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6

To answer all three questions it suffices to construct a field of any given infinite cardinality, since a field is a ring which is additively a group. To that effect, let $A$ be an infinite set. Then the field $\mathbb Q(A)$ of rational functions over $\mathbb Q$ using the elements of $A$ as indeterminates is a field with the same cardinality as $A$.


6

Yes, many results of this form follow from the Löwenheim–Skolem theorem, which asserts that if a first-order theory (a particular way to write down axioms something should satisfy; this includes groups, rings, fields, and more) has an infinite model then it has a model of every infinite cardinality. The Löwenheim–Skolem theorem has much weirder ...


4

Do this argument satisfy your needs? If $\mathbb{F}_{9}\subset\mathbb{F}_{27}$ then $\mathbb{F}_{27}$ would be a vector space over the field $\mathbb{F}_{9}$ (since it is an abelian group with compatible multiplication of $\mathbb{F}_9$). If $\{a_1,\dots,a_n \}$ is a basis for this vector space, then every element in $\mathbb{F}_{27}$ can be written ...


4

The claim is false. For $p \neq 2$, take $K = \mathbb{Q}_p$, $L = \mathbb{Q}_p(\zeta_p)$, and $x = \zeta_p$. For what it is worth, the extension $L/K$ is tamely totally ramified. For the exercise, there is a concrete description of tamely totally ramified extensions $L/K$ with degree $n$: $L = K(\sqrt[n]{\pi})$ for some uniformizer $\pi$ of $K$. The ...


4

If $\Bbb F_9 \subset \Bbb F_{27}$, then the groups of invertible elements should be in the same relation, i.e. $\Bbb F_9 ^* \subset \Bbb F_{27} ^*$. But $\Bbb F_9 ^*$ has $8$ elements, while $\Bbb F_{27} ^*$ has $26$, and $8 \nmid 26$, so $\Bbb F_9 \not\subset \Bbb F_{27}$. In fact, the argument can be extended: if $\Bbb F_{p^m} \subset \Bbb F_{p^n}$, then ...


3

This is not true: $\Bbb Q(\sqrt 2) = \Bbb Q(1+\sqrt 2)$ even if the minimal polynomial $x^2-2$ of $\sqrt 2$ is different from the minimal polynomial $x^2-2x-1$ of $1+\sqrt 2$. The true result is: if $a$ and $b$ have the same minimal polynomial $f$ over $\Bbb Q$, then $\Bbb Q(a) \cong \Bbb Q(b)$, because they are both isomorphic to $\Bbb Q[X]/(f)$. You ...


3

Let we take $x=\exp\left(W(\log 2)\right)$, i.e. a solution of $x^x=2$. Step 1. $x\not\in\mathbb{Q}$. Assuming $x=\frac{p}{q}$ with $\gcd(p,q)=1$, we have $p^p=2^q\cdot q^p$, absurd. Step 2. $x$ is not an algebraic number. Assuming that $x$ is in algebraic number, the Gelfond-Schneider theorem gives that $2$ is a trascendental number. It is not, so: $\...


3

An algebraic closure must be a splitting field of all polynomials over itself (coefficients from the algebraic closure). So existence and uniqueness of a splitting field of all polynomials over a field $K$ does not trivially imply existence and uniqueness of an algebraic closure of $K$. However, it is not hard to prove that an algebraic extension of an ...


3

There are many standard ways of showing the general result. Jef Laga's is a textbook method for proving one direction. Here's something specifice to your construction. Let use denote by $\alpha$ the coset $x+\langle f_1(x)\rangle\in\Bbb{F}_3[x]/\langle f_1(x)\rangle$. Because $\alpha$ is a zero of $f_1$, we see that $\alpha^2=\alpha^2-f_1(\alpha)=-1$. ...


3

Any proper subfield of $\;\Bbb Z_p\;$ would have to be, in particular, a sugbroup of its additive group. But this is a group of order a prime $\;p\;$ , and as such it has no proper divisors, so by Lagrange's Theorem that's impossible.


2

It’s a proper subset, but it’s not a subfield at all: for example, $2^2=1$ in $\Bbb F_3$, and $2^2=4\ne 1$ in $\Bbb F_5$, so they don’t have the same operations. Similarly, $1+2=0$ in $\Bbb F_3$, but $1+2=3\ne 0$ in $\Bbb F_5$.


2

When you take the splitting field of all irreducible polynomials in a given field $K$ then all those polynomials split in the bigger field, but there is no guarantee that all polynomials in the bigger field split since there are many more new polynomials now. The difficulty with proving an algebraic closure exists is precisely this: it is easy to add the ...


1

Here is a very elementary solution which shows the power given to us by all those nice results the other solutions use. Any morphism of unitary rings $$ f\colon \mathbf F_3[x]/(x^2+1)\rightarrow \mathbf F_3[x]/(x^3+2x^2+1) $$ is dominated by a morphism $$ g\colon \mathbf F_3[x]\rightarrow \mathbf F_3[x] $$ in the following sense: the diagram $$ \begin{...


1

Your $f(x)$ is not a polynomial, which must have a fixed, but arbitrary degree $n\ge 0$. It is a theorem then, that every polynomial of degree $n$ over a field $F$ has at most $n$ zeros. This is no longer true over, say, a skew-field. Consider the polynomial $x^2+1$ over the quaternion algebra, as an example - see here. It has infinitely many roots.


1

The degree must be finite, otherwise $\sum \limits _{n = 0} ^\infty \frac {x^n} {n!}$ would be a polynomial - which clearly it isn't, being $\Bbb e^x$. In fact, a polynomial is just a function from $\Bbb N$ to $F$ with finite support.


1

If you talk about the existence of field contain strictly F as sub field , the answer is yes, as indicated in the comments, with for example $K = F (t)$ rational field of polinomial ring $F[t]$. If you talk about finite extension $K$ over $F$,then $K$ exist if and only if $F$ is not algebraically closed. if you talk about a proper sub extension $K$ of $...


1

$x^3+a$ with $a \in \mathbb F_{11}$ is always reducible because $x \mapsto x^3$ is a bijection in $\mathbb F_{11}^\times$ since $\gcd(3,10)=1$. $x^3+x^2+2$ is irreducible over $\mathbb F_{11}$.


1

The reason why we care about real closed fields is that these have very tame geometry. This does include quantifier elimination, which in this case tells us that a projection of a semialgebraic set is semialgebraic. But there are many other nice properties stemming from the elementary equivalence. For example, you can apply more or less any theorem true ...


1

The difference between real closed and algebraically closed fields isn't really rooted in model theory. Recall that a field $K$ is algebraically closed if every nontrivial polynomial over $K$ has a root in $K$. Note that if $K$ is actually an ordered field, $K$ can't be algebraically closed, since $x^2 = -1$ can't have a solution (exercise: the square of any ...


1

Yes, there are proper connected subfields of $\mathbb{C}$ other than $\mathbb{R}$. A construction can be found in a paper by J.Dieudonné: J.Dieudonné, Sur les corps topologiques connexes, C.R. Acad. Sci. Paris vol. 221 (1945) pp. 396-398. I don't have access to the above paper at the moment. It can in fact be proved that the only proper path-connected ...



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