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The standard argument for this requires a bit of machinery, namely the following theorem (the formulation below is from Atiyah-MacDonald, Proposition 5.1). Theorem. Let $B$ be a (commutative) ring and $A$ a subring and $x \in B$. Then the following four statements are equivalent. $x$ is integral over $B$; $A[x]$ is a finitely generated $A$-module; $A[x]$ ...

2

For any $r,s\in R$, we can write $r=r_i+r_j$ and $s=s_i+s_j$, where $r_i,s_i\in I$, etc. Let $a=r_i+s_j$, so $a-r=s_j-r_j\in J$, and $a-s=r_i-s_i\in I$. In particular, $a+I=s+I$, and $a+J=r+J$. This shows that the map $$\varphi\colon R\to R/I\times R/J:a\mapsto (a+I,a+J)$$ is surjective. Verify it is a homomorphism, find its kernel, and apply the ...

2

Here's the proof that I was given in my course. It uses the fact that $\displaystyle \sum_{d \mid n} \phi(d) = n$, which can be proved using elementary principles, but is slightly tricky. We show that if there is an element of order $d$, then there are exactly $\phi(d)$ elements of order $d$. Let $a$ be an element of order $d$. The argument you provided ...

2

An injective object in a locally small category $\mathcal{C}$ is an object $I$ such that the functor $\mathcal{C} (-, I) : \mathcal{C}^\mathrm{op} \to \mathbf{Set}$ sends monomorphisms in $\mathcal{C}$ to surjections. To put it in simpler terms, $I$ is injective if, for every monomorphism $f : A \to B$ and every morphism $a : A \to I$, there is a morphism $b ... 1 Well the obvious definition works: If$(A,\sim_A)$,$(B,\sim_B)$are sets with equivalence relations, then define a morphism$(A,\sim_A)\to(B,\sim_B)$to be a map$f:A\to B$which respects the relations, i.e.$x\sim_A y \implies f(x) \sim_B f(y)$. It is readily verified that the composition of such morphisms is again a morphism and that the identity on$A$is ... 1 Let$g(X,Y) = (X-Y-Y^{p-1})(X-Y^2-Y^{p-2})\cdots (X-Y^{\frac{p-1}{2}}-Y^{\frac{p+1}{2}}) \in \mathbb{Q}[X,Y]$, so that$g(X,\zeta)=f(X) \in \mathbb{Q}[X]\subset\mathbb{Q}(\zeta)[X]$. We can identify$\mathbb{Q}(\zeta)(X)$with$\mathbb{Q}(X)[Y]/\Phi_p (Y)$, where$\Phi_p (Y) = (Y^p-1)/(Y-1)$is the p-th cyclotomic polynomial. Since the image of$g$in the ... 1 There is a way more concrete than the proof from Atiyah-MacDonald, that lends further insight. Namely one can represent algebraic integers as eigenvalues of square-integer matrices. Such matrices have monic integral characteristic polynomials, so their eigenvalues are algebraic integers. Conversely,$\alpha\,$is an eigenvector of its companion matrix, i.e. ... 1 There are two directions. When you're asking for help with an "if and only if" problem, 99% of the time you should say which direction you're having trouble with (and usually, you should be even more specific than that). Starting with an isomorphism, to find a change of basis: First of all, an isomorphism$(V_1,T_1)\to(V_2,T_2)$is a function$V_1\to V_2\$ ...

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