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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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