2.8. Polynomial Rings (for Galois Theory).

Say $(R, +, \cdot)$ is a ring with identity $1 \ne 0$. We will always assume $R$ is of this form.

1. If $x$ is indeterminate (i.e. a variable) a polynomial in $x$ is a formal sum

\[ a_dx^d + \cdots + a_1x + a_0. \]

with $a_k \in R$. * 2. We say the degree $\deg(p) = d$ if $a_d$. Otherwise, set $\deg(p) = -\infty$ if $a_d = \cdots = a_0 = 0$. * 3. Let $R[x]$ be the collectionn of all such $p(x)$. This is the polynomial ring over $R$. * 4. More generally, say $\{x_1, x_2, \dots, x_n\}$ is a collection of $n$ indeterminates. Inductively define $R[x_1, x_2, \dots, x_n] = S[x_n]$ where $S = R[x_1, x_2, \dots, x_{n-1}]$. A typical element is in the form

\[ p(x_1, \cdots, x_n) = \sum_{i_1 = 1}^{d_1}\cdots\sum_{i_n = 1}^{d_n}a(i_1, \cdots, i_n)x_1^{i_1}\cdots x_n^{i_n}. \]

We then say that this $p(x_1, \dots, x_n)$ is a polynomial in $n$ variables. * 5. Assuming that the highest coefficient \ $a(d_1, \dots, d_n) \ne 0$, we then say the degree is

\[ \deg(p) = \max\{i_1 + \cdots + i_n\} \text{ (fix here)}. \]

Otherwise, if all terms are zero, i.e., if all $a(i_1, \dots, i_n) = 0$, then we set $\deg(p) = -\infty$.

Example. Consider the polynomial ring $R[x, y]$ as the polynomial ring in $n = 2$ variables. Let $p(x, y) = 1 + x^3 + y^3$. Then $\deg(p) = 3$.

$(R[x_1, \cdots, x_n], + , \cdot)$ is a ring with $1 \ne 0$.

We'll showt this by induction:

\[\begin{align*} P(n) = "(R[x_1, \dots, x_n]) \text{ is a ring with } 1 \ne 0". \end{align*}\]

We have already shown the base case $P(1)$ is true. Assume that $P(n)$ is true for some $n_0$. We show $P(n)$ is true for $n = n_0 + 1$.

By definition, $R[x_1, \dots, x_n] = S[x_n]$ where $S = R[x_1, \dots, x_{n_0}]$. By our inductive hypothesis, we have that \ $(S, +, \cdot )$ is a ring with $1 \ne 0$. Hence by $P(1)$ we have that $(S[x_n], \cdot, +)$ is also a ring with $1 \ne 0$. Hence $P(n_0 + 1)$ is true.

Assume $R$ is an integral domain.

1. Suppose $p, q \in R[x_1, \cdots x_n]$. Then

$$ \deg(p \cdot q) = \deg(p) + \deg(q).
$$ * 2. $R[x_1, \dots, x_n]$ is also an integral domain. * 3. The units of $R[x_1, \dots, x_n]$ is $R^{\times}$.

1. If either $p = 0$ or $q = 0$, then $p \cdot q = 0$. Then observe that

\[ \deg(p \cdot q) = \deg(p) + \deg(q) \]

doesn't make sense unless we choose to set $\deg(0) = -\infty$. Hence we see that

\[ \deg(p \cdot q) = -\infty. \]

We could make it positive infinity, but we assume that it is negative for subtle reasons later on.

Now assume that $p, q \ne 0$. We write our polynomials

\[ p = \sum_{i_1, \dots, i_n}^{d_1, d_2, \cdots, d_n}a(i_1, \dots, i_n)x_1^{i_1}\cdots x_n^{i_n} \qquad q = \sum_{j_1, \dots, j_n}^{e_1, e_2, \cdots, e_n}a(j_1, \dots, j_n)x_1^{j_1}\cdots x_n^{j_n} \]

where $\deg(p) = d_1 + d_2 + \cdots + d_n$ and $\deg(q) = e_1 + e_2 + \cdots + e_n$. Then we see that

\[ p \cdot q = \sum_{k_1, \dots, k_n}^{d_1 + e_1, d_2 + e_2, \cdots, d_n + e_n}c(i_1, \dots, i_n)x_1^{k_1}\cdots x_n^{k_n} \]

where

\[ c(k_1, \dots, k_n) = \sum_{i_1+j_1 = k_1, \dots, i_n+j_n = k_n.} a(i_1, \dots i_n)b(j_1, \dots, j_n). \]

The leading term is $c(d_1 + e_1, \dots, d_n + e_n) = a(d_1, \dots, d_n )b(e_1, \dots, e_n)$. Hence, the leading term is also nonzero, so the degree must be

\[ d_1 + e_1 + \dots + d_n + e_n = \deg(p) + \deg(q). \]

There are actually many ways to define the degree of a polynomial in $R[x_1, \dots, x_n]$ when$n \ge 2$. * 2., 3. We can show both (2) and (3) at the same time via induction. Let

\[ P(n) = "R[x_1, \dots, x_n] \text{ is an integral domain and } R[x_1, \dots, x_n]^{\times} = R^{\times}." \]

We've done this in the one-variable case, so that $P(1)$ is true. We can invoke our inductive hypothesis to suppose that $P(n_0)$ is true for some $n_0$. We next show that $n = n_0 + 1$ is true.

By definition, our ring $R[x_1, \dots, x_{n_0}] = S[x_n]$ for

\[ S = R[x_1, \dots, x_{n_0}]. \]

By our inductive hypothesis, we know that (1) $S$ is an integral domain. Since $P(1)$ is true, we know that $S[x_{n_0}]$ is an integral domain. By (2), we know that $S^\times = R^\times$. By $P(1)$, we see that $S[x_n] = R^\times$. This show that $P(n)$ is true for all $n$.

Say $(R, +, \cdots)$ is a ring with $1 \ne 0$ and $I \subsetneqq R$ is a ideal. Denote $S = R[x_1, \dots, x_n]$.

1. $J = I[x_1, \dots, x_n]$ is a proper ideal of $S$
2. $S/J = (R/J)[x_1, \dots, x_n]$
3. Moreover, say $R$ is commutative. If $I \subset R$ is a prime ideal of $R$, then $J$ is a prime ideal of $S$.

1., 2. Denote $\overline{R} = R/I$ as a ring with identity $1 \ne 0$, which holds since we are working with a proper ideal. We know both $S$ and $\overline{R}[x_1, \dots, x_n]$ is also a ring with $1 \ne 0$.

Consider the following "reduction mod $I$" map:

\[ \phi: R[x_1,\dots, x_n] \to \overline{R}[x_1,\dots, x_n] \]

where if $\displaystyle p = \sum_{i_1, \dots, i_n}^{d_1, d_2, \cdots, d_n}a(i_1, \dots, i_n)x_1^{i_1}\cdots x_n^{i_n}$ then

\[ \overline{p} = \sum_{i_1, \dots, i_n}^{d_1, d_2, \cdots, d_n}\overline{a(i_1, \dots, i_n)}x_1^{i_1}\cdots x_n^{i_n} \]

where $\overline{a} = a + I$ in $R/I$. \ Claim: This is a ring homomorphism. In fact, this map is surjective. \ Neither of these are difficult to show.

Observe that

\[ \ker(\phi) = \Big\{p = \sum_{i_1, \dots, i_n}^{d_1, d_2, \cdots, d_n}a(i_1, \dots, i_n)x_1^{i_1}\cdots x_n^{i_n} mid a(i_1, \dots, i_n)\in I \Big\} = I[x_1, \dots, x_n]. \]

The First Isomorphism Theorem for Rings states that

\[\begin{align*} S/J &\cong R[x_1, \dots, x_n]/\ker(\phi)\\ &\cong \im(\phi)\\ &= (R/I)[x_1, \dots, x_n]. \end{align*}\]

At this point, we've shown (2). Now observe that $J \subsetneqq R$ since $(R/I)[x_1, \dots, x_n]$ is a ring with identity $1 \ne 0$. This proves (1). To show (3), say $R$ is a commutative ring and $I \subset R$ is a prime ideal. Since $I$ is prime, we see that $(R/I)$ is an integral domain. Hence we see that $(R/I)[x_1, \dots, x_n]$ is an integral domain. Since $S/J \cong (R/I)[x_1, \dots, x_n]$, and we know that $R[x_1, \dots, x_n]$ is also a commutative ring, we see that $J \subset S$ must be a prime ideal as well.