New answers tagged

0

The trick is to consider the conjugation-action over $G\times\wp(G)\to\wp(G)$.


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 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

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.


0

Partial answer from another angle: From the viewpoint of ring theory, the following is (obviously) true. Proposition. Let $R$ denote a ring. Suppose $f,g$ and $\epsilon$ are elements of $R$. If $\epsilon$ is quasiregular, then $$f = g + \epsilon f \;\rightarrow\; f = \frac{1}{1 - \epsilon} g.$$ If we can find a name for this proposition,...


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 ...


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 ...


0

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.


0

As discussed in the comments, this almost always fails. In particular, it fails in your example (as long as $V$ is nonzero): if $\alpha:k\to k$ is any automorphism, then you can define a ring-homomorphism $\varphi:k[x,y,z]\to \textrm{End}_k(V)$ by $\varphi(f(x,y,z))=\alpha(f(0,0,0))$ (where $\alpha(f(0,0,0))\in k$ acts on $V$ by scalar multiplication). ...


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

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 ...


0

Consider the short exact sequence $0\xrightarrow{}M_1\xrightarrow{}M_1\oplus M_2\xrightarrow{\pi}(M_1\oplus M_2)/M_1\xrightarrow{}0$, then $M_1$ and $(M_1\oplus M_2)/M_1$ are both Artinian. Let $N_1\supset N_2\supset\cdots$ be the descending chain, let $N_i' = N_i\cap M_1$, let $N_i'' = (N_i+M_2)/M_2$, then it is easy to see that $\{N_i'\}_{i=1}^\infty$ and ...


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 ...


0

That is a simple form of CRT = Chinese Remainder Theorem that I call Easy CRT. Below is one simple way to present it. You can find many example applications in prior posts here. Theorem $ $ (Easy CRT) $\rm\ \ $ If $\rm\ m,\:n\:$ are coprime integers then $$\rm \begin{eqnarray}\rm x&\equiv&\rm a\,\ (mod\ m) \\ \rm x&\equiv&\rm b\,\ (mod\ ...


0

The answer in the linked question still applies. Consider the matrix \begin{align*} y &= \begin{pmatrix} 0 & \cdots & 0 & 1 \\ 0 & \cdots & 0 & 0\\ \vdots & \ddots & 0 & 0\\ 0 & \cdots & 0 & 0 \end{pmatrix} \in M_n(\mathbb{R}). \end{align*} If $y = x^n$ for some $x\in M_n(\mathbb{R})$, then all the ...


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.


-2

Suppose $ I = (7, 3 + \sqrt{19}) = (\alpha) $. Then there exist $ r,s $ in the ring such that $ r \alpha = 7 $ and $ s \alpha = 3 + \sqrt{19} $ and taking norms gives $ N(\alpha)\mid 49 $ and $ N(\alpha) \mid -10 $, so $ |N(\alpha)| = 1 $, meaning $ \alpha $ is a unit, a contradiction. Then, a slightly more advanced approach would be to use the fact that a ...


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 ...


0

It's a local ring whose unique maximal ideal is the class of infinitesimal elements. In fact, I believe for any two ideals, you have either $I \subseteq J$ or $J \subseteq I$. (thus, you have a GCD domain) All of the non-infinitesimal elements are units, so the only nontrivial aspect of 'factorization' is how infinitesimal an element is. e.g. $\epsilon^2$ ...


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 ...


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 $\,D = \Bbb Z$.


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.


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

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 $D$.


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.


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 $...


0

For i), it means $P$ is divisible by $X^3-1$. In other words: $P(1)=0\;$ and $\;P(\mathrm e^{\tfrac{\mathrm i\pi}3})=0$. ii) Hint: use the Chinese remainder theorem. iii) Hint: think of the factorisation of $X^3-1$ into irreducible factors and use i).


0

The ring has to contain 0, 1, and the two zero-divisors, which we can call $x$ and $y$. Think about what $x^2$, $y^2$, and $x y$ could be. Then suppose it contains some other element $r$ which is not a zero-divisor, and think about what $r x$ and $r y$ could be.


0

Setup Following page 32 of Vakil's notes,let S be a multiplicative subring of a ring A; i.e., $1 ∈ S ∧ x,y ∈ S ⇒ x · y ∈ S$. Then we consider “formal fractions”, S⁻¹A ≔ { a / s ∣ a ∈ A , s ∈ S } The property we're interested in is 𝒫 : A-algebra → Bool 𝒫 f ≔ for every e in S, f e ∈ B is invertible Want to show: S⁻¹A is initial among A-algebras B ...


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 ...


1

Here is an option which takes place between your two suggestions, without appealing to choice: Direct limit of finitely generated subrings. We define $k[X]$ to be the direct limit of the system $\{ k[Y]\mid Y\in[X]^{<\omega}\}$ with inclusions as the embeddings. Now to check that the operations are well-defined we only need to check they restrict ...


0

Using the definition of "irreducible" you wrote in the question, the answer is no, as illustrated in Daniel's answer. However, the generally accepted definition of "irreducible" by those who do research on factorization in commutative rings with zero divisors is: $a$ is irreducible if $a = bc \Rightarrow (a) = (b)$ or $(a) = (c)$. Using this definition, ...


0

How about another proof? If $M$ is a free module over a commutative ring with identity $A$, then any two bases of $M$ have the same cardinality. This is a general fact about commutative rings ("invariant basis number"). That cardinality is called the rank of $M$. If $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ is an exact sequence of ...


5

Let $R$ be a ring and $r\in R$ be a nilpotent element i.e $r^n=0$. Then: $$(1-r)(r^{n-1}+r^{n-2}+...+r+1)=1-r^n=1$$ Hence $1-r$ is invertible. Thus also $1-(-r)=1+r$ is invertible, just because $-r$ is nilpotent. Your idea is not bad, but need slight modification. Pick $r\in R$ a nilpotent element. Then $rx\in R[x]$ is a nilpotent element. Hence: $$f(x)=1+...


1

Assume there are polynomials $f$ with $\phi(\overline f)=0$ and $f\notin (y-x^2,z-x^3)$. Among these. let $f_0$ have minimal degree in $z$. If $f_0$ has positive degree $k$ in $z$, say $f(x,y,z)=p_0(x,y)+p_1(x,y)z+\ldots +p_k(x,y)z^k$, then $f_0(x,y,z)-p(x,y)z^{k-1}x^3$ is equivalent to $f_0$ but of lower degree in $z$. We conclude that $f_0$ has degree $0$ ...


0

Those two rings are not isomorphic (the first is an integral domain but no field, the second one is a field), but indeed there is a connection between them, if you want to show that $(xy^2-1)$ is prime. $xy^2-1 \in (K[x])[y]$ is primitive, hence it is irreducible iff it is irreducible in $(K(x))[y]$ by the Gauß lemma. In that latter polynomial ring, we can ...


1

Since $1 \in A_0$, every $\delta_n$ with $n \ge 1$ must be zero. The highest degree $\delta_i$ will be the product of the highest degree $a_i$ and the highest degree $b_i$, and since $A$ is a domain this is non-zero. Therefore since the degrees add the highest $a_i$ and $b_i$ must both be in degree $0$, i.e. $a$ and $a^{-1}$ are homogeneous degree $0$.


1

Your $f(x)$ is not a polynomial, which must have a fixed, but arbitrary degree $n\ge 0$. It is a theorem then, that every polynomial of degree $n$ over a field $F$ has at most $n$ zeros. This is no longer true over, say, a skew-field. Consider the polynomial $x^2+1$ over the quaternion algebra, as an example - see here. It has infinitely many roots.


1

The degree must be finite, otherwise $\sum \limits _{n = 0} ^\infty \frac {x^n} {n!}$ would be a polynomial - which clearly it isn't, being $\Bbb e^x$. In fact, a polynomial is just a function from $\Bbb N$ to $F$ with finite support.


1

$$f(a)=0\implies \forall x\in A, ax=0\implies a=0.$$ Therefore, $\ker(f)=\{0\}$, and thus the claim follow.


15

Here's an example. Let $\zeta=e^{2\pi i/7}$ and let $p(x)=(\zeta-x)(\zeta^2-x)(\zeta^4-x)$. Since $\zeta$ is a root of $p$ but $\bar{\zeta}$ is not, $p$ does not have real coefficients. Now notice that $(\zeta^4)^2=\zeta$, so the squaring map permutes the roots of $p(x)$. We can use this to compute \begin{align*} p(x^2)&=(\zeta-x^2)(\zeta^2-x^2)(\...


4

"trivial in $D$" is bad use of the word. What you want to determine are the $D$-algebra automorphisms (i.e. ring automorphisms that are also $D$-linear). Such an automorphism is determined by the image of $X$. What you want to show first is that $\phi(X)$ is a linear polynomial. Consider degrees! $X$ must lie in the image!


0

The proof of the first statement: Let $q_i$ an isolated prime ideal. Then {$p_i$} is a isolated set because $p_i$ is minimal in the set of the associated primes. If you take S=A-$p_i$ then $q_i=S(a)$, where $S(a)=a^{ec}$. So, $q_i$ is uniquely determined by $a$.


3

It is true if the ring does not have divisor of zero. Suppose that $m<n$, write $n=pm+r$, you have $a^n=a^{pm+r}=b^{pm+r}=b^{pm}b^r=a^{pm}b^r$. You deduce that $a^{pm}a^r=a^{pm}b^r$ and $a^{pm}(a^r-b^r)$ and $a^r=b^r$. Thus the assertion is true for $(m,r)$ you can repeat this process until the rest is 1.


3

For part c, you proved $A$ was a domain, but the task at hand was to prove $A/I$ is a domain. Clearly, $A/I$ is not a domain because: $$(2,2)\cdot (2,3) \equiv (4,6) \equiv 0 \pmod{(4,6)}$$ For part d, your solution is correct, but we could also use our solution for part c here by showing that $(2,2)$ is a zero divisor, meaning that it can't be invertible.



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