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15

Many authors take the existence of $1$ as part of the definition of a ring. In fact, I would disagree with Alessandro's comment and claim that most authors take the existence of $1$ to be part of the definition of a ring. There is another object, often called a rng (pronounced "rung"), which is defined by taking all the axioms that define a ring except you ...

10

It makes a difference if you are considering irreducibility in $\mathbb{Z}[x]$ or in $\mathbb{Q}[x]$. A counterexample for the first case: $f(x) = x$ is irreducible in $\mathbb{Z}[x]$, but $f(2x) = 2\cdot x$ is not. In the second case your statement is true for all $n\neq 0$, since then $n$ is a unit in $\mathbb{Q}[X]$, so $f(x)\mapsto f(nx)$ is an ...

7

$\mathbb{C}[x,y] / \langle xy \rangle$ is not isomorphic to $\mathbb{C}[x] \oplus \mathbb{C}[y]$ because the first ring has no idempotents other than $0,1$, while the second does. In fact, $\mathbb{C}[x,y] / \langle xy \rangle$ cannot be written as a direct sum of any two nonzero rings: In a commutative ring with unity $R$, if $R = R_1 \oplus R_2$, then ...

7

$\mathbb{Z}$-modules are precisely abelian groups. As every ring is an abelian group, it is a $\mathbb{Z}$-module. It is entirely possible to be a module over more than one ring. For example, if $M$ is an $R$-module then it is also an $S$-module for any subring $S$ of $R$ (you seem to be interested in the case where $M = R$). Another example is given by ...

7

No, $IJ\subseteq I\cap J\subseteq I\cup J\subseteq I+J$. The middle inclusion is true for all sets. So you just need to prove that $IJ\subseteq I\cap J$ and $I\cup J\subseteq I+J$. It's easy to find counter-examples in $\mathbb Z$ to your inclusions: $$(6\mathbb Z)(4\mathbb Z)=24\mathbb Z\subsetneq 12\mathbb Z=6\mathbb Z\cap 4\mathbb Z$$ $$6\mathbb ... 6 I'm currently teaching out of the 4th edition of Stewart's Galois Theory textbook. Stewart defines a ring to be what other authors might call a commutative ring with unity. The reason is simple: in this book, there is not much call for noncommutative rings, nor for rings without unity, and it gets old writing "commutative ring with unity" over and over, when ... 6 xg(x) has no constant term, so the constant term of 2f(x)+xg(x) is twice the constant term of f. But it must be equal to 1, which is a contradiction because 1 is not divisible by 2. 4 Suppose that neither f nor g is the zero polynomial. Then there exist non-negative integers k and l and ring elements a_0,a_1,\dots, a_k, with a_k\ne 0, and b_0,b_1,\dots,b_l, with b_l\ne 0, such that$$f=a_0+a_1x+\cdots+a_kx^k \quad\text{and}\quad g=b_0+b_1x+\cdots+b_lx^l.$$The coefficient of x^{k+l} in the product fg is a_kb_l. ... 4 Suppose R\simeq K[T], where K is a commutative ring. Then K is an integral domain and since \dim K[T]=1 we get \dim K=0 (why?). It follows that K is a field. So k[X,X^{-1}]\simeq K[T]. Now use this answer. 3 Going off user26857's comment, we provide a counterexample for Proposition 3.14 in the case that M is not finitely generated. Hopefully you can use this to construct a counterexample for Corollary 3.15 as well. Take A = \mathbb{Z}, and let M be the direct sum of \mathbb{Z}/k\mathbb{Z} as k ranges through \mathbb{N}. This is not finitely ... 3 The idea is that computing ring operations in A, then applying f, is the same as first applying f, then computing ring operations in B. 2 You can use the quo-function, like this: Q<x0,y0,z0> := quo<P | x^2+y, y*z+1>; This gives you a new ring Q, which is isomorphic to P modulo the ideal generated by x^2+y and yz+1. This is documented in the subsection "Affine Algebras" of the section "Commutative Algebra" in the MAGMA handbook. Changing the normal form: I think (but ... 2 Let A = \mathbb Z and M = \mathbb Z/2\mathbb Z. 2$$(-a)x + ax = (-a + a)x = 0x = 0,$$so (-a)x is an additive inverse of ax. Now by the uniqueness of the additive inverse, (-a)x = -(ax). 2 I don't know of any general procedure. However, for the example you give,$$a^2-17b^2=\pm2$$implies$$a^2-b^2\equiv2\pmod4\ ,$$which is impossible since the squares modulo 4 are 0 and 1 only. Furthermore, as the product of norms is -8, if one of them is \pm4 then the other must be \mp2, and we have just shown that this is impossible. 2 Write f = a_0 + ... + a_mX^m and g = b_0 + ... + b_nX^n. It suffices to show that for each maximal ideal M, there is a coefficient of fg not in M. Let k and l be minimal with a_k,b_l \notin M. Check that the coefficient of X^{k+l} is not in M. 2$$f(r) = f(r.1) = f(r).f(1).$$f(1) = 0 \Rightarrow f(r) = 0 \forall r \in R 2 I'm not sure this answers your first question, but here's one way to prove k[x,y,z]/(y - x^2, z - x^3) \cong k[x]. The intuition is that$$ k[x,y,z]/(y - x^2, z - x^3) \cong k[x,x^2,x^3] = k[x] $$but I'm guessing you want something more rigorous. First, let's consider the following lemma. Lemma. Let R be a unital commutative ring and R[x] be the ... 2 There is a unique k-linear ring homomorphism \phi:k[x,y,z]\to k[t] such that \phi(x)=t, \phi(y)=t^2 and \phi(z)=t^3, and the ideal I=(y-x^2,z-y^3) is conttained in its kernel, as its generators are. If f=\sum_{i=0}^na_ix^i is an elemntt of of k[x] which is in I, then \phi(f) is zero. But \phi(f) is just \sum_{i=0}^na_it^i, whose ... 1 Let the "any constant" be \mu, \mu \geq 0. Then a^{2} - 17 \, b^{2} = \mu becomes$$x^{2} - 17 \, y^{2} = 1 \tag{1}where a = \sqrt{\mu} \, x and b = \sqrt{\mu} \, y. Equation (1) is a Pell equation, see Pell Equations, and has solutions \begin{align} x_{n} &= \frac{1}{2} \, \left[ (33+8 \, \sqrt{17})^{n} + (33 - 8 \, \sqrt{17})^{n} \right] ... 1 There may be simpler examples, but in Lemma 2.2 of "Centres and fixed-point rings of artinian rings" by Christian U. Jensen and Søren Jøndrup (Mathematische Zeitschrift 130, 189-197 (1973)) it's shown that if k is a field and V a k-vector space of dimension at least the cardinality of k, then the ring R=k\oplus V, where V is a square zero ideal, ... 1 Hint: Use distributivity to show that ax+(-(ax))=0, then apply uniqueness. 1 In the lines of the accepted answer, one can show that the global dimension of k[x,x^{-1}] is 1, so that if it is isomorphic to a polynomial ring, it must be isomorphic to one of the form D[y] with D a semisimple commutative domain, which is thus a field. 1 Note the following theorem of Jacobson (see, e.g., T.Y. Lam, A First Course in Noncommutative Rings, Theorem 12.10). Theorem Let A be a ring such that, for any x \in A there exists an integer n(x) > 1 such that x^{n(x)}=x. Then A is commutative. Now let A be a ring satisfying your property. Since x^5=x for all x \in A, the ring is ... 1 I think the most explicit description you can get is via partial fraction decompositions. Let P denote the set of monic irreducible polynomials over k. Then every f/g \in k(t) has a unique representation of the form\frac{f}{g}=\sum_{n \ge 0} a_{0,0,n} t^n +\sum_{p \in P}\sum_{r > 0} \frac{a_{p,r,0} + a_{p,r,1}t + \cdots ...

1

The Weyl algebra is a counterexample. It is a noncommutative ring with no proper nontrivial two-sided ideals that has no zero divisors and is not a division ring.

1

Over a field, a minimal polynomial can always be taken to be monic: just divide by the leading coefficient. Over a ring, there might be no nonzero monic polynomial at all with $\alpha$ as a root. Consider e.g. algebraic numbers that are not algebraic integers.

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