# Tag Info

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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 ...

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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.

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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 $a+b\alpha + c\beta$.

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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 ...

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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.

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$(ii)$ 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 ...

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

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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 ...

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

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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 ...

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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.

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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 ...

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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) ...

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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 ...

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

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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.

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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 $(1,0),(0,1),(0,i)$.

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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 ...

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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 $x$. ...

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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 $\ge 2$. ...

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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 ...

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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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