# Algebra exercises for High-School

Algebra exercises for high school students. Exercises and solutions in groups, rings, and fields. Our goal is to help high school students by giving them some basic algebra exercises. All solutions are detailed.

Exercise: Justify if the following applications are group or ring homomorphisms for the usual operations

• $f:\mathbb{C}\to\mathbb{R},\; x\mapsto |x|$.
• $g:\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z},\;x\mapsto \bar{x}$ .

Solution: 1) For the multiplication operation we have $|z_1z_2|=|z_1||z_2|$ for any $z_1,z_2\in \mathbb{C}$. So that $f(z_1z_2)=f(z_1)f(z_2)$ for any $z_1,z_2\in \mathbb{C}$. But neither $\mathbb{C}$ nor $\mathbb{R}$ are groups for the multiplication operation. Hence $f$ id not a morphism for groups for the multiplication operation. Let us now see what happen for the operation "+". In general we have $|z_1+z_2|\neq |z_1|+|z_2|$ if $z_1,z_2\in \mathbb{C}$ (for example $f(1+i)=\sqrt{2}\neq 2=f(1)+f(i)$). This shows that $f(z_1+z_2)\neq f(z_1)+f(z_2)$, in general. Hence $f:(\mathbb{C},+)\to (\mathbb{R},+)$ is not a morphism.

2) $g$ is a morphism of rings. In fact, we have $g(1)=\bar{1}$ (the property for identity elements holds). Moreover, for any $x,y\in \mathbb{Z}$ we have \begin{align*} g(x+y)=\bar{x+y}=\bar{x}+\bar{y}=g(x)+g(y)\end{align*} and \begin{align*} g(xy)=\bar{xy}=\bar{x}\times \bar{y}=g(x)g(y).\end{align*}

Exercise: Let $\mathbb{K}$ and $\mathbb{M}$ two fields and let $f:\mathbb{K}\to \mathbb{M}$ be an homomorphism of rings. Show that $f$ is injective.

Solution: Let an element $x\in \ker(f)$ (the kernel of $f$ ) such that $x\neq 0$. As $\mathbb{K}$ is a field then we can consider the inverse $x^{-1}$. As $f(x)=0,$ we have \begin{align*} 0=f(x)\times f(x^{-1})=f(x\times x^{-1})=f(1_{\mathbb{K}})=1_{\mathbb{M}}. \end{align*} Absurd!! Then $\ker(f)=\{0\},$ hence $f$ is injective.