# Algebra Review Nov 28 [TOC] ## 4.1: Homomorphisms and Ideals - What is a **ring homomorphism**? - Does $f(0) = 0$ for a ring morphism? - Is the image of a ring morphism still a ring? - Give several examples of rings and ring morphisms. > See p. 115: > > - The identity morphism $R\to R$ where $r\mapsto r$ > - The inclusion $\Z \to \Q$ where $n\mapsto n = {n\over 1}$. > - Reduction $\mod m$, i.e. $\Z \to \Z/m\Z$ where $n \mapsto \bar n$. > - Reduction of polynomials $\Z[x]\to \Z/m\Z[x]$ where $f(x) \mapsto \bar{f}(x)$ (check: what does this mean?) > - The evaluation morphism > - Complex conjugation > - $\Z/6\Z\to \Z/3\Z$ where $a\mod 6\mapsto a\mod 3$. Why is this well-defined? > - Conjugation: $R\to R$ where $r\mapsto srs^{-1}$ for some fixed $s\in R$. - What is the **kernel** of a ring morphism? - What is an **ideal**? - What is the **ideal generated by an element**? - What is $(2,3) \trianglelefteq \Z$? - What is $(2,4) \trianglelefteq \Z$ - What is $(a, b)\trianglelefteq \Z$? - Is it true that $(a) = (b) \implies a = b$? Give a proof or counterexample - > $(-2) = (2)$ but $-2 = u 2$ where $u\in \Z^\times = \pm 1$. - What is a **principal ideal**? - Given an example of an ideal that is principal and an ideal that is not principal. > $(3, x+2)\in \R[x]$, see 4.1 Example 4. - What is a **PID**? - Give some examples of PIDs. > $\Z, k[x]$ - Prove that your examples are PIDs. (See 4.1 Prop 1.2) - How can you find a generator for an ideal $I$ in $\Z$? What about $k[x]$? > Find a prime number resp. an irreducible polynomial. - What is the evaluation morphism $\mathrm{ev}_a: k[x] \to k$ for some $a\in k$? What is its kernel? - What is the **quotient ring** $R/I$? - What does it mean for elements to be equal in $R/I$? > $a+I = b+I \iff a-b\in I$. - What does it mean for an operation or a homomorphism to be **well-defined**? - Prove that $R/I$ is a well-defined ring. (See 4.1 Prop 1.5 and 1.6) - If $R$ is an integral domain, is $R/I$ always an integral domain? Give a proof or counterexample. - Let $R = \Z/6\Z$ and $I = (2)$ and describe $R/I$. - Let $R = {\Z\over 2\Z}[x]$ and $I = (x^2+x+1)$ and describe $R/I$. Hint: use that $x^2 \equiv x+1 \mod I$ since $-1 \equiv 1$ - How is $\Q[x]/(x^3+2)$ related to $\Q[2^{1\over 3}]$? # Problems ## Core Problems: - ###### 4.1.12 (Use bars correctly, and explain why your solutions are distinct.) ![image-20221127190458367](/home/zack/.config/Typora/typora-user-images/image-20221127190458367.png) - $R = \Z_5[x]/f = \left\{ a_0 + a_1 x \mid a_0, a_1 \in \Z_5 \right\}$, check that $\bar 2^2 = \bar 3^2 = \bar 4$, $\bar x ^2 = \overline{-1} = \bar 4$, and $(\overline{4x})^2 =\bar 4 ^2 \cdot\bar x^2 = \bar 1 \cdot \bar 4 = \bar 4$. Conclusions: $R$ can not be a field, since $y^2-4$ could only have at most 2 roots there. - 4.1.14a ![image-20221127190507204](/home/zack/.config/Typora/typora-user-images/image-20221127190507204.png) - a. $R = \left\{ a_1 x + a_0 \right\} = \left\{ 0,1,x,x+1\right\}$ for $\Z_2[x]/(x^2+x)$ | $+$ | $0$ | $1$ | $x$ | $x+1$ | | ----- | ---- | ---- | ----- | ----- | | $0$ | $0$ | $1$ | $x$ | $x+1$ | | $1$ | | $0$ | $x+1$ | $x$ | | $x$ | | | $0$ | $1$ | | $1+x$ | | | | $0$ | | $\times$ | $0$ | $1$ | $x$ | $x+1$ | | ----- | ---- | ---- | ---- | ----- | | $0$ | $0$ | $0$ | $0$ | $0$ | | $1$ | | $1$ | $x$ | $x+1$ | | $x$ | | | $x^2=-x=x$ | $x^2+x=0$ | | $1+x$ | | | | $x^2+2x+1 = x+0+1 =x+1$ | - b. $R = \left\{ a_1 x + a_0 \right\}$ for $R=\Z_3[x]/(x^2+x-1)$, so $x^2 = -x+1 = 2x+1$. Yields $\left\{ 0,1,2, x, x+1, x+2, 2x, 2x+1, 2x+2\right\}$ | $+$ | $0$ | $1$ | $2$ | $x$ | $x+1$ | $x+2$ | $2x$ | $2x+1$ | $2x+2$ | | ------ | ---- | ---- | ---- | -------- | ------ | ------ | ------ | ------ | ------ | | $0$ | $0$ | $1$ | $2$ | $\cdots$ | | | | | | | $1$ | | $2$ | $0$ | $x+1$ | $x+2$ | $x$ | $2x+1$ | $2x+2$ | $2x$ | | $2$ | | | $1$ | $x+2$ | $x$ | $x+1$ | $2x+2$ | $2x$ | $2x+1$ | | $x$ | | | | $2x$ | $2x+1$ | $2x+2$ | $0$ | $1$ | $2$ | | $x+1$ | | | | | $2x+2$ | $2x$ | $1$ | $2$ | $0$ | | $x+2$ | | | | | | $2x+1$ | $2$ | $0$ | $1$ | | $2x$ | | | | | | | $x$ | $x+1$ | $x+2$ | | $2x+1$ | | | | | | | | $x+2$ | $x$ | | $2x+2$ | | | | | | | | | $x+1$ | | $\times$ | $0$ | $1$ | $2$ | $x$ | $x+1$ | $x+2$ | $2x$ | $2x+1$ | $2x+2$ | | -------- | ---- | ---- | -------- | -------- | ------------------------------- | --------------------------- | ------------------------------ | ----------- | ----------- | | $0$ | $0$ | $0$ | $\cdots$ | | | | | | | | $1$ | | $1$ | $2$ | $\cdots$ | | | | | | | $2$ | | $2$ | $1$ | $2x$ | $2x+2$ | $2x+1$ | $x$ | $x+2$ | $x+1$ | | $x$ | | | | $2x$ | | | | | | | $x+1$ | | | | | $x^2+2x+1 = (2x+1) + 2x+1 =x+2$ | | | | | | $x+2$ | | | | | | $x^2+4x+4 = (2x+1) +x+1 =2$ | | | | | $2x$ | | | | | | | $4x^2 = 4(2x+1) = 8x+4 = 2x+1$ | | | | $2x+1$ | | | | | | | | $4x^2+4x+1$ | | | $2x+2$ | | | | | | | | | $4x^2+8x+4$ | - c. $R = \left\{ a_2 x^2 + a_1 x + a_0 \right\}$ - 4.1.16 (Some symbols are missing in some printings of \#16. See Ted Shifrin's webpage for corrections.) *Let $R,S$ be commutative rings and let $\phi: R\to S$ be a ring homomorphism.* a. *Given an ideal $J\subset S$, define $\mathcal J := \left\{ a\in R \mid \phi(a) \in J \right\} \subseteq R$. This is usually denoted $\phi^{-1}(J)$. Prove that $\mathcal J$ is an ideal.* *b. Given an ideal $I\subseteq R$, define $\mathcal I := \left\{ \phi(a) \mid a\in I \right\} \subseteq S$.* *Prove that $\mathcal I$ is an ideal, provided $\phi$ is surjective. Give an example to demonstrate that the latter hypothesis is necessary.* - a NTS: $\mathcal{J} + \mathcal{J} \subseteq \mathcal{J}$ and $R\mathcal{J} \subseteq \mathcal{J}$. $a_1, a_2\in \mathcal{J} \implies \phi(a_1), \phi(a_2)\in J \implies J \ni \phi(a_1) + \phi(a_2) = \phi(a_1 + a_2) \implies a_1 + a_2\in \mathcal J$. $r\in R, a\in \mathcal J \implies \phi(a)\in J \implies J\ni r\phi(a) = \phi(ra) \implies ra\in \mathcal J$. - b NTS: $\mathcal{I} + \mathcal{I} \subseteq \mathcal{I}$ and $R\mathcal{I} \subseteq \mathcal{I}$. $i_1, i_2\in \mathcal{I} \implies i_1 = \phi(a_1), i_2 = \phi(a_2)$ for some $a_1, a_2\in I$ $\implies i_1 + i_2 = \phi(a_1) + \phi(a_2) = \phi(a_1 + a_2)$ and $a_1 + a_2\in I$, so $a_1+a_2\mapsto i_1 + i_2$ and $i_1+i_2\in \mathcal{I}$. $i\in \mathcal{I}, r\in R \implies i=\phi(a), a\in I\implies ri = r\phi(a) = \phi(s) \phi(a) = \phi(sa)$ with $ra\in J$ so $ra\mapsto ri$ and $ri\in \mathcal{I}$. Here we've used surjectivity to pick $s$ such that $\phi(s) = r$. Easy counterexample: $\phi: \Z \to \Q$ with $n\mapsto {n\over 1}$, then $\Q$ has no nontrivial ideals so e.g. $\phi(2\Z)$ is not an ideal in $\Q$ since it's nontrivial. - 4.1.17 ![image-20221127190554416](/home/zack/.config/Typora/typora-user-images/image-20221127190554416.png) - a. Clear - b. For $\Z$, $n \in (a) \cap (b) \iff a\mid n \text{ and } b\mid n \iff \mathrm{lcm}(a,b) \mid n \iff n\in (\mathrm{lcm}(a, b))$, so $(a)\cap (b) = (\mathrm{lcm}(a, b))$. For example, $(2) \cap (3)$ are things that are multiples of 2 and 3, thus multiples of 6. $(2) \cap (4)$ are multiples of 2 and 4, so multiples of 2. OTOH $n \in (a) + (b) \iff n = ta + sb \iff \gcd(a, b)\mid n \iff n\in (\gcd(a, b))$, so $(a) + (b) = (\gcd(a, b))$. This holds verbatim for $F[x]$ since lcm and gcd exist there too. - c. LHS: $(a_1,\cdots, a_n) := \left\{ r_1 a_1 + \cdots + r_n a_n \right\}$ RHS: $(a_1) + \cdots + (a_n) = \left\{ \ell_1 + \cdots + \ell_n \mid \ell_1\in (a_i), \cdots, \ell_n\in (a_n) \right\}$ but $\ell_i\in (a_i) \iff \ell_i = r_i a_i$, QED. - 4.1.19 ![image-20221127190602199](/home/zack/.config/Typora/typora-user-images/image-20221127190602199.png) - Prove something strong: if $R$ is a PID, then every ideal in $S := R/I$ is principal (although this may not be a domain). Let $H$ be an ideal in $R/I$, let $J = \left\{ r\in R \mid r+I\in H \right\}$, i.e. $\phi^{-1}(H)$ as in the previous exercise. This is an ideal in $R$, so $J = (j)$ for some $j\in R$. Claim: $(j+I) = H$, i.e. reduce $j$ mod $I$ to get $H$. $(j+I) \subseteq H$: clear, since $j$ is an $r\in R$ such that $r+I\in H$, so $j+I \in H$. $H \subseteq (j+I)$: let $a+I\in H$ be a coset in $H$. Then $a\in R$ and in fact $a\in J$ by definition of $J$. So $a=rj$ for some $r\in R$, thus $a+I = rj + I = (r+I)(j+I)$, thus $j+I$ divides $a+I$, thus $a+I \in (j+I)$ since to divide is to contain. ## Advanced problems - 4.2.13 (In part (d), instead of using Exercise 12, use Exercise 4.1.21.) ![image-20221127204035999](/home/zack/.config/Typora/typora-user-images/image-20221127204035999.png) ![image-20221127204021346](/home/zack/.config/Typora/typora-user-images/image-20221127204021346.png) a. Maximal: $p\Z$ with $p\neq 0$. Primes: maximals along with $(0)$. Prove that $R/I$ is a field iff $I$ is maximal and $R/I$ is a domain iff $I$ is prime, then check $\Z[x]/(x)\cong \Z$ is a domain that is not a field so $(x)$ is a prime that is not maximal. b. Definitional: no nonzero zero divisors $\implies$ if $ab=0$ with $a\neq 0$ then $b=0$ $\implies (0)$ is prime, i.e. $ab\in (0), b\not\in (0) \implies b\in (0)$. c. $\implies$: Suppose $I$ is prime, WTS $(a+I)(b+I) = 0 + I, b+I\neq 0+I\implies a+I = 0 + I$. Check $(a+I)(b+I) = ab + I = 0 + I \iff ab\in I$ and $b\not\in I\implies a\in I$ since $I$ is prime, so $a + I = 0 + I$. $\impliedby:$ Same argument backwards. d. Fields iff no non proper nontrivial ideals. $\implies$: if $I$ is maximal then $R/I$ has no nontrivial proper ideals, since ideals of $R/I$ corresponds to ideals of $R$ containing $I$, of which there are non by maximality. $\impliedby$: If $R/I$ is a field then there are no ideals of $R$ containing $I$, so $I$ is maximal.