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Field theory problems and solution in abstract algebra


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  • If the Quotient Ring is a Field, then the Ideal is MaximalIf the Quotient Ring is a Field, then the Ideal is Maximal Let $R$ be a ring with unit $1\neq 0$. Prove that if $M$ is an ideal of $R$ such that $R/M$ is a field, then $M$ is a maximal ideal of $R$. (Do not assume that the ring $R$ is commutative.)   Proof. Let $I$ be an ideal of $R$ such that \[M \subset I \subset […]
  • Prove the Cauchy-Schwarz InequalityProve the Cauchy-Schwarz Inequality Let $\mathbf{a}, \mathbf{b}$ be vectors in $\R^n$. Prove the Cauchy-Schwarz inequality: \[|\mathbf{a}\cdot \mathbf{b}|\leq \|\mathbf{a}\|\,\|\mathbf{b}\|.\]   We give two proofs. Proof 1 Let $x$ be a variable and consider the length of the vector […]
  • Nilpotent Ideal and Surjective Module HomomorphismsNilpotent Ideal and Surjective Module Homomorphisms Let $R$ be a commutative ring and let $I$ be a nilpotent ideal of $R$. Let $M$ and $N$ be $R$-modules and let $\phi:M\to N$ be an $R$-module homomorphism. Prove that if the induced homomorphism $\bar{\phi}: M/IM \to N/IN$ is surjective, then $\phi$ is surjective.   […]
  • Null Space, Nullity, Range, Rank of a Projection Linear TransformationNull Space, Nullity, Range, Rank of a Projection Linear Transformation Let $\mathbf{u}=\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$ and $T:\R^3 \to \R^3$ be the linear transformation \[T(\mathbf{x})=\proj_{\mathbf{u}}\mathbf{x}=\left(\, \frac{\mathbf{u}\cdot \mathbf{x}}{\mathbf{u}\cdot \mathbf{u}} \,\right)\mathbf{u}.\] (a) […]
  • Quiz 12. Find Eigenvalues and their Algebraic and Geometric MultiplicitiesQuiz 12. Find Eigenvalues and their Algebraic and Geometric Multiplicities (a) Let \[A=\begin{bmatrix} 0 & 0 & 0 & 0 \\ 1 &1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{bmatrix}.\] Find the eigenvalues of the matrix $A$. Also give the algebraic multiplicity of each eigenvalue. (b) Let \[A=\begin{bmatrix} 0 & 0 & 0 & 0 […]
  • In a Field of Positive Characteristic, $A^p=I$ Does Not Imply that $A$ is Diagonalizable.In a Field of Positive Characteristic, $A^p=I$ Does Not Imply that $A$ is Diagonalizable. Show that the matrix $A=\begin{bmatrix} 1 & \alpha\\ 0& 1 \end{bmatrix}$, where $\alpha$ is an element of a field $F$ of characteristic $p>0$ satisfies $A^p=I$ and the matrix is not diagonalizable over $F$ if $\alpha \neq 0$. Comment. Remark that if $A$ is a square […]
  • The Product of a Subgroup and a Normal Subgroup is a SubgroupThe Product of a Subgroup and a Normal Subgroup is a Subgroup Let $G$ be a group. Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$. The product of $H$ and $N$ is defined to be the subset \[H\cdot N=\{hn\in G\mid h \in H, n\in N\}.\] Prove that the product $H\cdot N$ is a subgroup of […]
  • Quiz 9. Find a Basis of the Subspace Spanned by Four MatricesQuiz 9. Find a Basis of the Subspace Spanned by Four Matrices Let $V$ be the vector space of all $2\times 2$ real matrices. Let $S=\{A_1, A_2, A_3, A_4\}$, where \[A_1=\begin{bmatrix} 1 & 2\\ -1& 3 \end{bmatrix}, A_2=\begin{bmatrix} 0 & -1\\ 1& 4 \end{bmatrix}, A_3=\begin{bmatrix} -1 & 0\\ 1& -10 \end{bmatrix}, […]

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