How to go about proving that $-1+(x-4)(x-3)(x-2)(x-1)$ is irreducible in $\mathbb{Q}$?

How do you show that $-1+(x-4)(x-3)(x-2)(x-1)$ is irreducible in $\mathbb{Q}$? I don’t think you can use the eisenstein criterion here Answer Actually, the obvious generalization is also true. Let $P$ and $Q$ be polynomial factors, so that the given expression equals $PQ$. Then $PQ=-1$ at each of the integer values, $1,2,3,4$ for this case. So … Read more

Badly behaved, but easy-to-manipulate examples of rings to test hypotheses on

In calculating examples in mathematics it’s often useful to have a quite misbehaving but easy-to-manipulate object to test hypotheses on. Examples are the function f(x)={0 if x∈Q1 if x∉Q in analysis, or the Baumslag-Solitar groups B(n,m) in group theory. Do there exist rings that are like this? If so, which are your favourites? At the moment I tend to … Read more

Show R\Bbb R and R[X]\Bbb R[X] not isomorphic

Why are the rings R and R[x] not isomorphic to eachother ? Think it might have to do with multiplicative inverses but I’m not sure. Answer You are right: the element “x” has no multiplicative inverse, that is there is no polynomial p(x) such that x⋅p(x)=1. AttributionSource : Link , Question Author : jamie , … Read more

Generalization of Chinese Remainder Theorem

Q: With ring R, if I,J⊆R are ideals such that I+J=R, then the map R/(I∩J)→R/I×R/J given by a+(I∩J)↦(a+I,a+J) is an isomorphism, broadly generalizing the Chinese Remainder Theorem. Can someone help me get started on this one? Answer Consider the canonical maps πI:R→R/I and πI:R→R/J and form the morphism φ(r)=(πI(r),πJ(r)). As Censi LI explained in his … Read more

prove (n)⊇(m)⟺n∣m (n) \supseteq (m)\iff n\mid m\ (contains = divides for principal ideals)

For non-zero integers m and n, prove (m)⊂(n) iif n divides m, where (n) is the principal ideal. My attempt is following. For non-zero integers m and n, assume that (m)⊂(n). Then, mk∈(m) is also in (n). This means that ∃nh such that mk=nh. Then, we have m=nhk−1. Assume that n divides m for non-zero … Read more

Are fractions with zero divisors in the denominator never well defined?

Are fractions with zero divisors in the denominator never well defined? I know that for a fraction in modular arithmetic to be well defined, the denominator must not be a zero divisor, e.g: x \equiv \frac a b \pmod p Then b\neq k \cdot p, k\in \Bbb{Z}, but what about fractions in other rings? Answer … Read more