Can you give an example of a field of characteristic zero (other than the complex numbers, real numbers and rational numbers)?
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I was preparing for an area exam in analysis and came across a problem in the book Real Analysis by Haaser &amp; Sullivan. From p.34 Q 2.4.3, If the field F is isomorphic to the subset S' of F', show that S' is a subfield of F'. I would appreciate any hints on how to solve this problem as I'm stuck, but that's not my actual question.  I understand that for finite fields this implies that two sets of the same cardinality must have the same field structure, if any exists. The classification of finite fields answers the above question in a constructive manner.  What got me curious is the infinite case. Even in the finite case it's surprising to me that the field axioms are so "restrictive", in a sense, that alternate field structures are simply not possible on sets of equal cardinality. I then started looking for examples of fields with characteristic zero while thinking about this problem. I didn't find many. So far, I listed the rationals, algebraic numbers, real numbers, complex numbers and the p-adic fields. What are other examples? Is there an analogous classification for fields of characteristic zero?
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If $\mathbb{F}=\{ 0,1,x,y\}$ is a field with four elements, why is the characteristic 2?   In a field with four elements, the characteristic is 2 or 3(since it has to be a prime). Is there any way to determine precisely the characteristic of this field, and in general any finite field?
18171
Let $F$ be a field. Why does $F[x]$ have infinitely many irreducible elements?   For the case F has characteristic 0  Then x-a is irreducible for all a $\in F$ since x satisfies no non-trivial relations in F.  Obviously this argument fails for a finite field since there are only finitely many a to choose from.  So how may I construct irreducible polynomials in a finite field?  I figure it must involve higher powers of x, maybe $x^n-a$ ?
111341
how to show that between any two distinct real numbers there are infinitely many rational numbers?  do we need to use the fact that a real number is an equivalent class of cauchy sequence of rational numbers?
484
Let $\mathbb R$ be the field of real numbers. Its algebraic closure, the field $\mathbb C$, is a finite extension of $\mathbb R$, which has degree 2.   Are there other examples of fields (not algebraic closed) such that its algebraic closure is a finite extension?
8897
Let $K$ be a field. Then $K[x,y]$ is not a PID because $x$ is irreducible but the quotient $K[x,y]/(x)$ is isomorphic to $K[y]$, which is not a field.  So there must be an ideal in there which is not principal. What is an example of such an ideal?
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Is there any example of a field extension $K/F$ where degree of extension $[K:F]$ is finite  but the number of intermediate field is infinite ?
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How can I prove that $\frac{1}{a+b\sqrt[3]{2}+c\sqrt[3]{4}}$ is of the form $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ (i.e. that $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ is a field)  for all rational $a,b,c$ and $a+b\sqrt[3]{2}+c\sqrt[3]{4} \neq 0$?
2364
I shall assume all rings to be commutative in this question. The impatient can scroll down to the "blockquote" to read the actual question.  Whenever we have a polynomial over a ring, it defines a function from the ring to itself by evaluation. It's reasonable to ask when two different polynomials define the same function.  From the factor theorem it follows that an $n^\text{th}$ degree polynomial over an integral domain has at most $n$ roots. Then it's easy to show this:  Theorem. Let $R$ be an infinite integral domain and let $f \in R[X]$ such that $f(a)=0$ for all $a \in R$, then $f = 0$. Proof. $f$ has infinitely many roots, so it must be the zero polynomial. $\quad\square$  For finite rings a kind of opposite situation occurs:  Theorem. For any finite ring $R$ there are polynomials over $R$ that are different but agree on all elements. Proof. There are only finitely many functions from $R$ to itself, but $R[X]$ is infinite. $\quad\square$  If we make further assumptions it's of course possible to prove more, as Pete L. Clark wrote in this post: []  Then there is the question of infinite rings that are not integral domains. It's relatively easy to come up with examples of a ring $R$ with positive characteristic and a nonzero polynomial that evaluates to the zero function, e.g.: $$ R := \bigoplus_{n=1}^\infty \mathbb{Z}/6\mathbb{Z} \quad\text{and}\quad f(X) := X^3-X.$$  The Question:     This leaves open the case alluded to in this post's title: Is there a commutative ring of characteristic $0$ (hence infinite) such that a nonzero polynomial evaluates to the zero function?
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An $n$th-degree polynomial has at most $n$ distinct zeroes in the complex numbers.  But it may have an uncountable set of zeroes in the quaternions.  For example, $x^2+1$ has two zeroes in $\mathbb C$, but in $\mathbb H$, ${\bf i}\cos x + {\bf j}\sin x$ is a distinct zero of this polynomial for every $x$ in $[0, 2\pi)$, and obviously there are many other zeroes.  What is it about $\mathbb H$ that makes its behavior in this regard to be so different from the behavior of $\mathbb R$ and $\mathbb C$?  Is it simply because $\mathbb H$ is four-dimensional rather than two-dimensional? Are there any theorems that say when a ring will behave like $\mathbb H$ and when it will behave like $\mathbb C$?   Do all polynomials behave like this in $\mathbb H$? Or is this one unusual?
117070
Is $x^3+y^3+z^3-1$ irreducible over a field $k$ of characteristic $\neq 3$?   If I can show that $y^3+z^3-1$ is irreducible, I believe I can use Eisenstein's criterion. But I don't see how to show even this.  The other approach on my mind is to show that $k[x,y,z]/(x^3+y^3+z^3-1)$ is an integral domain, but this doesn't seem particularly promising either.
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The title pretty much says it all.  Of course, one answer (IMO unsatisfactory) to such questions goes something like "a definition is a definition, period."  In my experience, mathematical definitions are rarely completely arbitrary.  Therefore I figure there must be a good reason for insisting that a field be a non-trivial ring (among other things), but it's not obvious to me.  I realize that a trivial ring would make for a very boring field, but it is also a rather boring ring, and yet it is not disallowed as such.  EDIT: I found a hint of a rationale in the statement that "the zero ring ... does not behave like a finite field," but I could not find in the  for this assertion exactly how the zero ring fails to behave like a finite field.
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Is every monic real polynomial the characteristic polynomial of some real square matrix? I strongly suspect so, but I couldn't find a proof.
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Let $F$ be a field. When happens that the additive group of $F$ is isomorphic to the multiplicative group?  It is easily to work out that $F$ must have characteristic $0$, but then what?
80201
I have $2$ similar questions :    Is  the following set $K$ a field:  $$K=\underset{i=even}{\bigcup} \mathbb{F}_{2^{i}}$$ where $\mathbb{F}_{2^{i}}$  values are all finite fields?   Is so, what is the characteristic of this field?   Is $K \cup \mathbb{F}_{2^{3}}$ a field?    My answer is: yes both $1$ and $2$ are infinite fields by subfield criterion of finite fields, but I am not sure!  And what about the characteristic? Is it $2,$ which is the base prime, or is it $0,$ since it is infinite?
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Give an example of a UFD having a subring which is not a UFD.  I thought of $\mathbb{Z}[\sqrt{2},\sqrt{3}]$. Could you please explain my question. I am trying grasp the concepts, need help.
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Can you construct a field with 4 elements? can you help me think of any examples?