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8

Take $R = \mathbb{Z}[X]$. Let $p$ be a prime. Then $X^2+p$ is irreducible, and $(X^2+p)$ is prime. The ideal $(p)$ is also prime. The ideals are not co-prime. The ideal $(X^2+p) + (p)$ is not prime, as it contains $X \cdot X$ yet not $X$. Note that this domain has Krull dimension $2$, as has the example in the other answer. And, one needs Krull ...


6

Well, what's a next-worst scenario from a PID? Polynomial ring in two variables, i.e, for a field $k$, $k[X,Y]$, gives you an immediate counter-example. Let $k=\mathbb{R}$, $\mathfrak{p}=(X^2+Y^2)$, and $\mathfrak{q}=(X)$. Then $\mathfrak{p+q}=(X^2+Y^2,X)=(X,Y^2)$, which of course isn't prime.


5

Yes, that's right. This perspective leads to the idea that linear categories are a "many-object" generalization of rings, and naturally occurs in Morita theory.


5

The question is incorrect; $\mathbb{Z}[\sqrt{19}]$ is a UFD. The Minkowski bound for the field $\mathbb{Q}(\sqrt{19})$ is $\sqrt{19}$, and the ring of integers of $\mathbb{Q}(\sqrt{19})$ is $\mathbb{Z}[\sqrt{19}]$. It is a theorem that the ideal class group of a number field is generated by the ideals of norm less than the Minkowski bound. Hence, to ...


4

@Joel92 's comment suggests this counterexample: for a field $K$ all monic linear polynomials in $K[x]$ are prime. They all have degree 1, hence the same norm (the measure used for the "size of the remainder" in the Euclidean algorithm). No two are associates. @user26857 's comment offers $2 \pm i$ in the Gaussian integers. Each is prime since each has ...


3

Let $\rho \colon B \to A$ be a surjective ring homomorphism, and put $I = \ker(\rho)$, so $B/I \cong A$. If $\Phi \colon C \to B$ be an isomorphism, then the composite map $\rho \circ \Phi \colon C \to A$ is a surjective homomorphism with kernel $\Phi^{-1}(I)$, so $C/\Phi^{-1}(I) \cong A$, as you suspected. In fact, everything above holds if $\Phi$ is only ...


3

It is false. Consider the matrix.$$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ It is not the square of a complex matrix, so $p(X)=X^2$ is not surjective.


3

I think it follows from Gauss Lemma. $x^2-a$ is irreducible in $D$ and therefore it must be irreducible in $F$. In particular, it can't have any root.


2

As Joel92 notes, for some $p$, $\mathbb{Z}_p \subseteq R$. Let $a\in R\setminus \mathbb{Z}_p$. If $a$ is transcendental over $\mathbb{Z}_p$, then you have infinitely many subdomains of the form $ \mathbb{Z}_p[a^n] $. Thus $a$ is integral over $ \mathbb{Z}_p$. It means that $R$ is a finite field extension of $\mathbb{Z}_p$, and is in particular finite. ...


2

The standard proof of the Rational Root Test works in any UFD, so any root of $\,x^2-a\,$ in the fraction field of $D$ must be in $D,\,$ exactly as in the classical case $\,D = \Bbb Z$.


2

Yes, it's true. To see this, let $p$ an irreducible element in $D$. Consider the valuation $v_p$ on $F$. If we have $x=y^2$ in $F$, then $$v_p(x)=2v_p(y),$$ so every irreducible factor appears with an even exponent. As units can be incorporated into irreducible elements, this proves that, if $x$ is a square in $F$, it is a square in $D$.


2

All these can be found in the excellent classic `On ubiquity of Gorenstein rings' by Hyman Bass. So, let $R$ be a self injective local ring of finite length with $k$ the residue field. Then the functor $N\mapsto N^*=\operatorname{Hom}_R(N,R)$ is exact. This immediately implies $\ell(N^*)=e\ell(N)$ for all finite length modules where $\ell$ stands for length ...


2

The same logic from this solution to a special case applies here, except that you disregard the comments about $F_2$ and $F_3$ and just settle for all the quotients by prime ideals being fields. The battle plan is, briefly: The intersection of all prime ideals is the zero ideal The quotient by any prime ideal is in fact a field. The ring embeds into a ...


1

Let $R$ be a integral domain and $S$ a closed multiplicative subset. Then one defines $$ S^{-1}R=\{(r,s):r\in R,s\in S\}/\sim $$ where $(r,s)\sim(r',s')$ iff $rs'=sr'$. This set happens to be a ring with the operations $$ (r,s)+(r',s'):=(rs'+r's,ss'),\hspace{.3cm}(r,s)(r',s'):=(rr',ss'). $$ As you can see, you get nothing but elements of $S$ in "the ...


1

Question 1 is of course the same as asking whether $f \sim 0 \Longleftrightarrow F \in \langle n,x_i^{\phi(n)+1}-x_i, 1 \leq i \leq m \rangle$ holds. For composite $n$, this is even false in one variable. Consider $n=15$ and $f=(x^3-x)(x^5-x)$. Clearly $f(a)=0 \mod 15$ for any integer $a$, but $$(x^3-x)(x^5-x)=x^8-x^6-x^4+x^2 \notin (15,x^{9}-x).$$ Too ...


1

The infinitely many $7$-dimensional algebras described in Poonen's paper are all quotients of the finite dimensional algebra $A=k[w,x,y,z]/\mathfrak{m}^3$ (where $\mathfrak{m}$ is the ideal $(w,x,y,z)$). So $A$ is a $15$-dimensional counterexample.


1

The zero ring, a ring containing only one element, necessarily has $0 = 1$, since this ring satisfies all the ring axioms (except $0 \neq 1$), but only has one element. The zero ring is a bit pathological, so rather than writing "Let $R$ be any ring other than the zero ring", we exclude the zero ring via the "$0 \neq 1$" condition. There is also a notion "...


1

If $0 = 1$, then for any $x$ in the ring: $$x = x.1 = x.0 = 0$$ and therefore there is only one element in the ring.


1

The identity element under addition is denoted by $0$. It has the property that $0*a = 0 = a*0$ for all $a\in R$. The identity element under multiplication is denoted $1$. It has the property that $1*a = a = a*1$ for all $a \in R$. If $0 = 1$, then for all $a \in R$, $$ 0 = 0*a = 1*a = a.$$ So $R = \{0\}$. We don't like to call this a field as it doesn'...


1

In $R:=\mathbb{F}_p[\![x]\!]$, $(1-x)^p=1-x^p$. Because $1-x$ is an invertible element of $R$ (with inverse $1+x+x^2+\ldots$), $$\frac{1-x^p}{(1-x)^p}=1$$ in $R$. Therefore, [...]. The rest is up to you.


1

The "trivial ring" example (with all products zero) has already been given, but you can also make one with nontrivial multiplication by taking $\Bbb Z_5$ with normal multiplication and $\Bbb Z_3$ with zero multiplication and form the product ring to get an example with nonzero multiplication. You can also switch up which one uses zero multiplication, of ...


1

Hint: $R[x]/I \cong R$. 1- $I$ is prime iff $R$ is domain. 2- $I$ is maximal iff $R$ is field. 3- use (2) with the fact: $R$ is a commutative integral ring, $R[X]$ is a principal ideal domain imply $R$ is a field. 4- By the link above, $R$ is a field.


1

If the ring is finite, then it is an artinian reduced ring and therefore splits as the product of artinian local rings. Every term of the product must be reduced and an artinian local reduced ring is a field, proving your claim.


1

The following fact is useful and you should remember it: If an ideal $I \unlhd R$ contains $1 \in R$, then $I=R$. The proof is trivial: $1 \in I$ implies, by strong closure of multiplication, that $1 \cdot x \in I$ for any $x \in R$. But $1 \cdot x =x$, so $x \in I$. So $R \subset I$, and hence $I=R$. Now let $a$ be a unit, and define $I=(a)$. To ...


1

Since the question is already answered here, I will give you another sketch, which uses the fact that any regular local domain is factorial. This might be an overkill, but we already have an elementary solution anyway. The assumptions precisely state that $A$ is a one-dimensional regular local ring. In particular we have $A$ is factorial (a more or less ...


1

Yes. Rings that satisfy the ascending chain condition on ideals, are called Noetherian. A (commutative unital) ring is Noetherian if and only if every ideal is finitely generated. Proof of "If $R$ satisfies the ascending chain condition then every ideal is finitely generated". Take an ideal $I\subset R$. Suppose $R$ has ascending chain condition but $I$ is ...


1

the ring $\Bbb{R}[X]$ is principal ring because $\Bbb{R}$ is a field. the ideal in this quotient are in the forme classes of $\langle P(X)\rangle =\langle P(X)\rangle +\langle X^3-1\rangle =\{S(X)P(X)+T(X)(X^3-1), S,T\in \Bbb{R}[X]\}$ and better we can represent this class by class of $\langle R(X)\rangle$ where $R(X)$ is the rest by euclide division of $...


1

Here's a different answer that is, in some way, very discoverable and scalable. Let $F$ be a field not of characteristic 2. Then in $F^n$ (the product of n copies of $F$), every element with entries only from $\{1,-1\}$ satisfies $a^2=1$, and would be a root of $X^2-1$. This is quite different from the integers mod 8 since that ring is indecomposable, and ...


1

Your "functions with finite support" approach is the standard approach to take to this, and is very likely what any author who considers the question trivial has in mind. The correct notion of "finite" to take when defining monomials is the usual one, as you have done (this is necessary for the polynomial ring you get to be the free commutative $k$-algebra ...



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