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622
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location United Kingdom
age 22
visits member for 2 years, 5 months
seen 5 hours ago

Currently studying for an Msc in Mathematics


Mar
16
comment Sylow subgroups of soluble groups
mathoverflow.net/questions/24081/… may be of some interest?
Mar
14
accepted Question on Nilpotentcy of $G$ given $G/N$ nilpotent
Mar
14
accepted If $\kappa$ is an inaccessible cardinal then $|V_\kappa|=\kappa$
Mar
14
accepted Isomorphism of field of fractions
Mar
8
revised Isomorphism of field of fractions
updating questions due to errors
Mar
7
comment Jordan-Holder theorem for modules?
math.berkeley.edu/~serganov/math252/notes9.pdf might be helpful?
Mar
6
answered Odd Permutations
Mar
6
accepted Does $GL(n,K)$ act transitively on $1$-dim subspaces of $K$
Mar
6
comment Isomorphism of field of fractions
@Quimey Is my first solution then correct?
Mar
6
comment Isomorphism of field of fractions
@Quimey of course! I was just being a bit stupid, thanks for the help :)
Mar
6
comment Isomorphism of field of fractions
@Quimey I'm confused I could just be being stupid but if we have say $f:R\oplus R\rightarrow R$ such that $f((r_1,r_2))=(r_1)$ then this is a surjective homomorphism and has kernel $R$ so from the first isomorphism theorem $(R\oplus R)/R\cong R$? Have I made a silly mistake here?
Mar
6
asked Isomorphism of field of fractions
Mar
6
answered If G is a group of order $pq$, where $p$ and $q$ are primes. How do I prove that any nontrivial subgroup of $G$ must be cyclic?
Mar
5
comment Ring homomorphism and ideals
@terribleatmath it is just running essentially the same argument over and over that things in the image of $\phi$ must have a pre-image in $R$. For these sorts of questions when you say "I don't think I know anything about the inverse image"- you do. You need a bit more confidence to stick with it for longer. You know its the inverse image of a ring homomorphism, so the only thing that you can use is that information- you then just need to play around with it for a while and you will get there :)
Mar
5
comment Ring homomorphism and ideals
@terribleatmath then we can do a similar thing for additive homomorphism property. That is: if $a,b\in Im(\phi)$ then $exists x,y\in R$ such that $\phi(x)=a$ and $\phi(y)=b$ then $phi^{-1}(a+b)=\phi^{-1}(\phi(x)=\phi(y))$. Then as $\phi$ is a ring homomorphism we have that $(\phi(x)+\phi(y))=\phi(x+y)$ and so...
Mar
5
comment Ring homomorphism and ideals
@terribleatmath yes we do, for example as $\phi(0)=0$ that gives that $\phi^{-1}(0)=\phi^{-1}(\phi(0))=0$. The same argument runs for $1$.
Mar
5
comment If G is a group of order $pq$, where $p$ and $q$ are primes. How do I prove that any nontrivial subgroup of $G$ must be cyclic?
It's just Lagrange theorem that gives you that the order of a subgroup divides the order of a group and then that all subgroups of prime order are cyclic
Mar
5
comment Ring homomorphism and ideals
@terribleatmath in order to show this we simply need to show that $\phi^{-1}:Im(\phi)\rightarrow R$ is a ring homomorphism and then apply what we have already done, do you see? Any ideas how to proceed to show that $\phi^{-1}$ is a ring homomorphism?
Mar
5
comment Ring homomorphism and ideals
@terribleatmath what is afaik? Ok but from this definition we only have that $\phi^{-1}$ is defined on the image of $\phi$. So in your question when you say that $N$ is an ideal of $\phi(R)$ or $R'$, we may apply $\phi^{-1}$ to ideals of $\phi(R)$ otherwise this as this is where it is defined
Mar
5
comment Ring homomorphism and ideals
How are you defining $\phi^{-1}(N')$? If $N'\cap Im(R)=\{0\}$ what is $\phi^{-1}$ defined as? More specifically as $\phi$ is not an isomorphism is does not have a naturally defined inverse