I'd like to make textbook exercises, some as a refresher and others new to me, and I hope that mathematicians here will take a look and will point out when I miss something. Here is a first bunch.

Which of the following subsets of \(\mathbb C\) are fields with respect to the usual addition and multiplication of numbers:

(a) \(\mathbb Z\)?

Not a field. E.g., no inverse of 2.

(b) \(\{0,1\}\)?

Not a field. Not even a group.

(c) \(\{0\}\)?

Not a field. No multiplicative identity.

(d) \(\{a+b\sqrt 2; a, b \in \mathbb Q\}\)?

Yes.

(e) \(\{a+b\sqrt[3] 2; a, b \in \mathbb Q\}\)?

Not a field. Can't get inverse of \(\sqrt[3] 2\).

(f) \(\{a+b\sqrt[4] 2; a, b \in \mathbb Q\}\)?

Not a field. Can't get inverse of \(\sqrt[4] 2\).

(g) \(\{a+b\sqrt 2; a, b \in \mathbb Z\}\)?

Not a field. E.g., no inverse of 2.

(h) \(\{z \in \mathbb C : |z| \leq 1\}\)

Not a field. No inverses when \(|z| \lt 1\).

Just now, Genady said:

(b) {0,1} ?

Not a field. Not even a group.

this one is a bit tricky and yes can be defined as a (finite) field; it is in fact the smallest field.

0 + 0 = 0

0*0 = 0

1 + 1 = 0

1*1 = 1

1+0 = 0 + 1 = 1

1*0 = 0*1 = 0

 

 

39 minutes ago, studiot said:

this one is a bit tricky and yes can be defined as a (finite) field; it is in fact the smallest field.

0 + 0 = 0

0*0 = 0

1 + 1 = 0

1*1 = 1

1+0 = 0 + 1 = 1

1*0 = 0*1 = 0

 

 

Thank you. I don't think it fits in this exercise because 1 + 1 = 0 mod 2 while the exercise asks about "usual addition and multiplication of numbers".

Just now, Genady said:

Thank you. I don't think it fits in this exercise because 1 + 1 = 0 mod 2 while the exercise asks about "usual addition and multiplication of numbers".

2 and therefore mod2 does not exist in this field that has only two members.

19 minutes ago, studiot said:

2 and therefore mod2 does not exist in this field that has only two members.

Right. This set is not closed under usual addition. That is why it is not a field with respect to the usual addition.

Edited January 1Jan 1 by Genady

Just now, Genady said:

Right. This set is not closed under usual addition. That is why it is not a field with respect to the usual addition.

This makes little or no sense whatsoever.

The question can {0, 1} form a field ? makes sense.

The question can {0, 1} form a field according to the rules of a different set of numbers ? makes little sense.

The rules of forming a field do not include a limitation of the binary operations, but only require the existence of a suitable pair of such operations.

 

38 minutes ago, studiot said:

This makes little or no sense whatsoever.

The question can {0, 1} form a field ? makes sense.

The question can {0, 1} form a field according to the rules of a different set of numbers ? makes little sense.

The rules of forming a field do not include a limitation of the binary operations, but only require the existence of a suitable pair of such operations.

 

OK

Let's take a different exercise.

Show that any subfield of \(\mathbb C\) contains \(\mathbb Q\).

It has to contain 0 and 1. From those, with addition and subtraction, it has to contain all \(\mathbb Z\). From there, with multiplication and division, it has to contain all \(\mathbb Q\).

QED

Edited January 1Jan 1 by Genady

Just now, Genady said:

OK

Let's take a different exercise.

Show that any subfield of C contains Q .

It has to contain 0 and 1. From those, with addition and subtraction, it has to contain all Z . From there, with multiplication and division, it has to contain all Q .

QED

Interestingly I first thought that the imaginary axis is a counterexample but then I realised that it is not a field, although it does contain Q !

Not sure I accept your proof however since all members of C have the form (a + ib) whilst none of the members of Q have this form.

9 minutes ago, studiot said:

none of the members of Q have this form

They do, with \(b=0\) and a rational \(a\).

This was a challenging exercise.

Give an example of an infinite field with a finite subfield.

The simplest example I could come up with was a field of rational functions, \[\frac {a_n X^n+a_{n-1} X^{n-1}+...+a_0}{b_m X^m+b_{m-1} X^{m-1}+...+b_0}\] where all the coefficients a and b are from the field (0,1) as in the second post on the top of this thread, 

 

Edited January 1Jan 1 by Genady

Show that if the smallest subfield K of a field L has an order n then na=0 for all a in L.

The smallest subfield with an order n is {0, e, 2e, ... , (n-1)e} with ne=0. For any integer k, n(ke) = k(ne) = k(0) = 0.

For x in L\K, nx = n(ex/e) = (ne)x/e = 0(x/e) = 0

QED

This exercise generalizes the case of \( i^2=−1\).

Let \( L \supset K \) be a field extension and let \(\alpha \in L \setminus K, \alpha^2 \in K\). Show that \[K(\alpha)=\{a+b\alpha; a,b \in K\}.\]

Let's denote \(\alpha^2=r \in K\). 

Any polynomial \(a+b \alpha + c \alpha^2 +d \alpha^3 +... = a+b \alpha + c r +d r \alpha +... =(a+cr +...) + (b+dr+...)\alpha = s+t\alpha\), where \(s,t \in K\).

Inverse of a polynomial, \((a+b\alpha)^{-1}=(a-b\alpha)(a^2-b^2r)^{-1} = a(a^2-b^2r)^{-1} - b(a^2-b^2r)^{-1}\alpha= s+t\alpha\), where \(s,t \in K\).

QED

Edited January 2Jan 2 by Genady

Let \(K(\sqrt a)\) and \(K(\sqrt b)\), where \(a, b \in K; ab \neq 0\), be two field extensions of \(K\).  Show that \(K(\sqrt a) = K(\sqrt b)\) if and only if \(ab\) is a square in \(K\) (that is, there is a  \(c \in K\) such that \(ab = c^2\)). 

1. If \(K(\sqrt a) = K(\sqrt b)\), then \(\sqrt a=r+s \sqrt b\) for some \(r, s \in K\).
Then, \(\sqrt a - s \sqrt b =r\).
Squaring it, \(a + s^2 b - 2s \sqrt {ab} = r^2\).  
Or, \(2 s \sqrt {ab} = a + s^2 b - r^2\).
Since the RHS in the last equation is in \(K\), so must be its LHS. So, \(\sqrt {ab} \in K\) and thus, \(ab\) is a square in \(K\).

2. If \(ab = c^2\), then \(\sqrt {ab} = c, \sqrt a = c (\sqrt b)^{-1}\) and thus, \(K(\sqrt a) = K(\sqrt b)\).

QED

Edited January 2Jan 2 by Genady

This is the last exercise on this topic.

Let \(K \subseteq L\) be a field extension and let \(M_1,M_2\) be two fields containing \(K\) and contained in \(L\). Show that \(M_1M_2\) consists of all quotients of finite sums \(\sum a_ib_i\) where \(a_i \in M_1, b_i \in M_2\).

Any such quotient is constructed from elements of \(M_1, M_2\) by addition, multiplication, and inverses and thus it \(\in M_1M_2\). OTOH, a set of all such quotients constitutes a field that contains elements of \(M_1, M_2\) and thus must include \(M_1M_2\) because the latter is a smallest such field. Thus, they are equal.