# 2012

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UPSC Algebra

$$cos\:3\theta =4\:cos^3\:\theta -3\:cos\:\theta$$

### 1(a) How many elements of order 2 are there in the group of order 16 generated by $$a$$ and $$b$$ such that the order of $$a$$ is 8, the order of $$b$$ is 2, and $$b a b^{-1}=a^{-1}$$?

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
In a group $G$$G$GG$G$ generated by $a$$a$aa$a$ and $b$$b$bb$b$ with the given properties, we can describe the elements of $G$$G$GG$G$ in terms of $a$$a$aa$a$ and $b$$b$bb$b$. Specifically, any element $g$$g$gg$g$ in $G$$G$GG$G$ can be written as $g={a}^{m}{b}^{n}$$g={a}^{m}{b}^{n}$g=a^(m)b^(n)g = a^m b^n$g={a}^{m}{b}^{n}$, where $0\le m<8$$0\le m<8$0 <= m < 80 \leq m < 8$0\le m<8$ and $n=0,1$$n=0,1$n=0,1n = 0, 1$n=0,1$.
Step 1: Identify Elements of Order 2
An element $g$$g$gg$g$ has order 2 if ${g}^{2}=e$${g}^{2}=e$g^(2)=eg^2 = e${g}^{2}=e$, where $e$$e$ee$e$ is the identity element of the group.
Let’s consider the elements ${a}^{m}{b}^{n}$${a}^{m}{b}^{n}$a^(m)b^(n)a^m b^n${a}^{m}{b}^{n}$ and find those that satisfy $\left({a}^{m}{b}^{n}{\right)}^{2}=e$$\left({a}^{m}{b}^{n}{\right)}^{2}=e$(a^(m)b^(n))^(2)=e(a^m b^n)^2 = e$\left({a}^{m}{b}^{n}{\right)}^{2}=e$.
Case 1: $n=0$$n=0$n=0n = 0$n=0$
In this case, $\left({a}^{m}{\right)}^{2}={a}^{2m}=e$$\left({a}^{m}{\right)}^{2}={a}^{2m}=e$(a^(m))^(2)=a^(2m)=e(a^m)^2 = a^{2m} = e$\left({a}^{m}{\right)}^{2}={a}^{2m}=e$. Since $a$$a$aa$a$ has order 8, this is only possible if $m=4$$m=4$m=4m = 4$m=4$. So, ${a}^{4}$${a}^{4}$a^(4)a^4${a}^{4}$ is an element of order 2.
Case 2: $n=1$$n=1$n=1n = 1$n=1$
In this case, $\left({a}^{m}b{\right)}^{2}={a}^{m}b{a}^{m}b$$\left({a}^{m}b{\right)}^{2}={a}^{m}b{a}^{m}b$(a^(m)b)^(2)=a^(m)ba^(m)b(a^m b)^2 = a^m b a^m b$\left({a}^{m}b{\right)}^{2}={a}^{m}b{a}^{m}b$. Using the relation $ba{b}^{-1}={a}^{-1}$$ba{b}^{-1}={a}^{-1}$bab^(-1)=a^(-1)b a b^{-1} = a^{-1}$ba{b}^{-1}={a}^{-1}$, we get:
$\left({a}^{m}b{\right)}^{2}={a}^{m}b{a}^{m}b={a}^{m}b{b}^{-1}{a}^{-m}b={a}^{m}e{a}^{-m}b={a}^{m}{a}^{-m}b=eb=b$$\left({a}^{m}b{\right)}^{2}={a}^{m}b{a}^{m}b={a}^{m}b{b}^{-1}{a}^{-m}b={a}^{m}e{a}^{-m}b={a}^{m}{a}^{-m}b=eb=b$(a^(m)b)^(2)=a^(m)ba^(m)b=a^(m)bb^(-1)a^(-m)b=a^(m)ea^(-m)b=a^(m)a^(-m)b=eb=b(a^m b)^2 = a^m b a^m b = a^m b b^{-1} a^{-m} b = a^m e a^{-m} b = a^m a^{-m} b = e b = b$\left({a}^{m}b{\right)}^{2}={a}^{m}b{a}^{m}b={a}^{m}b{b}^{-1}{a}^{-m}b={a}^{m}e{a}^{-m}b={a}^{m}{a}^{-m}b=eb=b$
So, $b$$b$bb$b$ is also an element of order 2.
Case 3: $n=1$$n=1$n=1n = 1$n=1$ and $m\ne 0$$m\ne 0$m!=0m \neq 0$m\ne 0$
In this case, $\left({a}^{m}b{\right)}^{2}={a}^{m}b{a}^{m}b$$\left({a}^{m}b{\right)}^{2}={a}^{m}b{a}^{m}b$(a^(m)b)^(2)=a^(m)ba^(m)b(a^m b)^2 = a^m b a^m b$\left({a}^{m}b{\right)}^{2}={a}^{m}b{a}^{m}b$. Using the relation $ba{b}^{-1}={a}^{-1}$$ba{b}^{-1}={a}^{-1}$bab^(-1)=a^(-1)b a b^{-1} = a^{-1}$ba{b}^{-1}={a}^{-1}$, we get:
$\left({a}^{m}b{\right)}^{2}={a}^{m}b{a}^{m}b={a}^{m}b{b}^{-1}{a}^{-m}b={a}^{m}e{a}^{-m}b={a}^{m}{a}^{-m}b=eb=b$$\left({a}^{m}b{\right)}^{2}={a}^{m}b{a}^{m}b={a}^{m}b{b}^{-1}{a}^{-m}b={a}^{m}e{a}^{-m}b={a}^{m}{a}^{-m}b=eb=b$(a^(m)b)^(2)=a^(m)ba^(m)b=a^(m)bb^(-1)a^(-m)b=a^(m)ea^(-m)b=a^(m)a^(-m)b=eb=b(a^m b)^2 = a^m b a^m b = a^m b b^{-1} a^{-m} b = a^m e a^{-m} b = a^m a^{-m} b = e b = b$\left({a}^{m}b{\right)}^{2}={a}^{m}b{a}^{m}b={a}^{m}b{b}^{-1}{a}^{-m}b={a}^{m}e{a}^{-m}b={a}^{m}{a}^{-m}b=eb=b$
This equation simplifies to $b$$b$bb$b$, which means that $\left({a}^{m}b{\right)}^{2}\ne e$$\left({a}^{m}b{\right)}^{2}\ne e$(a^(m)b)^(2)!=e(a^m b)^2 \neq e$\left({a}^{m}b{\right)}^{2}\ne e$ unless $m=0$$m=0$m=0m = 0$m=0$.
Step 2: Count Elements of Order 2
From the above discussion, we find that there are only two elements of order 2: ${a}^{4}$${a}^{4}$a^(4)a^4${a}^{4}$ and $b$$b$bb$b$.
So, there are 2 elements of order 2 in the group $G$$G$GG$G$.
5/5
$$b=c\:cos\:A+a\:cos\:C$$
$$sin\left(\theta +\phi \right)=sin\:\theta \:cos\:\phi +cos\:\theta \:sin\:\phi$$

### 2(a) How many conjugacy classes does the permutation group $$S_5$$ of permutations of 5 numbers have? Write down one element in each class (preferably in terms of cycles).

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
In the symmetric group ${S}_{5}$${S}_{5}$S_(5)S_5${S}_{5}$, the conjugacy classes are determined by the cycle structure of the permutations. Two permutations are conjugate if and only if they have the same cycle structure.
Here are the different types of cycle structures that can appear in ${S}_{5}$${S}_{5}$S_(5)S_5${S}_{5}$:
1. Identity: $\left(1\right)$$\left(1\right)$(1)(1)$\left(1\right)$ (1 element)
2. Single transpositions: $\left(a\phantom{\rule{thinmathspace}{0ex}}b\right)$$\left(a\phantom{\rule{thinmathspace}{0ex}}b\right)$(ab)(a \, b)$\left(a\phantom{\rule{thinmathspace}{0ex}}b\right)$ (2-cycles) (10 elements)
3. Double transpositions: $\left(a\phantom{\rule{thinmathspace}{0ex}}b\right)\left(c\phantom{\rule{thinmathspace}{0ex}}d\right)$$\left(a\phantom{\rule{thinmathspace}{0ex}}b\right)\left(c\phantom{\rule{thinmathspace}{0ex}}d\right)$(ab)(cd)(a \, b)(c \, d)$\left(a\phantom{\rule{thinmathspace}{0ex}}b\right)\left(c\phantom{\rule{thinmathspace}{0ex}}d\right)$ (two 2-cycles) (15 elements)
4. 3-cycles: $\left(a\phantom{\rule{thinmathspace}{0ex}}b\phantom{\rule{thinmathspace}{0ex}}c\right)$$\left(a\phantom{\rule{thinmathspace}{0ex}}b\phantom{\rule{thinmathspace}{0ex}}c\right)$(abc)(a \, b \, c)$\left(a\phantom{\rule{thinmathspace}{0ex}}b\phantom{\rule{thinmathspace}{0ex}}c\right)$ (20 elements)
5. 4-cycles: $\left(a\phantom{\rule{thinmathspace}{0ex}}b\phantom{\rule{thinmathspace}{0ex}}c\phantom{\rule{thinmathspace}{0ex}}d\right)$$\left(a\phantom{\rule{thinmathspace}{0ex}}b\phantom{\rule{thinmathspace}{0ex}}c\phantom{\rule{thinmathspace}{0ex}}d\right)$(abcd)(a \, b \, c \, d)$\left(a\phantom{\rule{thinmathspace}{0ex}}b\phantom{\rule{thinmathspace}{0ex}}c\phantom{\rule{thinmathspace}{0ex}}d\right)$ (30 elements)
6. 5-cycles: $\left(a\phantom{\rule{thinmathspace}{0ex}}b\phantom{\rule{thinmathspace}{0ex}}c\phantom{\rule{thinmathspace}{0ex}}d\phantom{\rule{thinmathspace}{0ex}}e\right)$$\left(a\phantom{\rule{thinmathspace}{0ex}}b\phantom{\rule{thinmathspace}{0ex}}c\phantom{\rule{thinmathspace}{0ex}}d\phantom{\rule{thinmathspace}{0ex}}e\right)$(abcde)(a \, b \, c \, d \, e)$\left(a\phantom{\rule{thinmathspace}{0ex}}b\phantom{\rule{thinmathspace}{0ex}}c\phantom{\rule{thinmathspace}{0ex}}d\phantom{\rule{thinmathspace}{0ex}}e\right)$ (24 elements)
7. 3-cycle and a 2-cycle: $\left(a\phantom{\rule{thinmathspace}{0ex}}b\phantom{\rule{thinmathspace}{0ex}}c\right)\left(d\phantom{\rule{thinmathspace}{0ex}}e\right)$$\left(a\phantom{\rule{thinmathspace}{0ex}}b\phantom{\rule{thinmathspace}{0ex}}c\right)\left(d\phantom{\rule{thinmathspace}{0ex}}e\right)$(abc)(de)(a \, b \, c)(d \, e)$\left(a\phantom{\rule{thinmathspace}{0ex}}b\phantom{\rule{thinmathspace}{0ex}}c\right)\left(d\phantom{\rule{thinmathspace}{0ex}}e\right)$ (20 elements)
Each of these cycle structures represents a different conjugacy class. Therefore, ${S}_{5}$${S}_{5}$S_(5)S_5${S}_{5}$ has 7 conjugacy classes.
Here is one element from each conjugacy class, written in terms of cycles:
1. Identity: $\left(1\right)$$\left(1\right)$(1)(1)$\left(1\right)$
2. Single transpositions: $\left(1\phantom{\rule{thinmathspace}{0ex}}2\right)$$\left(1\phantom{\rule{thinmathspace}{0ex}}2\right)$(12)(1 \, 2)$\left(1\phantom{\rule{thinmathspace}{0ex}}2\right)$
3. Double transpositions: $\left(1\phantom{\rule{thinmathspace}{0ex}}2\right)\left(3\phantom{\rule{thinmathspace}{0ex}}4\right)$$\left(1\phantom{\rule{thinmathspace}{0ex}}2\right)\left(3\phantom{\rule{thinmathspace}{0ex}}4\right)$(12)(34)(1 \, 2)(3 \, 4)$\left(1\phantom{\rule{thinmathspace}{0ex}}2\right)\left(3\phantom{\rule{thinmathspace}{0ex}}4\right)$
4. 3-cycles: $\left(1\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}3\right)$$\left(1\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}3\right)$(123)(1 \, 2 \, 3)$\left(1\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}3\right)$
5. 4-cycles: $\left(1\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}3\phantom{\rule{thinmathspace}{0ex}}4\right)$$\left(1\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}3\phantom{\rule{thinmathspace}{0ex}}4\right)$(1234)(1 \, 2 \, 3 \, 4)$\left(1\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}3\phantom{\rule{thinmathspace}{0ex}}4\right)$
6. 5-cycles: $\left(1\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}3\phantom{\rule{thinmathspace}{0ex}}4\phantom{\rule{thinmathspace}{0ex}}5\right)$$\left(1\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}3\phantom{\rule{thinmathspace}{0ex}}4\phantom{\rule{thinmathspace}{0ex}}5\right)$(12345)(1 \, 2 \, 3 \, 4 \, 5)$\left(1\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}3\phantom{\rule{thinmathspace}{0ex}}4\phantom{\rule{thinmathspace}{0ex}}5\right)$
7. 3-cycle and a 2-cycle: $\left(1\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}3\right)\left(4\phantom{\rule{thinmathspace}{0ex}}5\right)$$\left(1\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}3\right)\left(4\phantom{\rule{thinmathspace}{0ex}}5\right)$(123)(45)(1 \, 2 \, 3)(4 \, 5)$\left(1\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}3\right)\left(4\phantom{\rule{thinmathspace}{0ex}}5\right)$
These elements are representatives of their respective conjugacy classes.
5/5
$$\frac{a}{sin\:A}=\frac{b}{sin\:B}=\frac{c}{sin\:C}$$
$$cos\:2\theta =cos^2\theta -sin^2\theta$$

### 3(a) Is the ideal generated by 2 and $$X$$ in the polynomial ring $$Z[X]$$ of polynomials in a single variable $$X$$ with coefficients in the ring of integers $$Z$$, a principal ideal? Justify your answer.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
In the polynomial ring $\mathbb{Z}\left[X\right]$$\mathbb{Z}\left[X\right]$Z[X]\mathbb{Z}[X]$\mathbb{Z}\left[X\right]$, consider the ideal $I$$I$II$I$ generated by 2 and $X$$X$XX$X$. We want to determine whether this ideal is a principal ideal, i.e., whether it can be generated by a single element.
Step 1: Describe the Ideal $I$$I$II$I$
The ideal $I$$I$II$I$ consists of all polynomials that can be formed as $f\left(X\right)=2a\left(X\right)+Xb\left(X\right)$$f\left(X\right)=2a\left(X\right)+Xb\left(X\right)$f(X)=2a(X)+Xb(X)f(X) = 2a(X) + Xb(X)$f\left(X\right)=2a\left(X\right)+Xb\left(X\right)$, where $a\left(X\right)$$a\left(X\right)$a(X)a(X)$a\left(X\right)$ and $b\left(X\right)$$b\left(X\right)$b(X)b(X)$b\left(X\right)$ are arbitrary polynomials in $\mathbb{Z}\left[X\right]$$\mathbb{Z}\left[X\right]$Z[X]\mathbb{Z}[X]$\mathbb{Z}\left[X\right]$.
Step 2: Check for Principal Ideal
To be a principal ideal, there must exist a single polynomial $g\left(X\right)$$g\left(X\right)$g(X)g(X)$g\left(X\right)$ in $\mathbb{Z}\left[X\right]$$\mathbb{Z}\left[X\right]$Z[X]\mathbb{Z}[X]$\mathbb{Z}\left[X\right]$ such that every element in $I$$I$II$I$ can be written as $g\left(X\right)\cdot h\left(X\right)$$g\left(X\right)\cdot h\left(X\right)$g(X)*h(X)g(X) \cdot h(X)$g\left(X\right)\cdot h\left(X\right)$, where $h\left(X\right)$$h\left(X\right)$h(X)h(X)$h\left(X\right)$ is another polynomial in $\mathbb{Z}\left[X\right]$$\mathbb{Z}\left[X\right]$Z[X]\mathbb{Z}[X]$\mathbb{Z}\left[X\right]$.
Let’s assume for the sake of contradiction that such a $g\left(X\right)$$g\left(X\right)$g(X)g(X)$g\left(X\right)$ exists. Then $g\left(X\right)$$g\left(X\right)$g(X)g(X)$g\left(X\right)$ must divide both 2 and $X$$X$XX$X$ in $\mathbb{Z}\left[X\right]$$\mathbb{Z}\left[X\right]$Z[X]\mathbb{Z}[X]$\mathbb{Z}\left[X\right]$.
1. $g\left(X\right)$$g\left(X\right)$g(X)g(X)$g\left(X\right)$ divides 2: The only polynomials that divide 2 are the constant polynomials $±1$$±1$+-1\pm 1$±1$ and $±2$$±2$+-2\pm 2$±2$.
2. $g\left(X\right)$$g\left(X\right)$g(X)g(X)$g\left(X\right)$ divides $X$$X$XX$X$: The polynomials that divide $X$$X$XX$X$ are $±1$$±1$+-1\pm 1$±1$ and $±X$$±X$+-X\pm X$±X$.
The only polynomials that satisfy both conditions are $±1$$±1$+-1\pm 1$±1$. If $g\left(X\right)=1$$g\left(X\right)=1$g(X)=1g(X) = 1$g\left(X\right)=1$ or $g\left(X\right)=-1$$g\left(X\right)=-1$g(X)=-1g(X) = -1$g\left(X\right)=-1$, then $I$$I$II$I$ would be the entire ring $\mathbb{Z}\left[X\right]$$\mathbb{Z}\left[X\right]$Z[X]\mathbb{Z}[X]$\mathbb{Z}\left[X\right]$, which it is not. Therefore, no such $g\left(X\right)$$g\left(X\right)$g(X)g(X)$g\left(X\right)$ can exist.
Conclusion
The ideal $I$$I$II$I$ generated by 2 and $X$$X$XX$X$ in $\mathbb{Z}\left[X\right]$$\mathbb{Z}\left[X\right]$Z[X]\mathbb{Z}[X]$\mathbb{Z}\left[X\right]$ is not a principal ideal.
5/5
$$cos\left(\theta -\phi \right)=cos\:\theta \:cos\:\phi +sin\:\theta \:sin\:\phi$$
$$c^2=a^2+b^2-2ab\:Cos\left(C\right)$$

### 4(a) Describe the maximal ideals in the ring of Gaussian integers $$\mathbb{Z}[i]=\{a+b i \mid a, b \in \mathbb{Z}\}$$.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
In the ring of Gaussian integers $\mathbb{Z}\left[i\right]$$\mathbb{Z}\left[i\right]$Z[i]\mathbb{Z}[i]$\mathbb{Z}\left[i\right]$, a maximal ideal is an ideal $M$$M$MM$M$ such that there are no other ideals $N$$N$NN$N$ with $M⊊N⊊\mathbb{Z}\left[i\right]$$M⊊N⊊\mathbb{Z}\left[i\right]$M⊊N⊊Z[i]M \subsetneq N \subsetneq \mathbb{Z}[i]$M⊊N⊊\mathbb{Z}\left[i\right]$.
Properties of Maximal Ideals in $\mathbb{Z}\left[i\right]$$\mathbb{Z}\left[i\right]$Z[i]\mathbb{Z}[i]$\mathbb{Z}\left[i\right]$
1. Nonzero Principal Ideals: In $\mathbb{Z}\left[i\right]$$\mathbb{Z}\left[i\right]$Z[i]\mathbb{Z}[i]$\mathbb{Z}\left[i\right]$, every maximal ideal is a nonzero principal ideal, meaning it is generated by a single Gaussian integer $a+bi$$a+bi$a+bia + bi$a+bi$ where $a+bi\ne 0$$a+bi\ne 0$a+bi!=0a + bi \neq 0$a+bi\ne 0$.
2. Generated by Irreducible Elements: In $\mathbb{Z}\left[i\right]$$\mathbb{Z}\left[i\right]$Z[i]\mathbb{Z}[i]$\mathbb{Z}\left[i\right]$, maximal ideals are generated by irreducible elements. An element $z$$z$zz$z$ in $\mathbb{Z}\left[i\right]$$\mathbb{Z}\left[i\right]$Z[i]\mathbb{Z}[i]$\mathbb{Z}\left[i\right]$ is irreducible if it is not a unit and cannot be expressed as a product of two non-units in $\mathbb{Z}\left[i\right]$$\mathbb{Z}\left[i\right]$Z[i]\mathbb{Z}[i]$\mathbb{Z}\left[i\right]$.
Classification of Maximal Ideals
The maximal ideals in $\mathbb{Z}\left[i\right]$$\mathbb{Z}\left[i\right]$Z[i]\mathbb{Z}[i]$\mathbb{Z}\left[i\right]$ can be classified as follows:
1. Ideals generated by primes congruent to 3 mod 4: For a prime $p$$p$pp$p$ in $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$ that is congruent to 3 mod 4, $\left(p\right)$$\left(p\right)$(p)(p)$\left(p\right)$ is a maximal ideal in $\mathbb{Z}\left[i\right]$$\mathbb{Z}\left[i\right]$Z[i]\mathbb{Z}[i]$\mathbb{Z}\left[i\right]$. This is because $p$$p$pp$p$ remains irreducible in $\mathbb{Z}\left[i\right]$$\mathbb{Z}\left[i\right]$Z[i]\mathbb{Z}[i]$\mathbb{Z}\left[i\right]$.
2. Ideals generated by $1+i$$1+i$1+i1 + i$1+i$: The ideal $\left(1+i\right)$$\left(1+i\right)$(1+i)(1 + i)$\left(1+i\right)$ is maximal. This is because $\mathbb{Z}\left[i\right]/\left(1+i\right)$$\mathbb{Z}\left[i\right]/\left(1+i\right)$Z[i]//(1+i)\mathbb{Z}[i]/(1+i)$\mathbb{Z}\left[i\right]/\left(1+i\right)$ is isomorphic to $\mathbb{Z}/\left(2\right)$$\mathbb{Z}/\left(2\right)$Z//(2)\mathbb{Z}/(2)$\mathbb{Z}/\left(2\right)$, which is a field.
3. Ideals generated by Gaussian primes not in $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$: A Gaussian integer $a+bi$$a+bi$a+bia + bi$a+bi$ is a Gaussian prime if it is irreducible in $\mathbb{Z}\left[i\right]$$\mathbb{Z}\left[i\right]$Z[i]\mathbb{Z}[i]$\mathbb{Z}\left[i\right]$. For example, $2+3i$$2+3i$2+3i2 + 3i$2+3i$ is a Gaussian prime, and $\left(2+3i\right)$$\left(2+3i\right)$(2+3i)(2 + 3i)$\left(2+3i\right)$ is a maximal ideal.
Summary
The maximal ideals in $\mathbb{Z}\left[i\right]$$\mathbb{Z}\left[i\right]$Z[i]\mathbb{Z}[i]$\mathbb{Z}\left[i\right]$ are generated by:
1. Primes in $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$ that are congruent to 3 mod 4.
2. The Gaussian integer $1+i$$1+i$1+i1 + i$1+i$.
3. Gaussian primes that are not in $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$.
Each of these generates a unique maximal ideal in $\mathbb{Z}\left[i\right]$$\mathbb{Z}\left[i\right]$Z[i]\mathbb{Z}[i]$\mathbb{Z}\left[i\right]$.
5/5
$$cot\:\theta =\frac{cos\:\theta }{sin\:\theta }$$

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