# Ring Homomorphism Textbook Question

Please help me understand the last three sentences in this paragraph from the Artin textbook. Where does this come from: "The monomials that appear in $r_0(t^2)$ have even degree, while those in $r_1(t^2)t^3$ have odd degree."

Let $\Phi: \mathbb{R}[x,y] \to \mathbb{R}[t]$ be the homomorphism that is the identity on the real numbers, and that sends $x \rightsquigarrow t^2, y \rightsquigarrow t^3$. Then it sends $g(x,y) \rightsquigarrow g(t^2,t^3)$. The polynomial $f(x,y) =y^2 - x^3$ is in the kernel of $\Phi$. We'll show that the kernel is the principal ideal $(f)$ generated by $f$, i.e., that if $g(x,y)$ is a polynomial and if $g(t^2,t^3) = 0$, then $f$ divides $g$. To show this, we regard f as a polynomial in $y$ whose coefficients are polynomials in $x$ (see (11.3.8)). It is a monic polynomial in $y$, so we can do division with remainder: $g = fq + r$, where $q$ and $r$ are polynomials, and where the remainder $r$, if not zero, has degree at most $1$ in $y$. We write the remainder as a polynomial in $y: r(x,y) = r_1(x)y+r_0(x)$. If $g(t^2,t^3) = 0$, then both $g$ and $fq$ are in the kernel of $\Phi$, so $r$ is too: $r(t^2,t^3) = r_1(t^2)t^3+ r_0(t^2) = 0$. The monomials that appear in $r_0(t^2)$ have even degree, while those in $r_1(t^2)t^3$ have odd degree. Therefore, in order for $r(t^2,t^3)$ to be zero, $r_0(x)$ and $r_1(x)$ must both be zero. Since the remainder is zero, $f$ divides $g$.

If $p(x)$ is any polynomial in $x$, then $p(t^2)$ is a polynomial in $t$ whose terms all have even degree. If necessary, write this out explicitly:
$$p(x)=a_0+a_1x+\ldots+a_nx^n\;,$$
\begin{align*} p(t^2)&=a_0+a_1t^2+\ldots+a_k(t^2)^k+\ldots+a_n(t^2)^n\\ &=a_0+a_1t^2+\ldots+a_kt^{2k}+\ldots+a_nt^{2n}\;. \end{align*}
If you now multiply that by an odd power of $t$, like $t^3$, the typical term will go from $a_kt^{2k}$ to $a_kt^{2k+3}$, with an odd power of $t$.
Thus, $r_0(t^2)$ has only terms in even powers of $t$, and $r_1(t^2)t^3$ has only terms in odd powers of $t$.