Let $T = \{\frac ab \in \Bbb Q \mid \text{$a$ and $b$ are relatively prime and $5 \nmid b$}\}$ Let $T = \{\frac ab \in \Bbb Q \mid \text{$a$ and $b$ are relatively prime and $5 \nmid  b$}\}$ . Show that $T$ is a ring under usual addition and multiplication. Also prove that $I = \{\frac ab \in T : 5\mid a \}$ is an ideal of $T$ and the quotient ring $T/I$ is a field.
I am having a problem in doing the problem!!
Let $x,y \in T$, then $x = \frac ab$ and $y = \frac cd$ and $\gcd(a,b) =1$ and $\gcd(c,d) =1$.
Having problem in the part to show that the ring is closed with respect to addition!!
 A: Here are the things you need to do to cover all of those parts.


*

*Show that addition works at all.  In particular, that when you add two fractions whose denominators are not divisible by $5$, one obtains a fraction whose denominator is not divisible by $5$.

*Note that commutativity and associativity of addition are inherited from $\mathbb{Q}$.  And the additive identity $0\in T$ is $\frac{0}{1}$, for instance.

*Show that multiplication works at all.  Again, the product of two elements of $T$ has a denominator which is not divisible by $5$.

*Again, commutativity and associativity are inherited.  And the multiplicative identity $1\in T$ shows up as $\frac{1}{1}$, for instance.  Distributivity is also inherited.
These together prove $T$ is a ring.  Now for $I$ being an ideal.


*To show $I$ is closed under addition, one must show that the sum of two fractions whose numerators are divisible by $5$ is itself a fraction whose numerator is divisible by $5$.

*To show $I$ is closed under multiplication by $T$, one must show that the product's numerator is divisible by $5$.  For this, let both of the fractions be reduced, and show that the denominator is not divisible by $5$ (otherwise the numerator might lose its $5$'s!).
So $I$ is an ideal.


*The quotient $T/I$ being a field is equivalent to saying $I$ is a maximal ideal in $T$.  There are a number of ways to tackle this, one is to show that if $x\in T\setminus I$, then $I+xT=T$.  (In fact, this equation can be used to find a multiplicative inverse of $x$ for any $x$, thereby showing $T/I$ is a field directly, if you wish.)


Hopefully this is enough of an outline.
