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22

First of all, let me clear a confusion of yours. $\Bbb R$ has cardinality $2^{\aleph_0}$. This is not necessarily $\aleph_1$. It is consistent with the axioms of modern set theory that the answer is positive or negative. Similarly $2^{2^{\aleph_0}}$ need not be equal to $\aleph_2$, and it could be much larger. Now that we cleared that issue. What is a ...


16

The surreal numbers are an ordered field too big even to be a set: they form a proper class that includes the ordinals.


8

Hint: $$\sqrt2=a+b\sqrt3\;,\;\;a,b\in\Bbb Q\implies 2=a^2+3b^2+2ab\sqrt3\implies\sqrt3=\frac{2-a^2-3b^2}{2ab}$$ assuming $\;ab\neq0\;$...but what happens if $\;ab=0\;$ ?


7

In $\mathbb{F}_{q^n}$ every element $x$ satisfies $x^{q^n}=x$. In fact, for $x=0$ this is clear, and for $x \neq 0$ it follows from Lagrange's theorem applied to the multiplicative group $\mathbb{F}_{q^n}^\times$. In particular, every element of $\mathbb{F}_{2^n}$ is a square.


6

The notation with the rectangular brackets is at first a ring: $\mathbb{Q}$-linear combinations of $1$ and $\sqrt{2}$. However this ring is actually a field since $\frac{1}{a+b\sqrt{2}}= \frac{a-b\sqrt{2}}{a^2-2b^2}$ is again a $\mathbb{Q}$-linear combination of $1$ and $\sqrt{2}$. So every non-zero element has an inverse and hence we have a field. The ...


5

Consider the map $\phi : \mathbb{F}_{2^n} \to \mathbb{F}_{2^n}$ given by $\phi(x) = x^2$. You can show that this is a homomorphism between rings (use the fact that $\operatorname{char}(\mathbb{F}_{2^n}) = 2$), so its kernel is an ideal in $\mathbb{F}_{2^n}$. As $\mathbb{F}_{2^n}$ is a field, $\ker\phi = \{0\}$ or $\ker\phi = \mathbb{F}_{2^n}$, but you should ...


4

$\mathbb{Q}[\sqrt{2}]$ gives you all polynomials with $\sqrt{2}$ as the variable, including $a+b\sqrt{2}+c\sqrt{2}^2+d\sqrt{2}^3$. This is the same as the set of $a+b\sqrt{2}$ of course because $\sqrt{2}^2$ is rational. $\mathbb{Q}(\sqrt{2})$ is the set of rational functions, which are ratios of polynomials of $\sqrt{2}$. That is ...


3

It is false. Take $F=\mathbb Q$ the rationals, and $\omega=e^{\frac{2\pi i}{12}}$. $\omega$ is a primitive $12$-root of unity, and we know that in this case $[\mathbb Q(\omega):\mathbb Q]=\phi(12)=4$. (If you do not know what this means, search for Euler totient.). Now see $\omega^3=e^{\frac{3(2\pi i)}{12}}=e^{\frac{2\pi i}{4}}$, so $[\mathbb ...


3

Hint. What can you say about the degree of the minimal polynomial of a non-separable element?


3

Fact: If $L,L'$ are field extensions of a field $K$ with $L \subseteq L'$, then $[L:K]$ divides $[L':K]$. In fact, we have $[L':K]=[L':L][L:K]$. If $\mathbb{Q}(\sqrt[m]{p}) \subseteq \mathbb{Q}(\sqrt[n]{q})$, we would get that $m | n$.


3

Suppose $A = \mathbb{Z}[x]/(x^4 + 9x + 6)$ is a field. Now $2 \in A$ and $2 \neq 0$ in $A$. So there exists a polynomial $p(x) \in \mathbb{Z}[x]$ of degree $\leq 3$ such that $2p(x) - 1 \in (x^4 + 9x + 6),$ which is clearly not possible unless $2p(x) -1 = 0,$ in which case $2$ is invertible in the ring $\mathbb{Z}[x].$


2

The algebraic closure has no universal property at all (and therefore one should rather speak of an algebraic closure). Uniqueness of extended homomorphisms fails, and this is what Galois theory is all about. (I really wonder how such statements can find their way into an advanced book on homotopy theory published by the AMS.) By the way, Grothendieck's ...


2

A subspace must contain the zero vector.


2

Consider the map $\phi : \mathbb{F}_{2^n} \to \mathbb{F}_{2^n}$ given by $\phi(x) = x^2$. Then, $\phi$ is injective because $x^2=y^2$ iff $0=x^2-y^2=(x+y)(x-y)=(x-y)^2$. Since there are no zero divisors in a field, $x-y=0$ and so $x=y$. Since the field is finite, $\phi$ is surjective and every element is a square.


2

Let $\;r^2\;$ be a zero of $\;f(x)\in F[x]\;$ : $$f(x)=\sum_{k=0}^n a_kx^k\implies 0=f(r^2)=\sum_{k=0}^na_k(r^2)^k=\sum_{k=0}^na_kr^{2k}$$


1

1) You must see your previus post, I fixed it, and the question turned out to be false, sorry for the mistake. In this case, the question is again false(F(r)=F(r^2)), and the same example works again, the same $\omega$, but instead to take $w^3$ you must take $w^2=e^{\frac{2\pi i}{6}}$ and then $[Q(\omega^2):Q]=\phi(6)=2$. Hint: 2,3,4,5 take $F=Q$. 4 must ...


1

Hint: look for a tower of subfields of $F$ whose union is the field you're looking for.


1

This is really the sort of Löwenheim-Skolem type argument. Take the union of the fields $K$, it is a countable set. Now define by induction a sequence of sets, whose union is a field containing $K$. At each step you gather all the products and sums and inverses of elements from the previous step. Now since $K$ is countable the first step of the induction ...


1

Hints: Notice that $\alpha = \sqrt{5}$ is a root of $f(x) = x^2 - 5$. Now take any $g(x) \in \mathbb{Q}[x]$ and divide it by $f (x)= x^2-5$. What is the remainder? Next, take $g(\alpha)$, and notice that it can be written in terms of $a + b\alpha$. Use the Eiseinstein Criterion to determine whether $f(x) = x^2-5$ is irreducible over $\mathbb{Q}$, from ...


1

Overkill answer: An abstract idea using Galois theory is to consider $C:=(K^*)^2\cap\mathbb{Q}^*$, where $K=\mathbb{Q}(\sqrt{2})$. Clearly $S:=(\mathbb{Q}^*)^2\cup 2(\mathbb{Q}^*)^2\subseteq C$, but every element $[c]$ of the quotient group $C/(\mathbb{Q}^*)^2$ induces a homomorphism $Gal(K/\mathbb{Q})\to\{\pm 1\}$ with ...


1

A number $p>1$ is prime if and only if the only solutions of the equation $x^2\equiv 1\ (\ mod\ p\ )$ are {$-1,1$} If only a weak pseudoprimetest is made (check, if $a^{p-1}\equiv 1\ (\ mod\ p\ )$) and $p$ passes this test, but the rabin-test fails for this base, then a nontrivial solution of $x^2\equiv 1\ (\ mod\ p\ )$ can be derived (showing that ...



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