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{\bf Math 113 Homework 10, due 4/9/2019}
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\medskip


\noindent {\bf 1.} (Note: in this exercise, please don't call the additive and
multiplicative identity elements 0 and 1 if there is any risk of confusion.)

(a) Let $F$ be the set of all functions $f:\R\to\R$, equipped with the
operations of 
function addition $(f+g)(x)=f(x)+g(x)$ and composition $(f\circ g)(x)=
f(g(x))$. Show that $(F,+,\circ)$ satisfies all the axioms of a ring with
unity, with just one exception -- which one?

(b) Let $\bar\R=\R\cup\{\infty\}$ be the set formed by adjoining an element
called ``$\infty$'' to $\R$, and consider the operations $a \oplus b =
\min(a,b)$ (= the lesser of $a$ and $b$; with the convention that $\infty$
is greater than any real number), and $a \otimes b=a+b$ (with the convention
that $a+\infty=\infty+a=\infty$ for all $a\in\R$, and
$\infty+\infty=\infty$). Show that $(\bar\R,\oplus,\otimes)$ satisfies all
the axioms of a field, with just one exception -- which one?

(c) Show that $\{a+b\sqrt{5}\,|\,a,b\in\Q\}$ with the usual
addition and multiplication is a field.
\bigskip

\noindent
{\bf 2.} True or false? (As usual, justify your answers)

(a) The set of all pure imaginary complex numbers $\{ai\,|\,a\in\R\}$ with
the usual addition and multiplication is a ring.

(b) If $R'$ is a subring of a field $K$, then $R'$ is also a field.

(c) If $K$ is a field then the equation $x^2=x$ has exactly two solutions
in $K$.

(d) If $K$ is a field with multiplicative identity $1$ and $K'$ is a subfield of $K$
with multiplicative identity $1'$, then $1'=1$. (Hint: use (c)).

(e) If $R$ is a ring with multiplicative identity $1\neq 0$ and $R'$ is a subring with multiplicative identity $1'\neq 0$, then $1'=1$. (Hint: consider a direct product.)

(f) The direct product of two fields is a field.

\bigskip

\noindent {\bf 3.} 18.20. The ring $M_2(\Z_2)$ is the ring of matrices $M = \begin{pmatrix}
a&b\\c&d
\end{pmatrix},$ where $a, b, c, d\in \Z_2$ are residues modulo $2$. You may use that a matrix $M$ is invertible if and only if $|M| = ad-bc$ is not zero (modulo $2$).

\bigskip 

\noindent {\bf 4.} 19.27 (remember: ``unity'' means multiplicative identity and ``unit'' means invertible element). 



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