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(Proposition 4.11 and its proof are attached below.) In the proof, I'm not sure why $S[\alpha_1]=S[x]/(g)$. As I understand it, we can define the surjective map $S[x]\to S[\alpha_1]$ by sending $x$ to $\alpha_1$. But couldn't the kernel of this map be much larger than $(g)$?

Proposition 4.11. Let $R\subset S$ be rings, and suppose that $f\in R[x]$ is a monic polynomial. If $f$ factors in $S[x]$ as $f=gh$, with $g$ and $h$ monic, then the coefficients of $g$ and $h$ are integral over $R$.

Proof. Adjoining a root $\alpha_1$ of $g$ to $S$ and using long division in the ring $S[\alpha_1]=S[x]/(g)$, we see that $g$ factors as $(x-\alpha_1)g_1$, where the degree of $g_1$ is one less than the degree of $g$. Repeating this process inductively, we may find an extension ring $T$ of $S$ and elements $\alpha_i$, and $\beta_j$ of $T$ such that $g=\prod(x-\alpha_i)$, $h=\prod(x-\beta_j)$ in $T[x]$. Since each $\alpha_i$ and $\beta_j$ is a root of the monic polynomial $f$, the subring $T'$ of $T$ generated as an $R$-algebra by the $\alpha_i$, and $\beta_j$ is integral over $R$. Since the coefficients of $g$ and $h$ are the elementary symmetric functions in the $\alpha_i$, and the $\beta_j$, respectively, they too are integral over $R$. $\qquad\square$

Juan Castaño
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klein4
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1 Answers1

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They are defining the notation $S[\alpha_1]$ to mean $S[x]/(g)$, i.e. they are formally adjoining a root of $g$ which is $x+(g)$. This is not taking place in some pre-given bigger ring.

For example, take the polynomial $x^3 - 2$ over $\Bbb R$ and adjoin a formal root $\alpha_1$ by considering the ring $\Bbb R[x]/(x^3-2)$ (which is not an integral domain or a field), and say that $\alpha_1 := x + (x^3-2)$. Note that $\alpha_1 \ne \sqrt[3]2$, and the minimal polynomial of $\alpha_1$ is $x^3-2$, not $x-\sqrt[3]2$. In the same scenario, we would have $x^3-2 = (x - \alpha_1)(x^2 + \alpha_1 x + \alpha_1^2)$.

Kenny Lau
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  • Oh, I see. Thanks! In your example, why is $\alpha_1\neq \sqrt[3]{2}$? – klein4 Mar 11 '21 at 01:29
  • Because $\alpha_1$ is $x + (x^3-2)$ in $\Bbb R[x]/(x^3-2)$ and $\sqrt[3]2$ is $\sqrt[3]2 + (x^3-2)$. Note that ${1, x, x^2}$ is an $\Bbb R$-basis of $\Bbb R[x]/(x^3-2)$. Phrased differently, the ring in question $\Bbb R[x]/(x^3-2)$ can be thought of as ${p + q\alpha_1 + r\alpha_1^2 \mid p,q,r \in \Bbb R}$ with the only constraint that $\alpha_1^3 = 2$. You cannot conclude that $\alpha_1 = \sqrt[3]2$. – Kenny Lau Mar 11 '21 at 14:05
  • Does it matter if the polynomial we quotient by isn't minimal? – Charles Hudgins May 07 '21 at 15:15
  • @CharlesHudgins Does it matter for what? – Kenny Lau May 08 '21 at 04:36
  • I remember learning that $S[\alpha] = S[x] / (m_\alpha$, where $m_\alpha$ is the minimal polynomial for $\alpha$. But it seems like you're saying $S[\alpha]$ can be any quotient by a polynomial with $\alpha$ as a root. Won't you get different (i.e. non-isomorphic) $S[\alpha]$ when you quotient by different polynomials with $\alpha$ as a root? – Charles Hudgins May 08 '21 at 12:22
  • @CharlesHudgins What you cited is true in a different context: if $\alpha$ is an integral element over $S$ (in some ambient ring $T$) then $S[\alpha]$, the subring of $T$, is isomorphic to $S[x]/m_\alpha$ where $m_\alpha$ is the minpoly of $\alpha$. What is different is that in your context $S[\alpha]$ is the "subring of $T$ generated by $S$ and $\alpha$", whereas in the context of the OP's question $S[\alpha]$ is some formally defined ring, not the subring generated by some subset. – Kenny Lau May 08 '21 at 12:48
  • That makes sense. Without an ambient ring, it doesn't really make sense to talk about "minimal." With the notation $S[\alpha]$ are we just trying to say "this is a ring which has a root for $f(x)$"? – Charles Hudgins May 08 '21 at 13:09