ISI MMath Solutions and Discussion 2017 : PMB

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ISI MMath PMB 2017 Subjective Questions along with hints and solutions and discussions

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Group A

Problem 1.

Let $a_n\in\mathbb{R}$ , such that $\sum^\infty_{n=1}|a_n|=\infty$ and $\sum^m_{n=1}a_n\to a\in\mathbb{R}$ as $m\to\infty$ . Let $a^+_n=max \{a_n,0\}$ . Show that $\sum^{\infty}_{n=1}a^+_n=\infty$ .

Topic: Real Analysis, Difficulty level: Medium

Indian Statistical Institute, ISI MMath 2017 Problem 1 Hints along with Solution & Discussion

Hint 1

Use the representation

$a^+_n=\frac{a_n+|a_n|}{2}.$

Fix any $C>0.$ There exists $m_0$ such that for all $m\geq m_0$

$\left|\sum^m_{n=1}a_n-a\right|<1$

Hint 2

$\sum^m_{n=1}|a_n|>2C+|a|+1.$

Hint 3

Then for all $m\geq m_0$ we have

$\sum^m_{n=1}a^+_n=\frac{1}{2}\sum^m_{n=1}a_n+\frac{1}{2}\sum^m_{n=1}|a_n|$

Full Solution

We use the representation

$a^+_n=\frac{a_n+|a_n|}{2}.$

Fix any $C>0.$ There exists $m_0$ such that for all $m\geq m_0$

$\left|\sum^m_{n=1}a_n-a\right|<1, \mbox{ and }$

$\sum^m_{n=1}|a_n|>2C+|a|+1.$

Then for all $m\geq m_0$ we have

$\sum^m_{n=1}a^+_n=\frac{1}{2}\sum^m_{n=1}a_n+\frac{1}{2}\sum^m_{n=1}|a_n|=$

$=\frac{1}{2}a+\frac{1}{2}\left(\sum^m_{n=1}a_n-a\right)+\frac{1}{2}\sum^m_{n=1}|a_n|\geq$

$\geq \frac{1}{2}\sum^m_{n=1}|a_n|-\frac{1}{2}|a|-\frac{1}{2}\left|\sum^m_{n=1}a_n-a\right|>$

$>\frac{1}{2}(2C+|a|+1)-\frac{1}{2}|a|-\frac{1}{2}=C.$

For any $C>0$ there exists $m_0$ such that for all $m\geq m_0$

$\sum^m_{n=1}a^+_n>C.$

In other words, $\sum^\infty_{n=1}a^+_n=\infty.$

Problem 2.

Let $E=$ $\{(x,y,z)\in\mathbb{R}^3|x,y,z>0,xy+yz+zx=1\}$ . Prove that there exists $(a,b,c)\in E$ such that $abc\geq xyz,$ for all $(x,y,z)\in E$ .

Topic: Real Analysis, Difficulty Level: Medium

Indian Statistical Institute, ISI MMath 2017 Problem 2 Hints along with Solution & Discussion

Hint 1

Apply the arithmetic mean-geometric mean inequality:

for all $n\geq 1$ and $a_1,\ldots,a_n\geq 0$

$(a_1\ldots a_n)^{1/n}\leq \frac{a_1+\ldots+a_n}{n}$

Hint 2

Let $n=3,$ $a_1=xy,$ $a_2=yz,$ $a_3=xz,$ where $(x,y,z)\in E.$ Then

$(a_1a_2a_3)^{1/3}=(xyz)^{2/3}\leq \frac{a_1+a_2+a_3}{3}=\frac{xy+yz+xz}{3}=\frac{1}{3}.$

Hint 3

Take

$(a,b,c)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)\in E.$

Full Solution

We apply the arithmetic mean-geometric mean inequality:

for all $n\geq 1$ and $a_1,\ldots,a_n\geq 0$

$(a_1\ldots a_n)^{1/n}\leq \frac{a_1+\ldots+a_n}{n}$

Let $n=3,$ $a_1=xy,$ $a_2=yz,$ $a_3=xz,$ where $(x,y,z)\in E.$ Then

$(a_1a_2a_3)^{1/3}=(xyz)^{2/3}\leq \frac{a_1+a_2+a_3}{3}=\frac{xy+yz+xz}{3}=\frac{1}{3}.$

So, for any $(x,y,z)\in E$ we have

$xyz\leq \frac{1}{3^{3/2}}.$

Take

$(a,b,c)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)\in E.$

Then

$abc=\frac{1}{3^{3/2}}\geq xyz$

for any $(x,y,z)\in E.$ The answer is

$(a,b,c)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right).$

Problem 3.

Let $f:\mathbb{R\to\mathbb{R}}$ be an increasing function. Suppose there are sequences $(x_n)$ and $(y_n)$ such that $x_n<0<y_n$ for all $n\geq 1$ and $f(y_n)-f(x_n)\to 0$ as $n\to\infty$ . Prove that $f$ is continuous at $0$ .

Topic: Real Analysis, Difficulty Level: Medium

Indian Statistical Institute, ISI MMath 2017 Problem 3 Hints along with Solution & Discussion

Hint 1

Try to form the mathematical expression with $\epsilon >0$ and $\epsilon \in \mathbb{R}$ from this given statement:
“Suppose there are sequences $(x_n)$ and $(y_n)$ such that $x_n<0<y_n$ for all $n\geq 1$ and $f(y_n)-f(x_n)\to 0$ as $n\to\infty$ “.

Hint 2

Let $\epsilon>0.$ There exists $n_0$ such that for all $n\geq n_0$

$f(y_n)-f(x_n)<\epsilon.$

Hint 3

Fix such $n\geq n_0$ and take $\delta=\min(y_n,-x_n)>0.$ Consider arbitrary $x\in (-\delta,\delta),$ $x\ne 0.$ If $x>0,$ then $x_n<0<x<\delta\leq y_n,$

$f(x)\leq f(y_n), f(0)\geq f(x_n)$

and

$|f(x)-f(0)|=f(x)-f(0)\leq f(y_n)-f(x_n)<\epsilon.$

If $x<0,$ then $x_n\leq -\delta<x<0<y_n,$

$f(x)\geq f(x_n), f(0)\leq f(y_n)$

Full Solution

Let $\epsilon>0.$ There exists $n_0$ such that for all $n\geq n_0$

$f(y_n)-f(x_n)<\epsilon.$

Fix such $n\geq n_0$ and take $\delta=\min(y_n,-x_n)>0.$ Consider arbitrary $x\in (-\delta,\delta),$ $x\ne 0.$ If $x>0,$ then $x_n<0<x<\delta\leq y_n,$

$f(x)\leq f(y_n), f(0)\geq f(x_n)$

and

$|f(x)-f(0)|=f(x)-f(0)\leq f(y_n)-f(x_n)<\epsilon.$

If $x<0,$ then $x_n\leq -\delta<x<0<y_n,$

$f(x)\geq f(x_n), f(0)\leq f(y_n)$

and

$|f(x)-f(0)|=f(0)-f(x)\leq f(y_n)-f(x_n)<\epsilon.$

In any case,

$|f(x)-f(0)|<\epsilon$

as soon as $|x|<\delta.$ $f$ is continuous at $0.$

Problem 4.

Do there exist continuous functions $P$ and $Q$ on [0,1] such that $y(t)=\sin(t^2)$ is a solution to $y''+Py'+Qy=0$ on $[\frac{1}{n},1]$ for all $n\geq 1 ?$ justify your answer.

Topic: Elements of Ordinary Differential Equations, Difficulty Level: Medium

Indian Statistical Institute, ISI MMath 2017 Problem 4 Hints along with Solution & Discussion

Hint 1

Assume that such functions exist and then proceed.

Hint 2

Derivatives of $y(t)=\sin(t^2)$ are equal to

$y'(t)=2t\cos(t^2), \ y''(t)=2\cos(t^2)-4t^2\sin(t^2).$

Hint 3

In particular,

$\lim_{t\to 0}y(t)=0, \lim_{t\to 0}y'(t)=0, \lim_{t\to 0}y''(t)=2.$

Full Solution

Assume that such functions exist. Derivatives of $y(t)=\sin(t^2)$ are equal to

$y'(t)=2t\cos(t^2), \ y''(t)=2\cos(t^2)-4t^2\sin(t^2).$

In particular,

$\lim_{t\to 0}y(t)=0, \lim_{t\to 0}y'(t)=0, \lim_{t\to 0}y''(t)=2.$

If for all $t\in (0,1]$ the equality

$y''(t)+P(t)y'(t)+Q(t)y(t)=0,$

then taking the limit as $t\to 0$ we get

$2+P(0)\cdot 0+Q(0)\cdot 0=0,$

which is impossible. So, such functions $P$ and $Q$ do not exist.

Problem 5.

Let $f:\mathbb{R}\to\mathbb{R}$ be defined by

$f(x)=\int^{1+e^{x^3+x}}_{e^{x^3+x}} e^{r^2} dr$

for all $x\in\mathbb{R}$ . Prove that $f$ is monotone.

Topic: Real Analysis, Difficulty Level: Easy

Indian Statistical Institute, ISI MMath 2017 Problem 5 Hints along with Solution & Discussion

Hint 1

Observe the relation

$f(x)=h(g(x)),$

where $g(x)=e^{x^3+x},$ $h(y)=\int^{1+y}_ye^{r^2}dr.$

Hint 2

We have

$g'(x)=e^{x^3+x}(3x^2+1)>0$

Hint 3

$h'(y)=e^{(1+y)^2}-e^{y^2}=e^{y^2}(e^{1+2y}-1).$

Full Solution

Observe the relation

$f(x)=h(g(x)),$

where $g(x)=e^{x^3+x},$ $h(y)=\int^{1+y}_ye^{r^2}dr.$ We have

$g'(x)=e^{x^3+x}(3x^2+1)>0$

$h'(y)=e^{(1+y)^2}-e^{y^2}=e^{y^2}(e^{1+2y}-1).$

In particular, $h'(y)>0$ for $y>-1/2.$ Since $g(x)>0,$ we have $h'(g(x))>0,$ $g'(x)>0$ and

$f'(x)=h'(g(x))g'(x)>0.$

The function $f(x)$ is strictly increasing.

Problem 6.

Let $w=\{w(i,j)\}_{1\leq i,j\leq m}$ be an $m\times m$ symmetric matrix with non-negative real entries such that $w(i,j)=0$ if and only if $i=j$ . Show that $d(i,j)=\min \{\sum^{k-1}_{j=0}w(i_j,i_{j+1})|k\geq 1,i_0=i,i_k=j,i_j\in \{1,\dots,m\}\}$ is a metric on $\{1,\dots,m\}$ .

Topic: Metric Space, Difficulty Level: High

Indian Statistical Institute, ISI MMath 2017 Problem 6 Hints along with Solution & Discussion

Hint 1

At first we verify that the minimum is achieved. Consider a path $i_0,i_1,i_2,\ldots,i_k$ between $i$ and $j$ of length $k+1\geq m+1.$ Among elements

$i_0,i_1,i_2,\ldots,i_k$

there are at least two identical elements: $i_r=i_s$ for some $r<s.$ Then

$i_0,i_1,\ldots,i_{r-1},i_s,\ldots,i_k$

is a path of length $<k+1.$ Also, $w(i_{r-1},i_r)=w(i_{r-1},i_s)$ and

$\sum^{k-1}_{l=0}w(i_l,i_{l+1})\geq \sum^{r-2}_{l=0}w(i_l,i_{l+1})+w(i_{r-1},i_s)+\sum^{k-1}_{l=s}w(i_l,i_{l+1}).$

It means that it is enough to consider only paths of length $\leq m.$ There are only finitely many such paths, hence the minimum in the definition of $d(i,j)$ is achieved. We check axioms of a metric.

Hint 2

For any path $i_0=i,i_1,\ldots,i_{k-1},i_k=j$ between $i$ and $j$ we have

$\sum^{k-1}_{l=0}w(i_l,i_{l+1})\geq 0.$

Hence, $d(i,j)\geq 0.$ Assume $d(i,j)=0.$ Since the minimum in the definition of $d(i,j)$ is achieved, we have

$\sum^{k-1}_{l=0}w(i_l,i_{l+1})= 0 \eqno(*)$

for some path $i_0=i,i_1,\ldots,i_{k-1},i_k=j$ between $i$ and $j.$ Since all terms in the left-hand side of (*) are non-negative, we deduce that

$w(i_l,i_{l+1})=0, \ l=0,\ldots,k-1.$

By definition of $w,$ it follows that

$i_0=i_1=\ldots=i_{k-1}=i_k.$

In particular, $i=i_0=i_k=j.$ We’ve proved that $d(i,j)\geq 0$ and $d(i,j)=0\Rightarrow i=j.$

Hint 3

Consider a path $i_0=i,i_1=i.$ Then $d(i,i)\leq w(i,i)=0.$ But $d(i,i)\geq 0,$ hence

$d(i,i)=0.$

This proves that $d(i,j)=0\Leftrightarrow i=j.$

Full Solution

At first we verify that the minimum is achieved. Consider a path $i_0,i_1,i_2,\ldots,i_k$ between $i$ and $j$ of length $k+1\geq m+1.$ Among elements

$i_0,i_1,i_2,\ldots,i_k$

there are at least two identical elements: $i_r=i_s$ for some $r<s.$ Then

$i_0,i_1,\ldots,i_{r-1},i_s,\ldots,i_k$

is a path of length $<k+1.$ Also, $w(i_{r-1},i_r)=w(i_{r-1},i_s)$ and

$\sum^{k-1}_{l=0}w(i_l,i_{l+1})\geq \sum^{r-2}_{l=0}w(i_l,i_{l+1})+w(i_{r-1},i_s)+\sum^{k-1}_{l=s}w(i_l,i_{l+1}).$

For any path $i_0=i,i_1,\ldots,i_{k-1},i_k=j$ between $i$ and $j$ we have

$\sum^{k-1}_{l=0}w(i_l,i_{l+1})\geq 0.$

Hence, $d(i,j)\geq 0.$ Assume $d(i,j)=0.$ Since the minimum in the definition of $d(i,j)$ is achieved, we have

$\sum^{k-1}_{l=0}w(i_l,i_{l+1})= 0 \eqno(*)$

for some path $i_0=i,i_1,\ldots,i_{k-1},i_k=j$ between $i$ and $j.$ Since all terms in the left-hand side of (*) are non-negative, we deduce that

$w(i_l,i_{l+1})=0, \ l=0,\ldots,k-1.$

By definition of $w,$ it follows that

$i_0=i_1=\ldots=i_{k-1}=i_k.$

In particular, $i=i_0=i_k=j.$ We’ve proved that $d(i,j)\geq 0$ and $d(i,j)=0\Rightarrow i=j.$

Consider a path $i_0=i,i_1=i.$ Then $d(i,i)\leq w(i,i)=0.$ But $d(i,i)\geq 0,$ hence

$d(i,i)=0.$

This proves that $d(i,j)=0\Leftrightarrow i=j.$

Consider a path $i_0=i,i_1,\ldots,i_{k-1},i_k=j$ between $i$ and $j,$ such that

$d(i,j)=\sum^{k-1}_{l=0}w(i_l,i_{l+1}).$

Observe that $i_k=j,i_{k-1},\ldots,i_1,i_0=i$ is a path between $j$ and $i.$ So,

$d(j,i)\leq \sum^{k-1}_{l=0}w(i_{k-l},i_{k-l+1})=\sum^{k-1}_{l=0}w(i_{l+1},i_{l})=\sum^{k-1}_{l=0}w(i_{l},i_{l+1})=d(i,j).$

We’ve proved that

$d(j,i)\leq d(i,j).$

Since this is true for all $i,j,$ we also get

$d(i,j)\leq d(j,i).$

Finally, $d(i,j)=d(j,i).$

Consider two paths $i_0=i,i_1,\ldots,i_k=j,$ $r_0=j,r_1,\ldots,r_s=r,$ such that

$d(i,j)=\sum^{k-1}_{l=0}w(i_l,i_{l+1}), d(j,r)=\sum^{s-1}_{p=0}w(r_p,r_{p+1}).$

Observe that $i_0,i_1,\ldots,i_k,r_1,\ldots,r_s$ is a path between $i$ and $r.$ Hence

$d(i,r)\leq \sum^{k-1}_{l=0}w(i_j,i_{j+1})+w(i_k,r_1)+\sum^{s-1}_{p=1}w(r_p,r_{p+1})=$

use that $w(i_k,r_1)=w(j,r_1)=w(r_0,r_1)$

$=\sum^{k-1}_{l=0}w(i_j,i_{j+1})+\sum^{s-1}_{p=0}w(r_p,r_{p+1})=d(i,j)+d(j,r).$

The triangle inequality is proved.

$d$ is a metric on $\{1,\ldots,m\}.$

Group B

Problem 7.

Factory $A$ produces $1$ bad watch in $100$ and factory $B$ produces $1$ bad watch in $200$ . You are given two watches from one of the factories and you don’t know which one.

What is the probability that the second watch works?
Given that the first watch works, what is the probability that the second watch works $?$

Topic: Probability Theory, Difficulty Level: Medium

Indian Statistical Institute, ISI MMath 2017 Problem 7 Hints along with Solution & Discussion

Hint 1

Introduce two hypothesis

$H_1=\{\mbox{ the watches are from the Factory } A\},$

$H_2=\{\mbox{ the watches are from the Factory } B\}.$

Hint 2

Then $P(H_1)=P(H_2)=\frac{1}{2}.$ Consider events

$E_1=\{\mbox{ the first watch works }\}$

$E_2=\{\mbox{ the second watch works }\}.$

It is given that $P(E_j|H_1)=0.99,$ $P(E_j|H_2)=0.995.$

Hint 3

By the law of total probability

$P(E_2)=P(E_2|H_1)P(H_1)+P(E_2|H_2)P(H_2)=\frac{0.99+0.995}{2}=0.9925$

Full Solution

Introduce two hypothesis

$H_1=\{\mbox{ the watches are from the Factory } A\},$

$H_2=\{\mbox{ the watches are from the Factory } B\}.$

Then $P(H_1)=P(H_2)=\frac{1}{2}.$ Consider events

$E_1=\{\mbox{ the first watch works }\}$

$E_2=\{\mbox{ the second watch works }\}.$

It is given that $P(E_j|H_1)=0.99,$ $P(E_j|H_2)=0.995.$

By the law of total probability

$P(E_2)=P(E_2|H_1)P(H_1)+P(E_2|H_2)P(H_2)=\frac{0.99+0.995}{2}=0.9925$

We compute the probability $P(E_2|E_1)$ by the definition:

$P(E_2|E_1)=\frac{P(E_1\cap E_2}{P(E_1)}$

From the previous part, we know $P(E_1)=0.9925.$ By the law of total probability

$P(E_1\cap E_2)=P(E_1\cap E_2|H_1)P(H_1)+P(E_1\cap E_2|H_2)P(H_2)=\frac{0.99^2+0.995^2}{2}=0.9850625$

So,

$P(E_2|E_1)=\frac{0.9850625}{0.9925}=0.9925063$

Problem 8.

Let $R$ be a commutative ring containing a field $k$ as a sub-ring. Assume that $R$ is a finite dimensional $k$ -vector space. Show that every prime ideal of $R$ is maximal.

Topic: Abstract Algebra- Ring and Field Theory, Difficulty Level: Medium

Indian Statistical Institute, ISI MMath 2017 Problem 8 Hints along with Solution & Discussion

Hint 1

Let $I$ be a prime ideal in $R$ . It is enough to check that $R/I$ is a field

Hint 2

The infinite sequence $1,a,a^2,\ldots$ is linearly dependent and there exists samllest $m$ such that

$c_0+c_1a+c_2a^2+\ldots+c_ma^m=0$

for some non-zero coefficients $c_0,c_1,\ldots,c_m\in k.$ If $c_0=0,$

Hint 3

Then

$a(c_1+c_2a+\ldots+c_ma^{m-1})=0.$

Full Solution

Let $I$ be a prime ideal in $R$ . It is enough to check that $R/I$ is a field, i.e. every nonzero element of $a\in R/I$ has a multiplicative inverse. $R/I$ is a finite-dimensional $k$ -vector space. Hence the infinite sequence $1,a,a^2,\ldots$ is linearly dependent and there exists smallest $m$ such that

$c_0+c_1a+c_2a^2+\ldots+c_ma^m=0$

for some non-zero coefficients $c_0,c_1,\ldots,c_m\in k.$ If $c_0=0,$ then

$a(c_1+c_2a+\ldots+c_ma^{m-1})=0.$

Let $x\in R$ be a representative of $a.$ Then

$x(c_1+c_2x+\ldots+c_mx^{m-1})\in I.$

$a\in R/I\setminus \{0\},$ hence $x\not\in I.$ The ideal $I$ is prime, so

$c_1+c_2x+\ldots+c_mx^{m-1}\in I$

and

$c_1+c_2a+\ldots+c_ma^{m-1}=0.$

The latter contradicts minimality of $m.$ So, $c_0\ne 0,$ and

$1=a\left(-\frac{c_1}{c_0}-\frac{c_2}{c_0}a-\ldots-\frac{c_m}{c_0}a^{m-1}\right)$

$a\in R/I$ has a multiplicative inverse. $R/I$ is a field and $I$ is a maximal ideal.

Problem 9.

Let $p,q$ be prime numbers and $n\in\mathbb{N}$ such that $p \nmid n-1. \mbox{ If }p\mid n^q-1$ then show that $q\mid p-1$ .

Topic: Number Theory, Difficulty Level: Easy

Indian Statistical Institute, ISI MMath 2017 Problem 9 Hints along with Solution & Discussion

Hint 1

By assumptions of the problem $(n,p)=1.$

Hint 2

Let $k$ be the order of $n$ modulo $p,$ i.e. $k\geq 2$ is the smallest positive integer such that $p\mid n^k-1.$

Hint 3

From $p\mid n^q-1$ we deduce that $k\mid q.$ $q$ is prime, so $k=q.$
Now use “Fermat’s Little Theorem”.

Full Solution

By assumptions of the problem $(n,p)=1.$ Let $k$ be the order of $n$ modulo $p,$ i.e. $k\geq 2$ is the smallest positive integer such that $p\mid n^k-1.$ From $p\mid n^q-1$ we deduce that $k\mid q.$ $q$ is prime, so $k=q.$ By the Fermat Little Theorem,

$p\mid n^{p-1}-1.$

Since $q$ is the order of $n$ modulo $p,$ it follows that $q\mid p-1.$

Problem 10.

Determine all finite groups which have exactly $3$ conjugacy classes.

Topic: Group Theory, Difficulty Level: Hard

Indian Statistical Institute, ISI MMath 2017 Problem 10 Hints along with Solution & Discussion

Hint 1

Let $G$ be a finite group with exactly $3$ conjugacy classes. One conjugacy class is always the identity class $\{e\}$ .

Hint 2

Let $m$ and $n$ be sizes of other two conjugacy classes. Conjugacy classes form a partition of a group, so

$|G|=1+m+n.$

Hint 3

The size of a conjugacy class of $g\in G$ equals to the index of a centralizer of $g,$

Full Solution

Let $G$ be a finite group with exactly $3$ conjugacy classes. One conjugacy class is always the identity class $\{e\}.$ Let $m$ and $n$ be sizes of other two conjugacy classes. Conjugacy classes form a partition of a group, so

$|G|=1+m+n.$

The size of a conjugacy class of $g\in G$ equals to the index of a centralizer of $g,$ i.e. it divides $|G|.$ So,

$m\mid |G|, \ n\mid |G|.$

Further,

$m\mid n+1=|G|-m, \ n\mid m+1=|G|-n.$

Assume that $m=n.$ Then $m\mid m+1,$ i.e. $m=n=1,$ $|G|=3.$ The only group of order $3$ is the cyclic group $Z_3.$ It is an abelian group, hence it has exactly 3 conjugacy classes.

Assume that $m<n.$ Then $m+1\leq n,$ hence $n=m+1,$ $m\mid 2=n+1-m.$ If $m=1,$ then $n=2$ and $|G|=4.$ The only groups of order 4 are the Klein 4 group or the cyclic group $Z_4.$ Both are abelian and thus have 4 conjugacy classes.

If $m=2,$ then $n=3$ and $|G|=6.$ There are two groups of order $6:$ an abelian group $Z_6,$ and a nonabelian group $S_3$ (permutations on three elements). We check that $S_3$ has three conjugacy classes:

$\{()\}, \{(12),(13),(23)\}, \{(123),(132)\}$

The only finite groups with three conjugacy classes are $Z_3$ and $S_3.$

Problem 11.

Let $F$ be a field, $a\in F, p$ a prime integer. Suppose the polynomial $x^p-a$ is reducible in $F[x]$ . Prove that this polynomial has a root in $F$ .

Topic: Field Theory, Difficulty Level: Medium

Indian Statistical Institute, ISI MMath 2017 Problem 11 Hints along with Solution & Discussion

Hint 1

In the case $a=0$ obviously $x=0$ is the needed root.

Hint 2

We assume firther that $a\ne 0.$ Let $K$ be the splitting field of $x^p-a$ over $F.$

Hint 3

Then

$x^p-a=(x-b_1)\ldots(x-b_p)$

for some $b_1,\ldots,b_p\in K.$

Full Solution

In the case $a=0$ obviously $x=0$ is the needed root. We assume firther that $a\ne 0.$ Let $K$ be the splitting field of $x^p-a$ over $F.$ Then

$x^p-a=(x-b_1)\ldots(x-b_p)$

for some $b_1,\ldots,b_p\in K.$ Since the polynomial $x^p-a$ is reducible, we may relabel its roots in such a way that for some $n<p$ a polynomial

$f(x)=(x-b_1)\ldots(x-b_n)$

is a factor of $x^p-a,$ i.e. $f(x)$ has coefficients in $F$ . Observe that for each $i,$ $b^p_i=a.$ Consider the free term of $f(x).$ It is of the form $\pm b_1b_2\ldots b_n.$ Hence, $b=b_1b_2\ldots b_n\in F.$ Further,

$b^p=(b_1b_2\ldots b_n)^p=b^p_1\ldots b^p_n=a^n.$

Since $(n,p)=1$ we can find integers $r,s$ such that

$1=rn+sp.$

We have $a^sb^r\in F,$ and

$(a^sb^r)^p=a^{sp}b^{rp}=a^{sp+rn}=a$

$a^sb^r$ is a root of $x^p-a$ in $F$ .

Problem 12.

Let $V$ be a finite-dimensional vector space over a field $F$ and let $T:V\to V$ be a linear transformation. Let $W\subseteq V$ be a subspace such that $T(W)\subseteq W$ . Suppose $T$ is diagonalizable. Is $T$ restricted to $W$ also diagonalizable?

Topic: Linear Algebra, Difficulty Level: Hard

Indian Statistical Institute, ISI MMath 2017 Problem 12 Hints along with Solution & Discussion

Hint 1

By assumption, $V$ is a direct sum of subspaces

$V=V_1\oplus \ldots\oplus V_k,$

and on each subspace $V_k$ $T$ acts as a multiplication by $\lambda_k,$ all $\lambda_1,\ldots\lambda_k$ are distinct.

Hint 2

Consider $w\in W.$ Then

$w=v_1+\ldots+v_k$

for some $v_i\in V_i,$ $1\leq i\leq k.$ Let $m$ be the number of nonzero summands $v_1,\ldots,v_k.$

Hint 3

By induction on $m$ we will prove that all $v_1,\ldots,v_k\in W.$ If $m=0,$ then $v_1=\ldots=v_k=0$ and there is nothing to prove.

Full Solution

By assumption, $V$ is a direct sum of subspaces

$V=V_1\oplus \ldots\oplus V_k,$

and on each subspace $V_k$ $T$ acts as a multiplication by $\lambda_k,$ all $\lambda_1,\ldots\lambda_k$ are distinct.

Consider $w\in W.$ Then

$w=v_1+\ldots+v_k$

for some $v_i\in V_i,$ $1\leq i\leq k.$ Let $m$ be the number of nonzero summands $v_1,\ldots,v_k.$ By induction on $m$ we will prove that all $v_1,\ldots,v_k\in W.$ If $m=0,$ then $v_1=\ldots=v_k=0$ and there is nothing to prove. Assume the result is proved for number of nonzero summands $<m,$ and let $v_i\ne 0.$ Then

$Tw=\lambda_1 v_1+\ldots +\lambda_k v_k,$

$Tw-\lambda_i w=\sum_{j\ne i}(\lambda_j-\lambda_i)v_i\in W.$

By inductive assumption, for all $j\ne i$

$(\lambda_j-\lambda_i)v_i\in W.$

Eigenvalues $\lambda_1,\ldots,\lambda_k$ are distinct, so $v_j\in W$ for $j\ne i.$ But then

$v_i=w-\sum_{j\ne i}v_j\in W.$

This statement shows that in the decomposition

$w=v_1+\ldots+v_k$

all summands are in $W,$ i.e.

$W=(W\cap V_1)\oplus\ldots \oplus (W\cap V_k).$

The restriction of $T$ to $W$ is diagonalizable.

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6 Comments

Shagun

September 18, 2020 at 5:35 pm Reply

Hi! Thanks for your selfless help by putting out all the solutions free of cost on a public domain. I really appreciate it and I’m extremely thankful.

I have a doubt. Can you please explain the reasoning behind the first hint of problem 9?
- FractionsHub
  
  September 18, 2020 at 9:17 pm Reply
  
  This is trivially seen from the result $p\mid n^q-1$ . If $(p,n)\neq 1$ then it implies that $p\mid n$ and so $n^q-1\equiv-1(mod p)$ .
Shagun

September 18, 2020 at 8:31 pm Reply

In problem 10, in the case when m<n, I understood that m+1 should be less than or equal to n.

But how did we immediately get in the next step, that n = m+1 ?
- FractionsHub
  
  September 18, 2020 at 9:23 pm Reply
  
  Since $n\mid {m+1}$ then from that it implies $n \leq {m+1}$ and thus the conclusion.
  - Shagun
    
    September 18, 2020 at 10:32 pm Reply
    
    Understood both of them, thanks!
Mausami

July 17, 2021 at 4:40 pm Reply

In Q.12. Why have we assumed that all eigenvalues are distinct, when we may have a diagonalizable linear transformation on V which has repeated eigenvalues as long as the sum of dimensions of the eigenspaces is dim(V)?