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Questions tagged [abstract-algebra]

For questions about monoids, groups, rings, modules, fields, vector spaces, algebras over fields, various types of lattices, and other such algebraic objects. Associate with related tags like [group-theory], [ring-theory], [modules], etc. as necessary to clarify which topic of abstract algebra is most related to your question and help other users when searching.

Abstract algebra is the study of algebraic objects, i.e. sets endowed with one or more operations on the elements of those sets. In particular, the study of abstract algebra considers the algebraic structures and properties of which such operations induce. It can be considered as the generalization of the study of the algebraic structure of the integers and real numbers (arithmetic), or the study of matrices and vector spaces (linear algebra).

Some algebraic objects are monoids, groups, rings, fields, vector spaces, modules, algebras, and categories, among many other less prominent objects.

Examples

  1. The set of non-negative integers $\mathbb{N} = \{0,1,2,3,\dotsc\}$ is a monoid under the operation $+$.

  2. The integers $\mathbb{Z} = \{\dotsc,-1,0,1,\dotsc\}$ under the binary operation of $+$ form a group.

  3. Furthermore, $\mathbb{Z}$ has the structure of a ring when you consider it as being equipped with both addition and multiplication.

  4. The real numbers $\mathbb{R}$ with their usual addition and multiplication form a field.

  5. The set of $n\times n$ matrices with entries in $\mathbb{R}$ with matrix addition and multiplication form a ring.

  6. The set of $1\times n$ vectors over the real numbers, with vector addition, and multiplication by elements of the $n\times n$ real matrices on the right are an example of a module for the ring of matrices.

In addition to studying the objects themselves, abstract algebra considers homomorphisms between the objects and various constructions and tools, which are useful for studying the objects.

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An extension field of the field of rational numbers

How to prove that $\mathbb{Q} (\sqrt[3]{2}+\sqrt[3]{4})= \mathbb{Q} (\sqrt[3]{2})$. "$\subset$" Since $\sqrt[3]{2} \in \mathbb{Q} (\sqrt[3]{2}) $ and $\sqrt[3]{4} \in \mathbb{Q} (\sqrt[3]{2}) $ (because $\sqrt[3]{4} = (\sqrt[3]{2})^2$), thus the…
george_smile
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Why are Quotient Rings called Quotient Rings?

Let $R$ be a ring, and $I$ be an ideal of $R$. Let $a\in R$. Definition 1.1 : The coset of $I$ with respect to $a$ is defined to be $a+I=\{a+x:x \in I\}$ Definition 1.2 : The set of cosets of $I$ in $R$ is defined to be $R/I=\{a+I:a \in R\}$ with…
MTN
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Inverse of $f: \mathbb{Z}_{21} \times \mathbb{Z}_{10} \to \mathbb{Z}_{210}$ such that $f(a,b)= 10a +21b$

Let $f: \mathbb{Z}_{21} \times \mathbb{Z}_{10} \to \mathbb{Z}_{210}$ such that $$f(a,b)= 10a +21b.$$ We have that $f$ is an isomorphism, but how does one go about finding explicitly the inverse $f^{-1}$ of $f$?
user310769
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Prove that $K$ is a field

Let be $K$ the set of real numbers that can be written as $a+b\sqrt2$, with $a$ and $b$ rational numbers. Prove that $K$ is a field. I have already proved that $0$ and $1$ $\in K$, and that sum and product of two elements $\in K$. I have also…
Surfer on the fall
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Interpretation of $\sum_{i=1}^n i$

There is an obvious interpretation of $\sum_{i=1}^n i$ as just a function $\mathbb N \to \mathbb N$: $$n \mapsto \underbrace{1 + 2 + 3 + \ldots}_{n}$$ However since the sum in hand equals $$\frac{n(n+1)}{2}$$ which is polynomial, one can treat it…
Yrogirg
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Proof that conjugate subgroup has the same order as the subgroup

I want to prove that $|xHx^{-1}|=|H|$ where $H$ is a subgroup of $G$ and $x\in G$. I know how to prove that for each element in $H$, $|xhx^{-1}|=|h|$, but I'm not sure how to extend that to the whole subgroup.
BCho
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Understanding cofree coalgebras

Let $V$ be a vector space over $K$ and $T(V)$ denote the tensor algebra on $V$. It is well known that $T(V)$ is the free algebra on $V$. I've been told that it is also the cofree coalgebra on $V$ with comultiplication $\Delta:T(V)\rightarrow T(V)…
Seth
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A ring with a unique nonzero proper ideal?

Is there any commutative unital ring $R$ with a unique nonzero proper ideal? In particular, there must be a nonzero proper ideal, so fields don't work. Clearly, such a ring must be local, and the unique (maximal) ideal $I$ contains every non-unit.…
Nishant
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operation that is commutative but not associative

Does there exist an operation that is commutative but not associative? do you know what is operation?
ghazale
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Multiple roots of a polynomial over a field of characteristic $p$

I have to show for what value of the prime $p$ does the polynomial $ x ^4 + x + 6$ have a root of multiplicity $>1$ over the field of characteristic $p$. $ p=2, 3, 5, 7 $ Please help. For $F$ a field of characteristic $3$, $f(x)= x^4 + x =…
preeti
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Endomorphism Rings in Ring vs R-Mod

Let $M$ be an $R$-module. Considering $M$ as an abelian group, forgetting the $R$-action on $M$, we can unambiguously write $\mathsf{End}(M)$ to denote the endomorphism ring with addition defined as in $M$ and composition as the multiplication…
ItsNotObvious
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Orders of Elements in GL(2,R)

Let A = $$\begin{pmatrix} 0&1\\ -1&0\\ \end{pmatrix}$$ and B = $$\begin{pmatrix} 0&-1\\ 1&-1\\ \end{pmatrix}$$ be elements in $GL(2, R)$. Show that $A$ and $B$ have finite orders but AB does not. I know that $AB$…
TfwBear
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H char G, K char G, K subgroup of H implies K char H?

I'm working on a proof, and I would like to show that if every subgroup is characteristic in G, then any subgroup of a subgroup is also characteristic in that subgroup. In other words, $H char G, K char G$ and $K \leq H \implies K char H$. Now I…
j215c228
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The product of a subgroup and a normal subgroup is a subgroup

Let $G$ be a group, $H$ a subgroup of $G$, and $N$ a normal subgroup of $G$. Verify that $HN=\{hn\mid h \in H, n \in N\}$ is a subgroup of $G$.
anilorap
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If $N\trianglelefteq G$, then $\phi(N) \leq \phi(G)$, where $\phi(N)$ is the Frattini subgroup of $N$.

I was thinking somehow to use normality of N as follows Since N is normal, then $G/N$ will be a group, so we can consider the natural map $pi \ :\ G \rightarrow G/N$, where $g \mapsto gN$, $ker(\pi) = N$. So It is here enough to prove that a…
user111750
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