# Tag Info

8

Modulo $7$ you get the polynomial $x^7 - x - 1$. This is an Artin-Schreier polynomial, so it's irreducible. There are some proofs of irreducibility here.

6

The question is very difficult. We give a brief description of what is known. If $n$ is not a prime, then we cannot "divide by $2$" $n-1$ times. In more conventional language, if $n$ is not prime, then the order of $2$ modulo $n$ is not equal to $n-1$/ If $n$ is prime, it is possible for the order of $2$ to be $n-1$, that is, for $2$ to be a generator of ...

4

The $p$-adic completion of $\mathbf Z[[X]]$ is $\mathbf Z_p[[X]]$. To prove it, it suffices to prove that the map $\mathbf Z[[X]] \to\mathbf Z_p[[X]]$ induces an isomorphism on the $p$-adic completions. This is obvious because $$\mathbf Z[[X]]/p^n\mathbf Z[[X]] \to\mathbf Z_p[[X]]/p^n\mathbf Z_p[[X]]$$ is an isomorphism, both sides identifying with $(\mathbf ... 4 http://am-solutions.wikispaces.com/Solutions+to+Chapter+1 "Let$A$be a ring,$R$its nilradical. Show that the following are equivalent: 1)$A$has exactly one prime ideal; 2) every element of$A$is either a unit or nilpotent; 3)$A/R$is a field. Proof. 1) ⇒ 2). Observe that$R$, which is the intersection of the prime ideals, is equal to the given ... 3$F[x]/(x^3)$consists of units and nilpotent elements, but has four ideals, so this suggests you meant something more like unique prime ideal. This is indeed true for commutative rings. The hypothesis that nonunits are nilpotent means that the nilradical is a maximal ideal. But considering that all prime ideals contain the nilradical, the nilradical is ... 3 The completion of$\mathbb{Z}[X]$is isomorphic to the ring of formal power series with coefficients in$\mathbb{Z}_p$that tend to 0$p$-adically. Indeed, it is, by definition, the ring of equivalence classes of Cauchy sequences$(f_k)$of polynomials$f_k\in \mathbb{Z}[X]$. To be a Cauchy sequence means that eventually, the polynomials become congruent ... 3 As Jyrki Lahtonen points out in the comments, one way to determine whether or not a particular polynomial$p(x)$is a unit in the quotient is to compute the greatest common divisor of$p(x)$and$x^k+1$. If it's$1$, then it's a unit. Otherwise it will be something other than$1$, and will not be a unit. That can be computationally intense though, so I'm ... 2 Actually one can show that every finite ring satisfying this identity is a product of copies of$\mathbb{F}_2$and$\mathbb{F}_4$. As you have said, it is commutative, and it is also reduced and of dimension$0$. Hence, it is a product of finite fields. But finite field$F$has this property iff$|F^*|$divides$4-1=3$, i.e. iff$F=\mathbb{F}_2$or ... 2 The norm is$N(a+bi)=a^2+b^2$, and a proof is in many books on number theory. I recommend Ireland and Rosen, "A classical Introduction to Modern Number Theory, Proposition$1.4.1$, or http://homepage.univie.ac.at/dietrich.burde/papers/burde_37_comm_alg.pdf, Proposition$1.1.12$for$\mathbb{Z}[\sqrt{-2}]$,$\mathbb{Z}[i]$,$\mathbb{Z}[\sqrt{2}]$, and ... 2 There's a reason why you can't find one! If by "ring" you mean a ring with unity, then every non-zero element of a finite ring$R$is either a unit or a zero divisor. See this link for a proof. 2 When you adjoin an element$\alpha$to a ring$R$subject to the relations$f_1(\alpha)=\cdots=f_n(\alpha)=0$, this means you are forming the ring $$\frac{R[x]}{(f_1(x),\ldots,f_n(x))}$$ so in a) you are forming $$\frac{\mathbb Z[x]}{(2x-6,6x-15)}\cong \frac{\mathbb Z[x]}{(x,3)}\cong \frac{\mathbb Z}{(3)}=\mathbb Z_3$$ 2 nik's example of$\mathbb H[X]$, where$\mathbb H$is the quaternions is a good one. The set of invertible matrices is not a ring at all (it is not closed under addition). Another example is the Weyl algebra $$W_1 = k\langle x,y\rangle / (xy - y x=1)$$ Of course, in general if$D$is an arbitrary division ring, then$D[X]$is a domain but not a ... 2 Wow, three answers and a comment with the same example, and not even the easiest example, IMO! (It's still a good example, though!) Similarly, is there an example within the quaternions? Absolutely! There are lots of nonzero subrings of$\Bbb H$that would already work. Consider for example$\{a+bi+cj+dk\mid a,b,c,d\in \Bbb Z\}\subseteq\Bbb H$. It's ... 2 The answer is that such matrices are always similar over$\def\Z{\Bbb Z}\Z$(conjugate in$GL_2(\Bbb Z)$). The question is deeper however than it might look at first, and as far as I can see any solution requires some somewhat subtle arithmetic considerations. A few things are easy:$A,B$always have determinant$~1$(from the constant coefficients of the ... 2 You can also prove this by passing to$\mathbb{Z}[i]$as follows : If$p\mid k^2+1$, then$p$is composite in$\mathbb{Z}[i]$: If$p$were irreducible in$\mathbb{Z}[i]$, then$p\mid (k+i)(k-i)$, whence$p\mid (k-i)$or$p\mid (k+i)$in$\mathbb{Z}[i]$, and so$p\mid \pm 1$in$\mathbb{Z}$, which is absurd. Hence,$p$is composite in$\mathbb{Z}[i]$... 1 some math terminology is quaint and is often a survival from a time when conceptual clarity had not been fully attained in some area of study. in particular, in the 18th and 19th centuries the concept of a quotient ring had not been formulated. all the more credit to those whose intuition was strong enough to guide them to remarkable discoveries. for ... 1 Notice that every element of prime field of$k$, lets call it$P$, is of the form$1\frac nm$because its the smallest subset of$k$containing$1$and$0$closed on field operations. We know that$f(1)=1$and$f(1\frac nm )=\frac nm f(1)$. This means that image of$f(P)$is the prime field of$L$. 1 Yes, it is true (assuming$X$is not irreducible, of course). Let$\phi\in\mathbb{k}[X]$that is zero on some irreducible component$X_i$of$X$and let$Y_1,\ldots, Y_r$be the other irreducible components of$X$. Since$Y:=Y_1\cup\cdots\cup Y_r$is closed, there is a regular function$\psi$such that$\psi|_Y=0$and$\psi(x)\neq 0$for some$x\in ...

1

It would be a good choice to go for Abstract algebra text by Dummit Foote for the part which you have asked for . Simultaneously Keep doing problems (*in the same order as it is given) http://www.math.kent.edu/~white/qual/list/ring.pdf Do not take it much seriously but it would be a good idea if you can try looking at "Algebra : Michael Artin" mainly for ...

1

The standard equivalence relation $R$ defined on $\mathbb Q$ is $$[\frac {a}{b}]=[\frac {c}{d}]$$ iff $ad=bc$. Try showing that if $[\frac {a}{b}]=[\frac{c}{d}]$ , then $[f (\frac {a}{b})]=[f (\frac {c}{d})]$. So, for example, for 1): Assume [$\frac {a}{b}$]=$[\frac {c}{d}]$. Then $f(\frac {a}{b})=\frac {a+b}{b}$ . Is it equivalent to ...

1

The zero divisors of this ring are the elements that are zero in at least one place. The product of two elements always has at least as many zeros as the factors. If the product of two elements has a zero then, a product of three elements certainly will have more. Your example of $(1,1,0),(0,1,1),(1,0,1)$ will work provided you define away $0$ from being a ...

1

By quotienting $\mathbb{Z}[x]$ by the ideal generated by $x+1$, you are effectively dictating that $x+1=0$, or that $x=-1$. Therefore, $\mathbb{Z}[x]/(x+1) \cong \mathbb{Z}[-1] \cong \mathbb{Z}$. An isomorphism $\phi$is given by $\phi([f]) = f(-1)$, where $[f]$ is the coset of the polynomial $f$, or, more simply, evaluation of a polynomial at $-1$. This map ...

1

Let $a=\alpha_1+\alpha_2 i, b=\beta_1+\beta_2i$ where $\alpha_1,\alpha_2,\beta_1,\beta_2\in\Bbb Z$. Then $$\frac ab=\frac{\alpha_1+\alpha_2i}{\beta_1+\beta_2i}=\frac{(\alpha_1+\alpha_2i)(\beta_1-\beta_2i)}{N(b)}=\frac{(\alpha_1\beta_1+\alpha_2\beta_2)-(\alpha_1\beta_2-\alpha_2\beta_1)i}{N(b)}$$ By a modified form of the division algorithm on the integers, ...

1

So you're saying let's define $\Bbb{Q} \xrightarrow{\phi} \Bbb{Z}_p$, by taking the reduced $a/b \in \Bbb{Q}$, and mapping it to $a b^{-1} \pmod{p}$. You want to show that $\phi$ is onto. That's easily done by letting $b$ be fixed at $1$ and letting $a$ vary over $\Bbb{Z}$. These are all reduced and so they map as $x \mapsto x \pmod{p}$ which is indeed a ...

1

For $\alpha:=\sqrt5-2\sqrt3$, we can state the following: $$(\alpha+2\sqrt3)^2=5\,.$$ So, $\alpha^2+4\sqrt3\alpha+4\cdot 3=5$, that is, $$\alpha^2+7=-4\sqrt3\alpha$$ Now square it again to obtain a rational equation that $\alpha$ satisfies. Try similar method for $\beta:=\sqrt{\sqrt[3]2-i}$.

1

You have an example of something called a local ring, this is a ring with a unique maximal ideal. You are taking the ring $\mathbb{Z}$ and you are localizing at $(p)$. (In short, localization a ring $R$ at a prime ideal $I$ is just inverting everything that is outside of $I$. The fact that $I$ is prime tells us that $R\backslash I$ is multiplicatively ...

1

If an element $\frac{a}{b}\in R$, with $p\nmid b$, is invertible, then there exists $\frac{c}{d}\in R$, with $p\nmid d$, such that $\frac{a}{b}\frac{c}{d}=1$; this means $ac=bd$. Since $p\nmid bd$, it follows that $p\nmid a$. Conversely, if $p\nmid a$ and $p\nmid b$, then $\frac{b}{a}\in R$. Now we know the noninvertible elements: they are of the form ...

1

Hint for $\Leftarrow$: Every ring with identiy has a prime ideal. Let $P$ be a prime ideal of $R$. Then it contains all the nilpotent elements (why?). It does not contain a unit (why?), so it is in fact the set of nilpotent elements of $R$ and hence because of the condition the unique prime ideal.

1

The ring of integers, Dedekind domains, and a host of other rings from number theory and ideal theory that are Noetherian all evidence that they simply show up "in nature" a lot. Algebraic geometers care a great deal about polynomial rings $\Bbb F[x_1,\dots x_n]$ over fields, which are Noetherian. In a nice case like $\Bbb F=\Bbb C$, the resulting ...

1

The answer to this question (which I have also asked myself at various times) might also be a "No". Maybe Noetherian rings aren't that important after all, and their perceived ubiquity is a meme in commutative-algebraic literature? Let me quote a paragraph from the Foreword of the (so far unfinished) translation of Commutative algebra -- Constructive ...

Only top voted, non community-wiki answers of a minimum length are eligible