# Tag Info

4

How many polynomials are there of degree 2? How many reducible polynomials are there of degree 2? Let the leading coefficient in both cases be 1.

4

In a commutative ring $R$ we have $\frac{R}{I}$ is a field if and only if $I$ is a maximal ideal. When we have a field $F$ we have that $F[x]$ is a PID, so an ideal is maximal if and only if it is generated by an irreducible polynomial. In this case however $\mathbb Z$ is not a field, so we cannot conclude $\mathbb Z[x]$ is a PID. So we cannot conclude ...

3

Hint 1 : For your first question, notice that an element in $K$ is of the form $P(\theta)$, with $P(X) \in \Bbb Q[X]$ and that for any $g \in G$, $g(P(\theta)) = P(g(\theta))$. Answer 1 :

3

If $\;k(X)=k\left[ \frac{f_i(X)}{g_i(X)}\;,\;\;1\le i\le n\right]\;$ , then there is a finite number of prime elements that can appear as factors of the the denominator of any element generated as a polynomial in the above $\;n\;$ elements and coefficients in $\;k\;$ . It is enough then to show there is an infinite number of prime elements in $\;k[X]\;$, ...

3

To see this directly: If there is an element of order $2k$, then there is an element of order $2$. But $x^2-1 = (x-1)^2$ in a field of characteristic $2$, so there are no nontrivial square roots of $1$.

3

If $x^2 + xy + y^2=0$, then $x^3 - y^3 = (x-y)(x^2+xy+y^2)=0$. But $3$ does not divide $31$, so $a\mapsto a^3$ is injective, and therefore $x=y$.

3

Here’s another method. You know that your transformation $\phi$ is of order two, and that the “conjugate” of $X$ is $1-X$. The minimal polynomial for $X$ over the fixed field is accordingly $f(T)=T^2-T+(X(1-X))$. Here I’ve used the sum of the conjugates for the linear coefficient (with the necessary change of sign) and the product of the conjugates for the ...

2

In a finite group, the order of an element divides the order of the group. A finite field of characteristic $2$ has $2^n$ elements for some positive integer $n$, so its multiplicative group has odd order.

2

A number $a$ is called constructible here if there exists a classic geometric construction (that is: using straightedge and compasses [and for the sake of completeness: picking a generic point]) that can construct a line segment of length $a$ times as long as a single given line segment. Since addition, subtraction, multiplication (regarding the given length ...

2

Let $p=2$. Note that $x^2+x+1$ is irreducible over the two-element field, for it has no roots in that field. Now let $p$ be odd. Let $F_p$ be the $p$-element field. We have $(-a)^2=a^2$, and if $a\ne 0$ then $-a\ne a$. Thus there are at most $\frac{p-1}{2}$ non-zero elements of $F_p$ that are squares of elements of $F_p$. (Actually, there are exactly ...

2

Since $\alpha \in K$ is a root of the irreducible polynomial $f \in L[X]$, then $f$ is the minimal polynomial of $\alpha$ over $L$. The degree $d$ of $\alpha$ over $L$ is $≤2$, because $[K : L]=2$. If $d=1$, what can you conclude? If $d=2$, write $f(X)=X^2+aX+b=(X-\alpha)(X-\beta)$. What are the relations between $\alpha$ and $\beta$?

2

Well, if you've actually proved the biconditional statements you mentioned, then you're done. Alternatively, show that $$\Bbb Q\subseteq\Bbb Q(i)\cap\Bbb Q\bigl(\sqrt2\bigr),$$ which I leave to you. Then, suppose $z\in\Bbb Q(i)\cap\Bbb Q\bigl(\sqrt2\bigr).$ Since $z\in\Bbb Q\bigl(\sqrt2\bigr),$ then $z\in\Bbb R.$ From there, we can use the fact that ...

2

Let $\alpha$ be any transcendental. Then $\beta := \frac{1}{\alpha}$ is transcendental and $\alpha\cdot \beta =1$. Thus 1. is false. Consider $f \colon \mathbb Q(\alpha) \to \mathbb Q(\beta), \frac{x_0 + x_1 \alpha + \ldots + x_n \alpha^n}{y_0 + y_1 \alpha + \ldots + y_m \alpha^m} \mapsto \frac{x_0 + x_1 \beta + \ldots + x_n \beta^n}{y_0 + y_1 \beta + ... 2 Here is a proof that point 3 is false that doesn't use any difficult theorems. As$\gamma$ranges through all transcendental numbers, the number$2^\gamma$takes on uncountably many values. Since only countably many of these values can be algebraic, there must exist a transcendental$\gamma$for which$2^{\gamma}$is also transcendental. Now let$\alpha = ...

2

First, you can argue that any abelian extension of $F$ is a composite of cyclic extensions. Next, it suffices to show that any such cyclic extension $E/F$ is of the form $F(\sqrt[m]{a})$ for some $m$ and some $a \in F$. Let $m = [E : F]$, so the Galois group of $E/F$ is isomorphic to $\mathbb{Z}/m\mathbb{Z}$. Let $\sigma$ generate this Galois group, and ...

1

There is a bit of abuse of notation involved here, since $f,g$ are being reused to identify their compositions $f\circ h$ and $g \circ h$ as if they were the same rational expressions in $k(t)$ but "restricted" to $k(h(t))$. But the notation needs to be taken with a grain of salt. It really means the compositions $f(h(t)) = g(h(t)$ are equal as rational ...

1

Let $T$ denote the transposition transformation, and let $I$ denote the identity transformation. We note that $$(T - I)^2 = T^2 - 2T + I = T^2 - I = 0$$ It follows that the minimal polynomial of $T$ divides $x-1$, which is to say that $1$ is its only eigenvalue. If $T$ is diagonalizable, it must then be equal to the identity. Since $T$ is not the ...

1

I assume that you know how to show that $\bar L$ is a field. (Tell me if this is unclear for you). In order to show $\overline{\bar{L}}=\bar{L}$, you have to prove two parts : $\bar{L} \subseteq \overline{\bar{L}}$. This part is easy since for any subfield $M \subseteq K$, one has $M \subseteq \overline M$. $\overline{\bar{L}} \subseteq \bar{L}$. This ...

1

clearly the extension is of degree 2 and as it is finite extension of finite field so is galois (normal,seprable,finite). then galois group consist of two elements G=(1,a) where a is Automorphism of Field with 9 elements keeping base field fixed. orbit(x)={x,a(x)} for any x in F= field with 9 elements now a(x)=x for each element of the base field by ...

1

yes it is true $Q\sqrt2$ and $Q(i)$ are extension of degree $2$ of $Q$ so the degree of $Q(i)\cap Q(\sqrt2)$ is either 1 or $2$, if it is 2 it implies that $Q(i)=Q(\sqrt2)$ thus $\sqrt2=a+bi, a,b\in Q$, by writing $(\sqrt2-a)^2=-b^2$, you obtain $a=0$ unless $\sqrt2\in Q$ a fact which is not true. If $a=0$, $\sqrt2=bi$ thus the $2=-b^2$ this not true also.

1

You don't need to find the minimal polynomial. That's because whatever the polynomial is for $b$ over $K$, call it $p$, it is still a polynomial for $b$ over $K(a)$. Therefore, the minimal polynomial for $b$ over $K(a)$ divides $p$ and therefore has degree no greater than $\deg(p)=[K(b):K]$.

1

Hint $$x^3-y^3=(x-y)(x^2+xy+y^{2})$$ Hint 2 If $x \neq 0$ and $y \neq 0$ you get $$x^{3}=y^3 \\ x^{31}=y^{31}$$ From here you get immediately $x=y$, which you can plug in the original equation. The case $x=0$ OR $y=0$ is easy...

1

If $y = 0$, clearly also $x = 0$. So suppose $y \ne 0$, multiply by $y^{-2}$, and set $t = x y^{-1}$ to obtain $t^{2} + t + 1 = 0$. Now you should know that the solutions of the latter equations are the two elements different from $0, 1$ in the field of order $2^{2} = 4$. Since $32 = 2^{5}$, the field of order $4$ is not a subfield of $F$. So no solutions ...

1

If $\operatorname{char}(K)\neq 2$, then $X$ satisfies a polynomial of degree $2$ over $K(Y)$ (namely, $p(t)=(2t-1)^2-Y$), so $[K(X):K(Y)]\leq 2$. Since $K(Y)\subseteq L\subset K(X)$, it follows that $K(Y)=L$. If $\operatorname{char}(K)=2$, on the other hand, this doesn't work (the polynomial $p(t)$ used above is identically $0$). And in fact it is clear ...

Only top voted, non community-wiki answers of a minimum length are eligible