Exercise A0.V.2.05

Let $R$ be a subset of $\mathbb{Z}[t]$ consisting of polynomials with no degree-$1$ monomial in $t.$

Proposition A

$R$ is in fact a subring of $\mathbb{Z}[t].$

Proof

Let $f, g \in R.$ For a degree-$n$ monomial to be present in $f + g$ it must be that it is present at least in $f$ or in $g.$ So $f + g$ contains no degree-$1$ monomials.

For a degree-$n$ monomial to be present in $fg,$ there must exist a degree-$m_1$ monomial present in $f$ and a degree-$m_2$ monomial present in $g$ such that $m_1 + m_2 = n.$ In the case when $n = 1,$ it must be that one of the monomials has degree $1$ and another has degree $0.$ But neither $f$ nor $g$ has a degree $1$ term. Hence $fg \in R.$

Since degree-$0$ polynomials have no degree-$1$ monomials, $\mathbb{Z} \subseteq R.$ This is sufficient to show that $R$ is a subring of $\mathbb{Z}[t].$

$\blacksquare$

Proposition B

The common divisors of $t^5$ and $t^6$ in $R$ are $1, t^2, t^3, t^4.$

Proof

A divisor of $t^n$ in $R$ is also a divisor of $t^n$ in $\mathbb{Z}[t]$. Since $\mathbb{Z}[t]$ is an UFD we need only consider the factors of $t^n$ as candidate divisors (up to units).

This gives us a list $t^0, t^1, t^2, t^3, t^4, t^5.$ However, $t^1 \notin R$ and $t^5$ cannot be a divisor of $t^6$ since there is no $v \in R$ such that $\deg(v) = 1.$ Similarly, $t^4$ does not divide $t^5$ in $R.$ Thus the only common divisors are, up to units, $1, t^2, t^3.$

$\blacksquare$

Proposition C

There exists no $\gcd(t^5,t^6)$ in $R.$

Proof

Suppose that, up to units, $\gcd(t^5,t^6) = t^3.$ Then, since $t^2$ divides both, it should also divide $t^3$ but this cannot happen as $t \notin R.$ It is clear that $\pm t^2$ and $\pm 1$ cannot be $\gcd(t^5, t^6),$ since they are not divisible by $t^3.$ We conclude that no $\gcd(t^5,t^6)$ exists in $R.$

$\blacksquare$