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adjoin
adjoin, v. (əˈdʒɔɪn) Forms: 4 aioyne, ajoine; 5–6 adione, adjone; 5–7 adioyne, adjoyne; 7– adjoin. [a. OFr. ajoin-, ajoign-, stem of ajoindre, mod. adjoindre:—L. adjung-ĕre to join to; f. ad to + jung-ĕre to join.] † 1. trans. lit. To join on; to join or unite (a person or thing to or unto another)....
Oxford English Dictionary
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Adjoin Definition & Meaning - Merriam-Webster
The meaning of ADJOIN is to add or attach by joining. How to use adjoin in a sentence.
www.merriam-webster.com
adjoin
adjoin/əˈdʒɔɪn; ə`dʒɔɪn/ v[I, Tn]be next or nearest to and joined with (sth) 临近; 邻近; 接近; 毗连 We heard laughter in the adjoining room. 我们听到了邻屋的笑声. The playing-field adjoins the school. 运动场紧靠著学校.
牛津英汉双解词典
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What does the adjoining mean? What is the adjoin operation? The wikipedia link is pretty scant, but from eat it appears to be something along the lines as the smallest step towards the union of two sets? Are Adjunctio...
There are two kinds of adjunction. The easiest is when your field $K$ and your element $\alpha\notin K$ are both sitting inside a bigger field $\Omega$. Then $K(\alpha)$ may be defined to be the smallest subfield of $\Omega$ that contains both $K$ and $\alpha$: it’s the intersection of all subfields...
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What is the type of the "adjoin" operator on rings? **Context:** I am trying to formalize the definition of a polynomial ring in a programming language. One way to think about it, is to start by formulating the defini...
In the language of category theory, "adjoin" is more than a function that produces rings from rings: it is a functor from the category of rings to itself
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$a$ in ring $R$, adjoin $b$ such that we have relation $b=a$ then will get ring isomorphic to R I have been told that if we have a ring $R$ and $a$ is an element of $R$ and if we adjoin an element $b$ with the relatio...
Formally, adjoining $b=a$ is forming a quotient: $R[x]/(x-a)$. Then $b$ is the class of $x$. Now $p(x) \mapsto p(a)$ is a surjective homomorphism $R[x] \to R$ whose kernel is the ideal $(x-a)$ because $x-a$ is monic and we can divide $p(x)$ by $x-a$, obtaining $p(x)=(x-a)q(x)+p(a)$. Therefore, $R[x]...
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Can we adjoin the fourth roots of $x$? I'm wondering when, if ever, we can adjoin the four powers of $\sqrt[4]{x}$ to $\mathbb{Z} / p \mathbb{Z}$, for $p$ prime and $x$ a natural, to create a field with $p^4$ elements...
This is possible if and only if $p\equiv1\pmod4$. A) If $p\equiv-1\pmod 4$ then it is not possible. Assume that $x$ has multiplicative order $m$. We know that $m\mid p-1$. On the other hand $4\mid p+1$, so $4m\mid p^2-1$. Given that the multiplicative group of $\Bbb{F}_{p^2}$ is cyclic of order $p^2...
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Galois group of $x^8+x^4+1$ As the title suggests, my objective is to characterize the Galois group of the splitting field of the polynomial $x^8+x^4+1$ over $\mathbb{Q}$. That is, if $L$ is the splitting field of $x^...
Another way of looking at this: \begin{align} X^8+X^4+1&=(X^4-X^2+1)(X^2-X+1)(X^2+X+1)\\\ &=\Phi_{12}\Phi_6\Phi_3\,, \end{align} where $\Phi_n$ is the cyclotomic polynomial, each irreducible, with roots the primitive $n$-th roots of unity. So the splitting field is $\Bbb Q(\zeta_{12})$, Galois group...
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The real numbers are a field extension of the rationals? In preparing for an upcoming course in field theory I am reading a Wikipedia article on field extensions. It states that the complex numbers are a field extensi...
You have to adjoin _uncountably many_ elements to the rationals to get the reals.
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Adjoining a root of $p$ to $F_p$ is etale?! I'm confused about etale extensions of $F_p$. We know the etale extensions of a field are the products of separable finite field extensions. But if you take $F_p$ and adjoin...
$p$ is $0$, and if you take $\mathbb{F}_p$ and adjoin a root of $0$ you just get $\mathbb{F}_p$ again.
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tower of simple extensions of $\mathbb Z$$_p$ for which the full extension will not be simple Let $s$ and $t$ be independent variables and let $p$ be a prime. Show that in the tower $\mathbb Z$$_p$$(s^p,t^p)\lt\mathb...
Be careful that when you consider $K\lt F$, what you have in K are polynomials over polynomials like $f(s^p,t^p)/g(s^p,t^p)$. And, adjoining only $s$, you need to show you can not obtain $t$ (which indeed is in $F$) from such polynomials. And similarly, adjoining only $t$, you need to show you can n...
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Adding a new parameter to function Suppose we have a function $y=f(x)$. Is it always possible to "adjoin" a parameter $t$ i.e s.t $y(t)=f(x(t))$?
The correct definition of a function require not only a ''rule'' $f(\cdot)$ that gives a value $y=f(x)$, but also a set $X$ that is the domain of the function (and a set $Y$ that is its codomain) $$ f:X\to Y \qquad y=f(x) $$ now, if $x$ is a function of some other variable $t$ this function is: $$ g...
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Clarification on the proof of theorem 8 (Set Theory and Metric Spaces - Kaplansky) There are 2 parts (red underline and yellow highlight) in the proof below that I couldn’t follow. 1. Devise **a** one-to-one map o...
1. Yes, we are constructing _one_ function on $A$, but _separately_ on $D$ and (two parts of) $E$. Starting from the given example, $f(A)=D\cup\\{a_2,a_3,\dots\\}\cup\\{b_2,b_3,\dots\\}\cup \dots$, assume specifically that $C=f(A)\cup\\{b_1,c_1\\}$. Then set $g(x)=x$ if $x\in D$, $g(x)=x$ if $x\in \...
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Is $\Bbb Q[\sqrt2]$ cyclotomic? This overview of Galois Theory claims that a field extension of $F$ is cyclotomic if it's obtained by adjoining an $n$th root of _any_ element of $F$. Wikipedia claims you have to adjoi...
I think the correct definition is, that for any field $K$, the extension $K(\zeta_n)$ is called a cyclotomic extension of $K$, for $\zeta_n$ being a root of unity of order $n$. The word "cyclotomic" is used in this way for many other definitions, like the $n$-th cyclotomic polynomial $x^n-1$, and so...
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Comparing fields with same degree Two part question: Are the fields $\mathbb{Q} (\sqrt[3]{2}, i \sqrt{3})$ and $\mathbb{Q} (\sqrt[3]{2}, i, \sqrt{3})$ identical in algebraic structure? I have in notes that they both h...
For starters, neither $i$ nor $\sqrt{3}$ is in $\mathbb{Q}(\sqrt[3]{2},i\sqrt{3})$. And I think you're right to mistrust that the degrees are equal since $|\mathbb{Q}(\sqrt[3]{2},i,\sqrt{3})/\mathbb{Q}|=12$.
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