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Suppose $I$ is an ideal of $\mathbb{R}[x]$ properly containing $\langle x^2+1\rangle$ and let $f(x)\in I\setminus\langle x^2+1\rangle$. Then $$f(x)=q(x)(x^2+1)+ax+b$$ for some $q(x)\in\mathbb{R}[x]$, $a,b\in\mathbb{R}$. Since $f(x)\notin\langle x^2+1\rangle$, you have $ax+b\ne0$, that is, not both $a$ and $b$ are zero. If $a=0$, then $b\ne0$ and so ...

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Hint: $\mathbf R[x]$ is a P.I.D.. In a P.I.D., the ideal generated by an irreducible element is maximal.

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Hint: $R/I$ is isomorphic to the ring $S$ of $2\times2$ matrices with coefficients in $\mathbb{Z}/2\mathbb{Z}$. Can you find a surjective ring homomorphism $R\to S$? Can you tell what its kernel is?

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The morphism $x\rightarrow t, y\rightarrow t^2$, $z\rightarrow 1$ induces a morphism You have $f:k[x,y,z)/(x^2-yz,z-1)\rightarrow k[t]$ defined by $f([x])=t, f([y])=t^2, f([z])=1$ where $[x]$ is the class of $x$ Consider $g:k[t]\rightarrow k[x,y,z]/(x^2-yz,z-1)$ defined by $g(t)=[x]$, you have $f(g(t)=f([x])=t$. $g(f([x]))=g(t)=[x]$, ...

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Suppose that $p$ is in the kernel. Then by polynomial division you can write $p=q(X^2-Y^3)+r$ where no term of $r$ has degree $\ge 2$ in $X$, and clearly $r$ is then also in the kernel. Now $r$ consists of terms of the form $aY^n$ and terms of the form $aXY^n$. Under $f$ the former terms become $at^{2n}$ and the latter terms become $at^{2n+3}$. But $t^{2n}$ ...

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Hint: Ideals are closed under external multiplication. For all $r \in F$, $x \in I$, $rx \in I$. Just apply this to the elements you obtained. Complete Solution: Let $I$ be any non-zero ideal of a field $F$. Then, $I$ contains at least one non-zero element $x$. Since $F$ is a field, $x^{-1} \in F$, which implies that $x^{-1} x = 1 \in I$ (since $I$ is an ...

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There are plenty of cases where this happens. It's really only interesting when the module action interacts in an interesting way with the multiplication The most straightforward example, which works when $R$ is commutative, is the concept of an $R$-algebra, which I'll let you look up yourself. In this case it's easy to see what the action does. If you ...

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The upper triangular matrices form a subalgebra of $M_2(F)$, and they are an obvious counterexample to the proposition.

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To challenge this knot, we have no choice but to replace any specific matrices. As one of the approaches, The rotation matrix is a part of correct answer. Because the matrix has special characteristics. As reasons, variables and equations can be reduced and commutativity is allowed under any two dimension. For example, the following figure shows that even if ...

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Every finite dimensional $k$-algebra is isomorphic to a subalgebra of $M_n(k)$ for some $n$ (you can take $n=\dim A$, in fact) So no, for then all algebras would be semisimple!

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Such a ring $R$ is called a valuation ring, if it is not a field. For ideals $I$, $J$ in a valuation ring we have either $I\subseteq J$ or $J\subseteq I$. Indeed, Let $x\in I$ such that $x\not\in J$. For $y\in J$ with $y\neq 0$ we have $xy^{-1}\not\in R$, hence we must have $yx^{-1}\in R$. It follows that $y=(yx^{-1})x\in I$, hence $J\subseteq I$.

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Partial answer: Since $\mathbb{R}[x]$ is a PID, we can just focus on the prime/maximal elements. Define $$A=\{ax+b : a \in \mathbb{R}_{\neq 0}\}$$ $$B=\{ax^2+bx+c : a \in \mathbb{R}_{\neq 0}, b \in \mathbb{R},c\in \mathbb{R} ,b^2-4ac < 0\}$$ Then $A \cup B$ is the set of prime elements of the ring $\mathbb{R}[x]$. Claim 0. If $a \neq 0$, then ...

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Consider the fields $K = \Bbb{Q}(\sqrt{5})$ and $L = \Bbb{Q}(i\sqrt{5})$ and suppose that $\varphi$ is a field homomorphism. If we look at $\varphi(\sqrt{5}^2)$ then we have $$\varphi(\sqrt{5}^2) = \varphi(\sqrt{5})^2 = \varphi(5) = 5,$$ so for $\varphi$ to exist, there must exist an element $\varphi(\sqrt{5})$ in $L$ such that $\varphi(\sqrt{5})^2 - 5 = ... 6 They are not isomorphic as fields: While$\mathbb{Q}(i \sqrt{5})$contains a square root of$-5$(namely$i \sqrt{5}$), the field$\mathbb{Q}(\sqrt{5})$does not (as it is a subfield of$\mathbb{R}$). They are however isomorphic as$\mathbb{Q}$-vector spaces, as they are both two-dimensional over$\mathbb{Q}$(because$\mathbb{Q}(\sqrt{5})$has$\{1, ...

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The absorbing (or zero) element of $A/aA$ is $aA$. Therefore if $b+aA=aA$ you just get the trivial equality $0.(c+aA)=0$ in $A/aA$.

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For any $r\in R,$ and any $a\in I,$ we have $r.a\in I.$ Take $a=1.$ Then $r.1=r \in I;$ For any $r\in R.$

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Hints, as requested: Let $I = \bigcap\limits_{j \in J} I_j$ be an intersection of ideals $I_j$ (where $J$ is an indexing set, finite or otherwise). Show that for any $x, y \in I$, $x - y \in I$. Use the fact that $x, y \in I_j$, $\forall j \in J$. Show that for any $r \in R$ and $x \in I$, $rx \in I$. Again, use the fact that $x \in I_j$, $\forall j \in ... 0 We can use the following useful lemma:$I$is a prime ideal iff$R/I$is an integral domain. So if$(n)$is a prime ideal iff$Z/nZ$is an integral domain iff$n$is a prime. 0 Hint: The polynomials$p_1,\dots,p_n$must each annihilate a nontrivial submodule of$M$, so to identify these polynomials, try seeing if you can find elements of$\mathbb R[t]$that annihilate some non-zero element of$M$. 0 The notation$\langle f(x) \rangle$refers to the ideal generated by the polynomial$f(x)$and it is defined to be the set$\{ f(x) g(x): g(x) \in R[x]\}$of all multiples of$f(x)$in$R[x]$. For example,$\langle x \rangle$is the set of all polynomials of the form$x g(x)$, which is exactly the set of all polynomials in$R[x]$whose constant term is 0. ... 0 Whenever you're talking about an expression like$\langle x \rangle$in an algebraic structure, it is always with respect to the operations in the structure. If$x$is in a semigroup and you're only talking about a subsemigroup generated by$x$, then you only have multiplication, and$\langle x\rangle=\{x,x^2, x^3, x^3,\ldots \}$. If$x$is in a group and ... 1 In a ring$R$, if$a$is an element then$\langle a\rangle$usually denotes the ideal generated by$a$, that is the smallest ideal of$R$contaning$a$, which is$Ra$(in particular it contains all$a^n$, but not only). 0 I'll answer for the case of a group homomorphism. Sometimes it helps to draw a picture in order to clarify this idea of a kernel for quotient groups. A big circle for G and G/H with circles in each for the kernel, image, ect, and a lines denoting the mapping between the groups. I'll try my best to clarify it in words. In any group homomorphism, Φ:G→G/H, you ... 1 If we're talking of the field$\;\Bbb F_{p^2}\;$, then the multiplicative group of this,$\;\left(\Bbb F_{p^2}\right)^*:=\Bbb F_{p^2}\setminus\{0\}\;$is a cyclic group (as is any finite subgroup of the multiplicative group of any field), and in this case it is a group of order$\;p^2-1\;$, so you're looking for the number of generators this group has, and ... 3 If$p = 2$then$-1$is a square mod$p$, so we may now assume that$p$is odd. Suppose that$n^2 \equiv -1 \bmod p$for some$n \in \mathbb{N}$. Then, using that$\mathbb{Z}/p\mathbb{Z}$is a field, $$1 \equiv n^{p-1} \equiv (n^2)^{\frac{p-1}{2}} \equiv (-1)^{\frac{p - 1}{2}} \bmod p$$ which implies that$\frac{p-1}{2}$is even, thus, that$p \equiv 1 ...

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Let's do some experimentation. $p = 3$: no. $p = 5$: yes, $2^2 \equiv -1$. $p = 7$: no. $p = 11$: no. $p = 13$: yes, $5^2 \equiv -1$. $p = 17$: yes, $4^2 \equiv -1$. $p = 19$: no. $p = 23$: no. It appears that only those prime numbers which are congruent to $1$ modulo $4$ have this property.

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For $p>2$ prime, $-1$ is a quadratic residue modulo $p$ if and only if $p \equiv 1 \bmod 4$. So you can determine whether $-1$ is a quadratic residue modulo $p$ as quickly as you can find the remainder of $p$ upon division by $4$.

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You are right about $$R=\left\{\begin{bmatrix}\alpha & \beta \\ \bar\beta & \bar\alpha\end{bmatrix}\mid \alpha,\beta\in\mathbb{C}\right\}$$, especially since $e=\frac12\begin{bmatrix}1&1\\ 1&1\end{bmatrix}$ would create zero divisors : $e(1-e)=0$. But presumably what you're reading is about $$R=\left\{\begin{bmatrix}\alpha & \beta ... 0 A simple example is the product ring \mathbb Z_2\times \Bbb Z_2 (the Klein four group). Every proper nonzero subring is isomorphic to \Bbb Z_2, which is a division ring. 2 The answer is no. For example, let y be an element of F and let g(x) = (x-y)^2f(x), where f(x) is in F[x] and has all simple roots. Then the splitting field of g over F is the same as that of f. 1 "Reducible" is not synonim of "has a root". A polynomial can be reducible on a given field without having roots in that field, like the polynomial$$ p(x)=x^4+2 x^2+1= (x^2+1)\cdot(x^2+1), $$which by sure has no real roots. In other words having a root x_0 is equivalent to say that the polynomial is reducible and one factor has degree one, that is ... 7 For a polynomial of degree 3, if it's not irreducible then it must split in factors that have degree 2 or 1, and there must be a factor of degree 1 (so a root). Hence it suffices to check that there are roots. In degree 5, your polynomial can split in two factors of respective degree 2 and 3 : so you can be reducible without having roots. 3 Suppose we have a ring homomorphism f\colon R\to\mathbb{Z}. The composite of the homomorphism f following the homomorphism j\colon \mathbb{Z}\to R : n\mapsto n\cdot 1_R is the identity homomorphism f\circ j=\mathrm{id}_{\mathbb{Z}} (this because \mathrm{id}_{\mathbb{Z}} is the only endomorphism of the ring \mathbb{Z}), which implies that j is ... 1 \newcommand{\Tr}{\mathrm{Tr}}I will give you a hint in the case when R is a commutative ring (which I assume to be a tacit assumption). X is a root of the characteristic polynomial$$ z^{2} - \Tr(X) z + \det(X). $$Now$$ 2 (z^{2} - \Tr(X) z + \det(X)) = 2 z^{2} - 2 \Tr(X) z + (\Tr(X)^{2} - \Tr(X^{2})). $$Can you fill in the details, and proceed from ... 0 Let \mathbb{Z}_n be the set of integers \{0,1,\ldots,n-1\} equipped with the operations of addition mod n and multiplication mod n. It can be shown this structure is a ring. \mathbb{Z}_n[x] is defined as the set of polynomials of the form a_n x^n + \cdots + a_1 x + a_0, where a_i \in \mathbb{Z}_n equipped with the usual operations of addition ... 0 The quotient ring \mathbb{Q}[x]/(x^2+1) consists of the set of all cosets of the form ax+b+(x^2+1) for a,b \in \mathbb{Q}. The natural ring homomorphism from \mathbb{Q}[x] to \mathbb{Q}[x]/(x^2+1) is defined by f(x) \mapsto \hat{f}(x) + (x^2+1), where \hat{f} is the remainder when dividing f by x^2+1. The kernel of this homomorphism is ... 1 You are very close. Let's suppose that both i_1 and i_2 \in R are the multiplicative identity. If a\in R, then a\star i_1=a=i_2\star a by definition since i_1 and i_2 are both the identity. i_1=i_1\star i_2 since i_2 is the identity. But i_1\star i_2=i_2 since i_1 is also the identity. So, i_1=i_2 and the identity is unique. 0$$ i_2 = i_1 * i_2 = i_1 $$2 Yes, although you can trim it down a bit; for clarity's sake. If i_n is an additive identity, then for any a in the ring: ~a+i_n=a~ and i_n+a=a, by definition and the commutative property of addition in abelian groups. If any i_1, i_2 are both additive identities, then we have: i_1+i_2=i_1 and i_1+i_2=i_2. Thus for any additive ... 0 Let \alpha = a_1 + a_2 \sqrt{2} and \beta = b_1 + b_2 \sqrt{2} be elements of \mathbb{Z}[\sqrt{2}] with \beta \neq 0. We wish to show that there exist \gamma and \delta in \mathbb{Z}[\sqrt{2}] such that \alpha = \gamma\beta + \delta and N(\delta) \leq N(\beta). To that end, note that in \mathbb{Q}(\sqrt{2}) we have \frac{\alpha}{\beta} = ... 1 I name the injective and surjective homomorphisms as$$0\rightarrow A\overset{j}\rightarrow B\overset{f}{\rightarrow} C\rightarrow 0$$and$$0\rightarrow C\overset{g}\rightarrow D\overset{p}\rightarrow E\rightarrow 0.$$To show that 0\rightarrow A\overset{j}\rightarrow B\overset{gf}{\rightarrow} D\overset{p}\rightarrow E\rightarrow 0 is exact one ... 1 If I understand your definitions correctly, any finite product of fields, including the zero ring, has this property. 2 (This is a revised answer; thanks to Henning Makholm for his comment.) In the general case the statement is false; here we show a counter-example. Let$$ R=\mathbb{Z}[Y]/\left< 3Y, Y^3+Y\right>. $$The elements of this ring are of the form a+bY+cY^2 with a\in\mathbb{Z} and b,c\in\mathbb{Z}_3. Define the two polynomials as$$ f(x) = ...

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Yes, it has been studied before, and even appears in textbooks. A Google search of "irreducible element" with "commutative ring" will also lead you to papers like this and this and this which obviously consider the concept.

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(1) Is there any example of a ring which is not a division ring but any of its subring is a division ring: The answer is yes: Consider the zero ring $(0)$. It is not a division ring since a division ring is defined to be non-zero. It does not have any proper subrings. Hence your statement is vacuously true. But it is not very interesting. (2) Is there any ...

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Here I try to explain better of zcn's example. Let $E=\mathbb{Q}, F=\mathbb{Q}(\sqrt{2}), G=\mathbb{Q}(2^{1/4}), H=\mathbb{Q}(2^{1/4},i)$. Then we see that $$\textrm{Gal}(F|E)=\mathbb{Z}/2\mathbb{Z}, \textrm{Gal}(G|F)=\mathbb{Z}/2\mathbb{Z},\textrm{Gal}(H|E)=\mathbb{D}_{4}$$ where the generators for $\textrm{Gal}(H|E)$ are $g:i\rightarrow -i$ (complex ...

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For the first part note that a standard argument shows that $\phi(q)=q$ where $q \in \mathbb{Q}$ denotes the constant function $q$. If you show that this also holds for irrational numbers you are done for the first part. I'll give a sketch of the proof. Let $C=C(\mathbb{R})$ and $D=D(\mathbb{R})$. Define $i: \mathbb{R} \to C(\mathbb{R})$ by $i(x) = c_x$, ...

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This question investigates another algebraic object that acts like a ring without a distributive law. The main point of that thread is to discover another operation so that the distributive law does hold. This is different from your question, but does get to the heart of what a ring is. A ring is both an abelian group under addition, and a semigroup (a ...

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We can work in the setting of groups. Let $f\colon G\to H$ be a homomorphism that sends a specific element $g_0$ of $G$ to a specific element $h_0$ of $H$. Then, for all elements $x$ in the kernel of $f$, the elements $xg_0$ are mapped to the same $h_0$. And every element mapped to $h_0$ is of this form. In your case, for arbitrary $f(x)$, the elements ...

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You need to define a lineal aplication, say $\phi$, from $\mathbb{Z}[x]/(5x-1)$ onto $\mathbb{Z}[\frac{1}{5}]$. Then you prove that $\phi$ preserves the multiplication (that is, that $\phi(ab)=\phi(a)\phi(b)$ for every $a,b\in\mathbb{Z}[x]/(5x-1)$ ). At this point you've showed that $\phi$ is a surjective ring homomorphism, and you only need to check that ...

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