1
$\begingroup$

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 $g$.

$\endgroup$
0
2
$\begingroup$

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:

$$p(x)=a_0+a_1x+\ldots+a_nx^n\;,$$

say, so

$$\begin{align*} p(t^2)&=a_0+a_1t^2+\ldots+a_k(t^2)^k+\ldots+a_n(t^2)^n\\ &=a_0+a_1t^2+\ldots+a_kt^{2k}+\ldots+a_nt^{2n}\;. \end{align*}$$

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 $t$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.