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6h
revised Injective Ring Homomorphism
edited body
22h
answered Group actions: Why do we place the condition that $S$ be finite in the following theorem?
1d
comment Injective Ring Homomorphism
No, just any $p(x)$. The map take any polynomial and assigns to it its values in $0$, $1$ and $2$.
1d
answered Injective Ring Homomorphism
1d
comment Group actions: Why do we place the condition that $S$ be finite in the following theorem?
Of course the set $S$ can be infinite too, but in that case the equality becomes simply $\infty=\infty$ which is rather uninteresting. The basic result is actually the existence of a bijection $[a]\longleftrightarrow G/G_a$ (for left actions)
Dec
16
awarded  Good Answer
Dec
16
answered How many times does the derivative of this function meet the x=0 line?
Dec
16
comment Show $G=(\mathbb{Q},+)$ is not finitely generated
It's a contradiction with the fact that the picked set generates $\Bbb Q$. Whatever finite set you pick, it generates a proper subgroup of $\Bbb Q$.
Dec
15
awarded  Caucus
Dec
15
comment Measurability in the proof of Minkowski's Bound for calculating the Class Number?
You must have never heard of Tamagawa measures then. encyclopediaofmath.org/index.php/Tamagawa_measure
Dec
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answered If $A$ is isomorphic to $B$ and $B$ is a field, then $A$ is a field?
Dec
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comment Number of complex solutions
$1$ is a perfectly good complex number too
Dec
9
comment Let $F|K$ be a field extension and $a \in F $ such that $[K(a):K]$ is odd integer
@yiokk : Since $(a)^2=a^2$, the element $a$ always satisfies $X^2-a^2\in K(a^2)[X]$.
Dec
9
answered Let $F|K$ be a field extension and $a \in F $ such that $[K(a):K]$ is odd integer
Dec
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revised How to prove that there are infinitely many primes without using contradiction
added 2 characters in body
Dec
6
comment $A$ is a commutative ring, $P$ is a prime ideal. Prove $A_P$ is local ring
Look at the definition of $A_P$ and make your guess!
Dec
6
answered $A$ is a commutative ring, $P$ is a prime ideal. Prove $A_P$ is local ring
Dec
6
revised How to prove that there are infinitely many primes without using contradiction
added 449 characters in body
Dec
6
comment How to prove that there are infinitely many primes without using contradiction
For instance $n!+1$ must be divisible by a prime bigger than $n$
Dec
6
comment How to prove that there are infinitely many primes without using contradiction
No. A prime in $\cal P\cap{\rm supp}(a)$ divides $1$, thus there are no such. Don't use contradiction!