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Results tagged with abstract-algebra
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user 298680
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.
2
votes
Accepted
Is $\mathbb R \times \mathbb Q$ a principal ideal domain
If you read the answers given in the link, you will see that they apply to every PID$^{(1)}$ $A$ and $B$ and not only to $\Bbb Z \times \Bbb Z$.
So, every ideal in $R=\mathbb R \times \mathbb Q$ is a …
- 23.9k
1
vote
Element order in the group
In that context, the number $v_2(|g|)$ denotes the $2$-valuation of the order of $g$, i.e. the greatest $n \in \mathbb N$ such that $2^n$ divides the order of $g$.
- 23.9k
2
votes
Accepted
Well-defined function
Here is an example of a "function" which is not well-defined :
$$f : \Bbb Q \to \Bbb Q \quad;\quad \frac{n}{m} \mapsto n$$
Indeed : $2=2/1=4/2$ should imply $f(2)=f(2/1)=2=f(4/2)=4$ which is obvious …
- 23.9k
2
votes
Accepted
Understanding ideals generated by $\langle a_1, ..., a_n \rangle$
Your ideal $I$ is just the set of polynomials of the form
$$f(X)=XP(X)+2Q(X) \qquad P,Q \in \Bbb Z[X]$$
It is equal to
$$\{2a_0+a_1X+\cdots+a_nX^n \mid a_i \in \Bbb Z, n ≥ 0\}$$
Your polynomial $5X^3 …
- 23.9k
1
vote
Accepted
For a prime number p, show that the Frobenius map given by $Frob_p(a) = a^p$ is an isomorphi...
What you've done so far is great.
In order to show that the Frobenius morphism is bijective, it is sufficient to show it is injective, since $\Bbb Z/p\Bbb Z$ is a finite set (and every injective func …
- 23.9k
2
votes
Accepted
Splitting field for $x^2 + 1 \in \Bbb Q[x]$
Let $K$ be an extension of $\Bbb Q$ (with $j : \Bbb Q \to K$) such that $X^2+1$ splits over $K$, say $X^2+1=(X-a)(X-b)$ with $a,b \in K$. We know that $b=-a$ since $(-a)^2=a^2=1$.
Let $K' := j(\Bbb Q …
- 23.9k
1
vote
Accepted
All simple cyclic module over $k[x]$ is of the form $k[x]/(f)$ with $f$ prime.
Let $R=k[x]$.
You forgot to mention in the question that $M$ is simple, as $R$-module (you wrote it in the title, though). Otherwise this is wrong, since $R$ is cyclic but not of the form $R/(f)$ with …
- 23.9k
5
votes
Accepted
Can someone please explain why it is the *smallest* subfield?
First of all, $K=F[x]/(irr(\alpha,F))$ is isomorphic to $K'=\phi_{\alpha}(F[x])$ by the first isomorphism theorem. Since $irr(\alpha,F)$ is irreducible, $K$ is a field, and so is $K'$.
Let $L$ a subf …
- 23.9k
4
votes
Accepted
Why is $[\mathbb{F}_{p^n}:\mathbb{F}_{p}] = n$?
Let $d$ be the dimension of $\mathbb{F}_{p^n}$ over $\mathbb{F}_{p}$. Then, you have an isomorphism as vector spaces:
$$\mathbb{F}_{p^n} \cong \mathbb{F}_{p}^d$$
The cardinality of the first one is $p …
- 23.9k
6
votes
Prove that there exists subgroup of any order of any power of $p$ in a $p$-group
Let $p$ a prime number. We will prove by induction on $n$ that every group of order $p^n$ has a normal subgroup of order $p^m$ for $0 \le m \le n$. Let $G$ a group of order $p^n$.
If $n=1$, that's cl …
- 23.9k
4
votes
Accepted
What is the kernel of $\varphi : R[u,v]\to R[x,1/x]$ defined by $\varphi(p(u,v))=p(x,1/x)$?
Clearly $I:=(1-uv) \subset \ker(\phi)$.
Now, look at the quotient ring $R[u,v] / I$. We want to show that any polynomial $P \in \ker(\phi)$ is $0$ modulo $I$, that is : $P \equiv 0 \pmod I$.
The t …
- 23.9k
1
vote
Accepted
How to find the commutator group of $G= \left\{\begin{pmatrix} 1 & a & b\\0 & 1 & c\\0 & 0 &...
Pick a matrix $A=\begin{pmatrix} 1 & 0 & \alpha\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$ and find $g,h \in G$ such that $A=ghg^{-1}h^{-1}$. You can try a matrix $h$ with $a=b=c=1$. I let you find the suita …
- 23.9k
3
votes
Accepted
Let $R$ be a ring except that addition and multiplication in $R$ are not assumed to be commu...
Let $a,b \in R$. Look at $$x := (1+1) (a+b).$$
On the one hand, this is equal to $x=(1+1)a + (1+1)b$.
On the other hand, this is equal to $x=1(a+b) + 1(a+b)$.
If you develop these two expressions, yo …
- 23.9k
3
votes
Accepted
Why do polynomials and integers both have a long division algorithm?
I think the algebraic structures you are looking for are called Euclidean domains.
An Euclidean domain $R$ doesn't need to be a field, but only a ring with a "euclidean function" $v : R \setminus \{0 …
- 23.9k
1
vote
Accepted
Identifying two groups from an exact sequence
I assume that the sequence is exact.
The image of $0 \to A$ is then the kernel of $s$, which is $0$. Hence $s$ is injective and $A \cong \mathrm{im}(s) = \ker(\times \ell) = d\Bbb Z / k \Bbb Z$ for s …
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