Multivariate Polynomials

Let R be a ring and $X_1, ..., X_K$ be variables. We define $R[X_1,...,X_k] = R[X_1,...,X_{k-1}][X_k]$. An element of $R[X_1,...,X_k]$ is of the form $\sum a_{\alpha_1,...,\alpha_k}X_1^{\alpha_1},...,X_k^{\alpha_k}$ (finite sum).

Remark: the order of the variables does not matter $R[X_1, ..., X_k] \cong R[X_{\sigma(1)},...,X_{\sigma(k)}], \sigma \in S_n$. If $(X_i)_{i \in \mathbb{Z}_{>0}}$ are variables we can define the ring of polynomials on finitely many varibles $R[(x_i)_{i \in \mathbb{Z}_{>0}}] = \bigcup R[X_1,...,X_i]$. Each polynomial will only contain finitely many variables.

Proposition: if R is an integral domain then $R[X_1,..,X_n]$ is an integral domain.

Lemma: R is a ring, $I \subseteq R$ is an ideal. $\phi: R[x] \rightarrow R/I[x], \pi: R \rightarrow R/I, \sum a_iX^i \mapsto \sum \pi(a_i)X^i$. $\phi$ is a ring homomorphism and ker($\phi) = I[x] = I \cdot R[x] = \{ \sum a_iX^i: a_i \in I\}$.

Proof: $\phi(\sum_i a_i X^i \cdot \sum_j b_j X^j) =$ $\phi (\sum_k(\sum_{i+j = k}a_ib_j)x^k) =$ $\sum_k \pi(\sum_{i+j = k}a_ib_j)x^k =$ $\sum_k(\sum_{i+j=k}\pi(a_i)\pi(b_j))x^k =$ $\sum_i\pi(a_i)x^i \cdot \sum_j \pi(b_j)x^j =$ $\phi(\sum_i a_ix^i)\phi(\sum_j b_jx^j$ $ker(\phi) =$ $\{\sum a_ix^i : \sum \pi(a_i) x^i =0\}$. For all i, $\pi(a_i) = 0, a_i \in I$ so the kernel is equal to $\{\sum a_ix^i : a_i \in I \}$.

Corollary: Let R be a ring, $I \subseteq R$ a prime ideal, then $I \cdot R[x]$ is prime

Proof: by the first isomorphism theorem $R[x]/I \cdot R[x] \cong R/I[x]$. Since I is prime, R/I is integral so R/I[x] is integral so $R[x]/I \cdot R[x]$ is integral and hence $I \cdot R[x]$ is prime.

Proposition (Gauss’ Lemma): Let R be a UFD, $P \in R[x]$ and $A, B \in F[x]$ such that $P = AB$ (F = Frac(R)) Then there exists $c \in F$ such that $cA, c^{-1}B \in R[x]$.

Proof: $A = \sum a_i c_i^{-1}x^i, B = \sum b_i d_i^{-1}x^i$ where $a_i, b_i, c_i, d_i \in R$ and $b_i, d_i \neq 0$. Then $A = \prod c_i A^*, B = \prod d_i B^*$ where $A^*, B^* \in R[x]$. So $dP = A^*b^*$ where $d = \prod c_i \prod d_i$. Let p be irreducible and let p divide d (then $d = pd^*$). Then in $R/(p)[x], 0 = \bar{A^*}\bar{B^*}$ (reduction mod p). Since (p) is prime, $R/(p)[x]$ is integral so $\bar{A^*} = 0$ or $\bar{B^*} = 0$. Assume $\bar{A^*} = 0$ then all coefficients are in (p) so $A^* = pC, C \in R[x].$ now $pd^*P=pCB^*$ but R[x] is integral so $d^*P = cB^*$. By induction we may assume that $d \in R^*$, we can keep pulling out irreducible factors. Now $P = (d^{-1}A^*)\cdot B^*, P = (d^{-1}\prod c_i)A \cdot (\prod d_i)B$.

If $P \in R[x]$ is a polynomial we define c(P) to be the gcd of the $a_i, P = \sum a_ix^i$ c(P) divides $a_i$ for all i and if d divides $q_i$ for all i, d divides c(P). $P = c(P)P^*$, where $c(P^*) =1$.

Corollary to Gauss’ lemma: Let $P \in R[x]$, c(P) = 1. P is irreducible in $R[x]$ if and only if P is irreducible on $F[x]$ (here $F[x]$ is Frac(R)).