Consider the polynomial $p(x) = 1+x^5+x^{10}$ with binary coefficients. Consider the multiplicative group of $\mathbb{F}_{16}$, and let $p(x)$ be evaluated at each of these $15$ elements. The only possible evaluations are $0$ and $1$. I am looking for more such polynomials which have binary coefficients, which when evaluated on the elements of an extension field […]

Let $R$ be a ring, $S\subset R$ a multiplicative subset, and let $M$ be a Noetherian $S^{-1}R$-module, then does there exist some Noetherian $R$-module such that $S^{-1}N\cong M$? What about if we only consider localizations of the form $R_f$? What if we also require $R$ to be Noetherian? If we let $N={_R M}$ then clearly […]

Not sure where to go with this, but I don’t think it is cyclic..

In order to make a smaller example for my question Galois group of the field of all constructible complex numbers, I am posing this new question. I know already, that E is a galois extension of $\mathbb{Q}$ of dimension 16, it being a galois extension of the splitting field $L$ of the polynomial $X^4-2$ of […]

This question already has an answer here: Normalisation of $k[x,y]/(y^2-x^2(x-1))$ 1 answer

This question already has an answer here: Intermediate ring between a field and an algebraic extension. 3 answers

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 $G$.

Let $G$ be a finite Group, H be a subgroup of $G$ of index $2$, and $x\in H$. Denote by $cl_G(x)$ conjugacy class of $x$ in G and by $cl_H(x)$ the conjugacy class of $x$ in $H$. Question is : $(a)$ Show that if $C_G(x)\leq H$, then $|cl_H(x)|=\frac{1}{2}|cl_G(x)|$. $(b)$ Show that if $C_G(x)$ is not […]

I want prove (or disprove) that given a finite group G, any two maximal subgroups that are isomorphic to $PGL(2,p)$ where p is prime, then their intersection is isomorphic $PGL(2,q_0)$ (for some $q_0$). I am aware of the fact that intersection of subfields is also a subfield. However, the first statement is unclear to me. […]

Let $R$ be a ring such that $R$ has no non-trivial right ideals. If there exists a nonzero element $a \in R$ with $aR=0$, prove that $|R|= p$ where $p$ is prime.

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