leq k < m$. Let $u \in R$, then $u(\sum_{i=0}^n r_ia_is_i) = \sum_{i=0}^n (u+i)a_is_i \in RAR$ hence $A \subseteq RAR$ (becase $a = 1 \cdot a \cdot 1$). Conversely if $s_i, r_i \in R, a_i \in A$, $r_ia_is_i \in (A)$ and $\sum_{i=0}^n a_ir_is_i \in (A)$ hence $RAR \subseteq (A)$ so $(A) = RAR$.

Remarks: if R is commutative then (A) = RA = AR = RAR. If $A = \{a\}$ and R is commutative then (A) = (a) = Ra = $\{ra : r\in R\}$. If R is not commutative then this might not be true as $\{ras: r, s \in R\}$ might not be an ideal.

Notation: if $A = \{a_1,...,a_n\}$ is finite we usually write $(A) = (a_1,..,a_n)$ instead of $(\{a_1,..,a_n\})$.

An ideal generated by a finite set is called a finitely generated ideal. An ideal generated by 1 element is called a principal ideal. Examples: $\{0\} = (0)$ is a principal ideal. If R is unitary, $R = (1)$ is also principal. In $\mathbb{Z}$ every ideal is of the form $n\mathbb{Z}$ and principal.

Proposition: Let R be a unitary ring 1) if $I \subseteq R$ is an ideal, then I = R if and only if I contains a unit. 2) if R is commutative and $1 \neq 0$, R is a field if and only if every ieal in R is either $\{0\}$ or R.

Proof: 1) If I = R then $1 \in I$ so it has a unit. If $u \in R^*$ and $u \in I$ then $u^{-1} \cdot u \in I$ so $1 \in I$. Then for all $r \in R, r\cdot 1 = r \in I$ so R = I. 2) If R is a field and $I \subseteq R$ is an ideal, if $I = \{0\}$ then $I \cap R^* = T \cap R \backslash \{0\}$ is nonempty. Conversely assume every ideal is $\{0\}$ or R. Then for all $u \in R \backslash \{0\}, (u) = R$. Since $1 \in R$ this means there exists a v such that uv = 1 so $u \in R^*$, hence R is a field.

Corollary: let F be a field, R a ring and $f : F \rightarrow R$ a ring homomorphism. Then either $f(F) = \{0\}$ or f is injective.

Proof: ker(F) is an ideal of F so either ker(f) = F in which case f is trivial, or ker(f) = $\{0\}$ and f is injective. In particular if R is unitary and f is unitary then f is injective, because the kernel must be trivial.

Let R be a ring, $I \subset R$ a proper ideal. I is said to be maximal if the only ideals containing I are I and R.

Proposition: let R and S be rings, and $f: R \rightarrow S$ a ring homomorphism 1) if $I \subseteq S$ is an ideal then $f^{-1}(I)$ is an ideal 2) if f is surjective and I is an ideal in R then f(I) is an ideal.

Proof: 1) $f^{-1}(I)$ is an additive subgroup (because the inverse image of a group is a group by an earlier theorem). Let $x \in f^{-1}(I)$ and $y \in R$, then $f(xy) = f(x)f(y) \in I$ so $xy \in f^{-1}(I)$. Similarly $f(yx) = f(y)f(x) \in I$ so $yx \in f^{-1}(I)$. 2) f(I) is an additive subgroup. Let $x \in f(I)$ and $u \in S = f(R)$ (since S is surjective). Then $x = f(s)$ so $s \in I$ and $y = f(r)$ where $r \in R$. Then $xy = f(s)f(r) = f(sr) \in f(I)$ since $s \in I$ and $yx = f(r)f(s) = f(rs) \in f(I)$.

Proposition: Let R be a unitary commutative ring where $1 \neq 0$, then $I \subset R$ is maximal if and only if R/I is a field.

Proof: if I is maximal: let $J \subseteq R/I$ be an ideal, $\pi : R \rightarrow R/I$ be a ring homomorphism. Then $\pi^{-1}(J) \subseteq R$ is an ideal. Therefore $\pi^{-1}(J) =$ either R or J. If $\pi^{-1}(J) = R$ then $R/I = J$. If $\pi^{-1}(J) = I$ then $J = \{0\}$ so R/I is a field. Conversely if R/I is a field and $I \subseteq J \subseteq R$ where J is an ideal, $\pi(J)$ is either $\{0\}$ or R/I. If $\pi(J) = \{0\}$ then $J \subseteq I$ so J = I. If $\pi(J) = R/I$ then J = R. For all $a \in R, (a+I) \cap J$ is nonempty, so $a+I = c+I \in J$. Since $a \in J, J = R$ so I is maximal.

Examples: in a field the only maximal ideal is $\{0\}$. In $\mathbb{Z}, n\mathbb{Z}$ is maximal if and only if $\mathbb{Z}/n\mathbb{Z}$ is a field, which is the case only when n is prime. In $F[x], (x)$ is a maximal ideal.

Krull’s theorem: Let R be a unitary ring, then every proper ideal is contained in a maximal ideal

Let R be a commutative unitary ring, $I \subset R$ a proper ideal, I is prime if for all $x, y \in R$ if $xy \in I$ then $x \in I$ or $y \in I$. In $\mathbb{Z}, n\mathbb{Z}$ is prime if and only if for all x, y if $n \vert xy$ then $n \vert x$ or $n \vert y$ if and only if n is prime if and only if $n\mathbb{Z}$ is maximal.

Proposition: Let R be a commutative unitary ring where $1 \neq 0$ and $I \subset R$ is an ideal. I is prime if and only if R/I is an integral domain.

Proof: I is proper and prime so $1+I \neq 0+I$. Let x+I be a zero divisor in R/I, then there exists a $y \in R\backslash I$ such that $(y+I)(x+I) = yx+I$ then $yx \in I$ so $x \in I$ hence $x+I=I$ (x is the zero coset) which means R/I is an integral domain. Assume R/I is an integral domain. Then $1+I \neq 0+I$ so I is proper. Let $xy \in I$, then $(xy+I) = (x+I)(y+I) = I$. Since I is integral either $x+I = I$ which means $x \in I$ or $y+I = I$ so $y \in I$.

Corollary: let R be a commutative unitary ring where $1 \neq 0$, if $I \subseteq R$ is maximal then I is prime.

Proof: given tht I is maximal, R/I is a field, so R/I is an integral domain and therefore I is prime.