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1

For a commutative ring $A$ I note $P(A)$ the property "every flat $A$-module is projective". Facts (that I leave as exercises) : 1) $P(A)$ is true if and only if $P( A / \sqrt{(0)} )$ is true 2) If $P(A)$ is true, so is $P(A')$ for any subring $A'$ of $A$. 3) $P(A)$ is true for any local ring $A$ which implies 4) $P(A)$ is true for any semi-local ring ...


1

A element in an integral domain is irreductible if it is not invertible nor a product of non-invertibles. An element $p$ of a commutative ring $A$ (not necessarily an integral domain) is prime if $A / p A$ is an integral domain, that is, is non-zero (that is $p$ is not invertible) and has no zero-divisor, that, whenever $ab \in pA$ then $a$ or $b$ is in $p ...


6

By unique prime factorization of integers, the multiplicative group $\newcommand{\Q}{\mathbb{Q}}\Q^\times$ is isomorphic to the additive group of $\newcommand{\Z}{\mathbb{Z}}(\Z/2\Z) \times \bigoplus_{p \text{ prime}} \Z$. The isomorphism comes from the fact that every rational number can be written uniquely as $\pm \prod_{p \text{ prime}} p^{a_p}$ for some ...


2

Here's a simple example: let $\star$ be the operation defined by $ 2 \star 2 = 2 $ $ 2 \star p = p $ $ p \star p = 1 $ $ p \star q = 1 $ $ 2 \star -1 = -1 $ $ p \star -1 = 1$ $-1 \star -1 = 1$ and extended to all nonzero rational numbers by distributivity: e.g. $$\begin{align}6 \star 8 &= (2 \cdot 3) \star (2 \cdot 2 \cdot 2) \\&= (2 \star 2) ...


0

Take a bijection $\mathbf Q\to \mathbf Q^\times$ and transport the ring structure. You can even take a bijection from $\mathbf Z$. Any countably infinite ring works.


0

(1) In a Noetherian ring nilradical can be equal to Jacobson radical, even if it is not a Artin ring. $\mathbb{Z}$ is an example. In fact any Jacobson ring has this property. (2) is not true, because in this case the nilradical is the zero ideal, but Jacobson radical is a non-zero ideal (since it is not a field). Take any Noetherian local domain which is a ...


3

Given a ring $R$ that's Noetherian as a $\Bbb Z$-module, it's automatically a (left- or right- as desired) Noetherian ring, as ideals are by definition closed under sum of elements; so every increasing chain of ideals is an increasing chain of $\Bbb Z$-submodules of $R$, and thus stabilizes. $\Bbb Q$ is indeed a Noetherian ring (it only has two ideals) ...


1

Every field is a Noetherian ring, since the only ideals of a field are $(0)$ and the field itself.


1

Another example of how you can engineer a single ring for a given $n$: let $T_n(F)$ be the ring of $n\times n$ upper triangular matrices over a field, and let $I$ be the ideal of strictly upper triangular matrices. This ideal has the property you seek. Then you can look at $\prod_{n=1}^\infty T_n(F)$ to find a ring with an ideal for each $n$ with this ...


0

If I'm reading this right, you're looking for an MCS which every prime ideal intersects the MCS nontrivially. There is a very famous lemma that shows the answer is negative. Lemma: If $M$ is a multiplicatively closed (not containing $0$) subset of a commutative ring $R$ (with identity), then there is an ideal of $R$ maximal with respect to being ...


0

Let $R$ be a commutative ring with unity $1 \neq 0$. Then every proper ideal is contained in a maximal ideal. In particular, starting with the zero ideal we get that the ring $R$ has at least one prime ideal (may be the ideal (0) itself). Now in this case, $A$ has identity $1 \neq 0$ (we can assume it because if $1 = 0$ then the ring is the zero ring and in ...


1

The implication you're looking for is easy: if $a+b\sqrt{-k}=3(x+y\sqrt{-k})$, then $$ a=3x,\qquad b=3y $$ by equating the real and imaginary parts. More interesting is asking whether $3$ is irreducible. Suppose $$ (a+b\sqrt{-k})(c+d\sqrt{-k})=3 $$ with both elements in the left hand sides noninvertible. Then, by taking conjugates, $$ (a^2+b^2k)(c^2+d^2k)=9 ...


1

Since $M\in P$ hence $I\subset M$


2

Because $M$ is a maximal element of $P$, it is in particular an element of $P$, or in symbols, $M\in P$. By definition, $P$ is the collection of ideals that contain $I$. Therefore, $M$ contains $I$.


3

Look at $k[x]/\langle x^n \rangle$, and consider the principal ideal generated by $x+\langle x^n \rangle$.


1

$S^{-1}B$ and $f(S)^{-1} B$ are in fact isomorphic, simply because the action of $A$ on $B$ is defined via $f: A \rightarrow B$. To see this explicitly, note that when you gave the definition of $S^{-1}B$, you defined $(b,s)$ ~ $(b',s')$ if there exists $t \in S$ such that $t (s b' - s' b)=0$. But what does $s b'$ mean? What is the action of $s$ on $B$? This ...


2

For a counterexample take the first Weyl algebra $\mathbf C[p,q]$ where $pq-qp=1$, which is a subalgebra of ${\rm End}_{\mathbf C}\mathbf C[X]$. Here $p$ acts on polynomials by $p f(X)=f'(X)$ and $q$ acts by $qf(X)=Xf(X)$. It should be checked $q,p$ are not invertible, yet $pq-qp=1$ means $(p)=(q)=1$. The point is that in noncommutative cases, even principal ...


1

For sure $$\mathbb{Z}[a] \cong \mathbb{Z}[x] / \langle x^3 + x^2 - 2x + 8\rangle \cong_{\text{as } \mathbb{Z} \text{ module}} \langle 1, \alpha, \alpha^2 \rangle$$ If my calculations are correct $(\frac{\alpha + \alpha^2}{2})^2 = \frac{\alpha + \alpha^2}{2} - 2\alpha - 2$. Obviously $\langle 1, \alpha, \alpha^2 \rangle \subseteq \langle 1, \alpha, ...


5

If it were a PID, then every nonzero prime ideal would be maximal. But $\mathbb Z[X]/p \mathbb Z[X] = \mathbb (Z/p\mathbb Z)[X]$ is an integral domain which is not a field.


0

Yes. If $\beta = \frac{\alpha + \alpha^2}{2}$, the key fact that you need to verify is that $\alpha^2$, $\alpha \beta$, and $\beta^2$ can all be written in the form $a+b\alpha + c\beta$.


7

Let $R$ be a ring with unity, take $r\in R \setminus \{0\}$, and suppose that $(r) = R$ as ideals. In the commutative case, this means that $r$ is invertible. Why? Because every element of $(r)$ is of the form $ar$, so we have $ar=1$ for some $a\in R$. But in the non-commutative case, if we are looking at two-sided ideals, the elements of $(r)$ include ...


0

The evaluation map $e_a:p\mapsto p(a)$, where $p\in\mathbb{C}[x]$ and $a\in\mathbb{C}$ is a homomorphism $e_a\colon \mathbb{C}[x]\to\mathbb{C}$. By the properties of products, also $$ f=e_0\times e_1\times e_2\colon\mathbb{C}[x]\to \mathbb{C}\times\mathbb{C}\times\mathbb{C} $$ defined by $$ f(p)=(p(0),p(1),p(2)) $$ is a (ring) homomorphism. You can also ...


4

This follows from $(1,0)(0,1)=(0,0)$, i.e. $R\times R$ has zero divisors (in an integral domain $0\neq 1$). You only mention a special case, which is also more immediate from the one line argument I gave.


0

$(ii)$ is the UFD analog of the monic case of the Rational Root Test. The classical proof in $\,\Bbb Z\,$ immediately extends to any UFD (or GCD domain). $(i)$ can be done slickly using evaluation hom's. Below is a sketch a proof of a more general result using only high-school algebra. We show $\rm\:(2,x)\ =\ (f)\ $ in $\rm\:\mathbb Z[x]\: $ yields a ...


0

Another hint: consider $\phi(f(x,y))=f(y^{2},y)$ with kernel $(x-y^2)$


0

In general, to show that a $R/I$ is isomorphic to another ring $S$, it is easier to produce a surjective homomorphism $R \to S$ that has kernel $I$ rather than an explicit isomorphism $R/I \to S$. In your case, consider the homomorphism $$R[x,y]\to R[y]\\x\mapsto y^2\\y\mapsto y$$ Is it surjective? What is its kernel? Note that $R[x,y] = (R[y])[x]$, so ...


4

Hint. In general, $A[T]/(T-a)\simeq A$ for $a\in A$. (In your case consider $A=R[y]$ and $a=y^2$.)


0

This is a special case of the Chinese Remainder Theorem: If $R$ is a commutative ring with $I, J$ coprime ideals (i.e. $I+J = R$), then $IJ = I \cap J$ and $$\frac{R}{IJ}\cong \frac RI\times \frac RJ$$ via the homomorphism induced by $$R \to R/I \times R/J\\x\mapsto (x+I, x+J)$$ Note that $\mathbb C[X]/(x + \alpha)\cong \mathbb C$ for any $\alpha \in ...


4

HINT: Consider the map $$ \Bbb C[X]\longrightarrow\Bbb C\times\Bbb C\times\Bbb C $$ given by $q(X)\mapsto(q(0),q(1),q(2))$. Convince yourself that is a homomorphism, that is surjective, and compute the kernel.


1

Observe that $$\overline{x^2 + x + 1} = \overline 0$$ since $x^2 + x + 1 \in (x^2 + x + 1)$, which means that $$\overline{x^2} =\overline{ - (x + 1)},$$ but since $-1 \equiv 1 \pmod 2,$ we get $$\overline{x^2} = \overline{x + 1}.$$


1

This is a community wiki solution designed to remove this item from the unanswered queue. Edit: The question has received an answer at MO, which looks correct to me -- Todd Trimble.


0

Hint: $2\mid 4=(1+\sqrt{-3})(1-\sqrt{-3})$ So, using your method, can you find two non-zero elements in the quotient that multiply to make $0$? Your solution fails as whilst it's true that $4\times 2=0$ in the quotient, both $4$ and $2$ are zero in the quotient!


1

$$\mathbb{Z}[\sqrt{-3}]/(2)\simeq\mathbb Z[X]/(X^2+3,2)$$ But $(X^2+3,2)=(X^2+1,2)$, so $$\mathbb{Z}[\sqrt{-3}]/(2)\simeq\mathbb Z[X]/(X^2+1,2)\simeq(\mathbb Z/2\mathbb Z)[X]/(X^2+1)=(\mathbb Z/2\mathbb Z)[X]/(X+1)^2$$ which is not an integral domain, so $2$ is not prime. (If you don't want to identify the quotient $\mathbb{Z}[\sqrt{-3}]/(2)$, only to show ...


0

Hint $\rm\ \ d\mid p,q\,\Rightarrow\, d'\mid p',q'\Rightarrow\, dd'\mid pp',qq'\Rightarrow\, dd'\mid(pp',qq')$ Remark $\ $ It works in any UFD/GCD domain, for $\, x\mapsto x'$ generalized from conjugation to any multiplicative map, which necessarily preserves divisibility, enabling the first arrow above.


0

You're plugging in wrong: $n=2$ is not the case $x^2 = 1$, but rather the case $x^1 = 1$. But the special case isn't part of the proof, just an illustration. In general, the point is that if some power—any power—of $x$ equals $1$, then $x$ is invertible.


1

Probably what is throwing you off is that $x\in R/P$ and not $x\in R$. Of course, $x^2=x$ in $R$ without $x$ being $1$. But in the quotient, there are no zero divisors (because you're quotienting by a prime ideal.) So you must examine $x^2-x=0$.


0

Because as $n=2$ we have that $x^2 = x$ then $x(x-1) = 0$ and $x$ is not zero. In a integral Domain what happens to $x$?


2

By definition, we have $\gcd(p,q)\mid p$ and $\gcd(p,q)\mid q$. For any $r,s\in\mathbb{Z}[i]$, if $r\mid s$ then by definition $s=kr$ for some $k\in\mathbb{Z}[i]$, hence $\mathrm{N}(s)=\mathrm{N}(k)\mathrm{N}(r)$, so we must have $\mathrm{N}(r)\mid \mathrm{N}(s)$. Therefore, $\mathrm{N}(\gcd(p,q))\mid \mathrm{N}(p)$ and $\mathrm{N}(\gcd(p,q))\mid ...


2

a) $4=\frac{2}{1}\cdot\frac21$ and similarly for $10$. On the other side, $5$ is irreducible, $6=\frac21\cdot\frac31$ and $\frac31$ is invertible in $P$, so $6$ is also irreducible, and similarly for $15$. It remains $9=3^2$ which is invertible in $P$. b) $P$ is a UFD as a ring of fractions of a UFD; see here. We also have that $P[x]$ is a UFD; see here.


1

As Derek Holt hinted, you should consider the polynomial $h=f-g$. By hypothesis $h\ne 0$. Then consider the associated polynomial function of $h$, denoted here by $h_K$. Suppose the contrary that $h_K=0$, that is, $h_K(a)=0$ for all $a\in K$. This means that all elements of $K$ are roots of $h$. But over a field a polynomial can't have more (distinct) roots ...


1

Hints (depending on what you mean, and see for example here) (a) $\mathbb{Z}[i]/(i-5) \cong \mathbb{Z}[x]/(x^2+1, x-5) \cong \mathbb{Z}/26\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/13\mathbb{Z}$ (b) $\mathbb{Z}[i]/(3,i+5) \cong \mathbb{Z}[x]/(x^2+1, x+5,3) \cong \mathbb{Z}/(26,3)\cong 0$ (a) $\mathbb{Z}[i][x]/(x-5) \cong \mathbb{Z}[i] $ (b) ...


1

Regarding 1) Usually the subring test involves closure of multiplication and subtraction (but this isn't too different from addition here). Also $a_h+b_h$ is an element of the ring $R$, not of the group $H$, so your notation is a bit confusing. The point is that a general element of $R[H]$ is a formal sum of elements of $H$ with arbitrary coefficients in ...


1

$a^{-1} \in I \Rightarrow f(a^{-1})=f(a)^{-1} \in f(I)$ EDIT:(as Simon suggested) There is no need for inverse to show ideal


3

An easy example would be the ring $$\mathbb{F}_2[x,y]/(x^2,xy,y^2)=\mathbb{F}_2[\tilde{x},\tilde{y}]=\{0,\,1,\,\tilde{x},\,\tilde{y},\,1+\tilde{x},\,1+\tilde{y},\,\tilde{x}+\tilde{y},\,1+\tilde{x}+\tilde{y}\}$$ where the ideal $$I=(\tilde{x},\tilde{y})=\{0,\tilde{x},\tilde{y},\tilde{x}+\tilde{y}\}$$ cannot be generated by a single element.


1

They both have dimension 3 as $\mathbb{Q}$ vector spaces, so a surjective homomorphism is also injective. In fact, the first has basis $1,x,x^2$, while the second $(1,0),(0,1),(0,i)$.


1

Of course, for a fixed $g \in G,$ $(\Phi_1 \cdot\Phi_2)(g)$ is well defined. We have to show that $(\Phi_1 \cdot\Phi_2)(g) \neq 0$ for only finitely many $g \in G.$ To this end, put $$ M = \{h \in G | \Phi_1(h) \neq 0\} $$ and $$ N = \{h \in G | \Phi_2(h) \neq 0 \}. $$ Note that $M$ and $N$ are finite, by assumption. Also note that for $g,h \in G,$ we have ...


1

I claim that the set of automorphisms of $\mathbb{Z}[x]$ are the ring homomorphisms $\phi$ that satisfy $\phi(x) = \phi_0+\phi_1 x$ where $\phi_0 \in \mathbb{Z}$ is arbitrary and $\phi_1 \in \{\pm 1\}$. Suppose $\phi$ is an automorphism. As above, we have $\phi(n) = n$ for $n \in \mathbb{Z}$. Since $\phi$ is an automorphism, we can find an inverse for $x$. ...


1

If $f(x)=a_0+a_1x+\cdots+a_nx^n$, $a_n\ne0$, $n\ge1$, then $\phi(f(x))=a_0+a_1\phi(x)+\cdots+a_n\phi(x)^n$ (note that $a_i\in\mathbb Z$ and then $\phi(a_i)=a_i$) which has the degree equal to $\deg\phi(x)^n=dn\ge d$. Moreover, if $d\ge2$ then $\deg\phi(f(x))=dn\ge2n\ge 2$, so the image of $\phi$ consists of constants and of polynomials of degree $\ge 2$. ...


1

Since Najib Idrissi has given a direct method, I'll add an answer based on the Chinese remainder theorem. In fact, this is an immediate application of the theorem. We just need to show that the ideals $(x - 1)$ and $(x^2 + 4)$ are coprime. By the polynomial division algorithm, $$ x^2 + 4 = (x + 1)(x - 1) + 5. $$ Since $5$ is a unit in $\mathbb Q$, we are ...


1

It's clear that $\ker \varphi$ contains $(x-1)(x^2+4)$, so it contains the ideal generated by $(x-1)(x^2+4)$. Conversely let $P \in \ker \varphi$. Then since $\mathbb{Q}[x]$ is a Euclidean domain, there exist unique polynomials $Q,R \in \mathbb{Q}[x]$ such that $$P = (x-1) (x^2+4) Q + R,$$ with $\deg R < 3$. Now $P(1) = 0 = R(1)$ and $P(\pm 2i) = 0 = ...



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