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\[\sum_{i=0}^\infty a_ix^i = a_0 + a_1x + \cdots + a_nx^n + \cdots,\]Let \(R\) be a ring. A polynomial \(f(x)\) over \(R\) is a formal sum
where \(a_i \in R\), and \(a_i = 0\) for all but a finite number of values of \(i\).
Formal sum 代表「\(+\)」只是形式上的，不能實際代值到 indeterminate \(x\) 中，也就是不能 evaluate。
”\(a_i = 0\) for all but a finite number of values of \(i\)” ：除了有限個 \(i\) 之外，其他 \(a_i\) 都等於 \(0\)；意思就是幾乎 for all。
\[f(x) + g(x) = \sum_{i\ge 0} (a_i+b_i)x^i, \\ f(x)g(x) = \sum_{n\ge 0}\Big(\sum_{i=0}^na_ib_{n-i} \Big)x^n.\]Let \(R\) be a ring. Let \(f(x) = \sum_{i=0}^\infty a_ix^i\) and \(g(x) = \sum_{i=0}^\infty b_ix^i\) be two polynomials over \(R\). Define addition and multiplication by
Then the set \(R[x]\) of all polynomials over \(R\) is a ring under these two operations. Furthermore, if \(R\) is commutative, then so is \(R[x]\) (and vice versa); if \(R\) is a ring with unity, then \(R[x]\) also has the multiplicative identity \(1_{R[x]}\), which is identical to \(1_R\).
If \(D\) is an integral domain, then so is \(D[x]\).
For two polynomials \(f(x),g(x)\in D\), if \(f(x)g(x) = 0\), then
\[S_n = \sum_{i=0}^na_ib_{n-i}=0,\ \forall n \ge 0,\]where \(a_i\) and \(b_i\) are the coefficients of \(f\) and \(g\), respectively.
Let’s start off with \(n=0\): \(a_0b_0=0\). Since \(D\) has no zero divisors, this implies that either \(a_0\) or \(b_0\) is zero. If we continue evaluating \(S_n\) from smaller \(n\) to larger ones, we can reduce \(S_n\) to the following form:
\[S_n = a_kb_{n-k}, \text{ for some } k, 0\le k \le n,\]since there must be \(n\) zeros among \(\{a_0,\cdots,a_{n-1}\}\) and \(\{b_0, \cdots,b_{n-1} \}\).
Now suppose neither \(f(x)\) nor \(g(x)\) is a zero polynomial. For \(i,j \in \mathbb{N}\), there exist \(a_i\) and \(b_j\) such that \(a_i \not=0\) and \(b_j\not =0\). Then from the above observation, we must have
\[S_{i+j} = a_ib_j = 0.\]However, this violates the fact that \(D\) is an integral domain since both \(a_i\) and \(b_j\) are in \(D\). This contradiction means that either \(f\) or \(g\) is zero, i.e. \(D[x]\) is an integral domain. ◼
First define that the degree of the zero polynomial is \(-\infty\).
Suppose \(\deg f(x)=n\) and \(\deg g(x)=m\). Then by this theorem, we have
\[\deg f(x)g(x) = \deg f(x) + \deg g(x),\]which holds even for the zero polynomial.
Now we can see that \(f(x)g(x)\) is the zero polynomial only if at least one of them is the zero polynomial. Thus the statement holds. ◼
The ring \(R[x_1,\cdots,x_n]\) of polynomials in \(n\) indetreminates over \(R\) can be defined as:
For all \(f(x_1,\cdots,x_n) \in R[x_1,\cdots,x_n]\),
\[f(x_1,\cdots x_n) = \sum_{i \ge 0}a_i\Big(\sum_{d_1+\cdots+d_n=i}x_1^{d_1}\cdots x_n^{d_n} \Big).\]Let \(F\) be a field. The set \(F[x]\) is not a field since the inverse of \(x\), \(1/x\), is not in \(F[x]\). On the other hand, we can construct a field containing \(F[x]\) as a subdomain: The field of quotients of \(F[x]\) is \(F(x) = \{f(x)/g(x): f(x),g(x)\in F[x], g(x)\not=0 \}\), called the field of rational functions in \(x\) over \(F\).
注意 notation！\(F(x)\) 是 field，而 \(F[x]\) 只是 integral domain（因為 \(F\) 是 integral domain；by theorem and corollary）。
The addition and multiplication defined above still work well for polynomials of infinite degree, which are called formal power series.
For \(f(x), g(x)\in\mathbb{R}[x]\), we have
\[\deg f(x)g(x) = \deg f(x) + \deg g(x);\]however when \(R = \mathbb{Z}_4\), the above equation is not necessarily valid.
\[\deg f(x)g(x) = \deg f(x) + \deg g(x)\]If \(R\) is an integral domain, then
for all \(f(x), g(x)\in R[x]\).
Proof
Suppose \(\deg f(x) = n\) and \(\deg g(x) = m\). This means that \(a_n\not=0\) and \(b_m\not=0\), where \(a_n\) and \(b_n\) are the coefficients of \(f(x)\) and \(g(x)\), respectively. Since \(R\) is an integral domain, \(a_nb_m \not=0\), and thus the equation holds. ◼
Let \(\phi: R\to R'\) be a ring homomorphism. Then \(R/\text{ker}\phi\) has a canonical ring structure such that \(R/\text{ker}\phi \cong \phi(R)\).
The kernel of a ring homomorphism is called the ideal.
Suppose \(R\) is a subring of \(R'\), and \(R'\) is commutative. Then for \(\alpha \in R'\), the ring homomorphism \(\phi_\alpha: R[x]\to R'\) given by
\[\phi_\alpha(a_0+a_1x+\cdots+a_nx^n) = a_0+a_1\alpha +\cdots+a_n\alpha^n\]is called the evaluation homomorphism. By the First Ring Isomorphism Theorem, we have
\[R[x]/\text{ker}\phi_{\alpha} \cong \phi(R[x]) = R[\alpha].\]為什麼 \(R\) 要是 \(R'\) 的 subring？因為這樣 \(a_0+a_1\alpha +\cdots+a_n\alpha^n\) 才保證在 \(R'\) 中。
Example
Let \(R'\) be a commutative ring with unity and characteristic \(n\), and let \(R = \mathbb{Z}_n \cong \langle 1_{R'} \rangle\). Then
\[\mathbb{Z}_n/\text{ker}\phi_\alpha \cong \langle 1_{R'}, \alpha \rangle.\]見此。
One can regard \(R[\alpha]\) as an extension of \(R\) by plugging in an element \(\alpha\) from a larger ring \(R'\). Then the above result just shows that every ring extension of \(R\) by a single element is isomorphic to \(R[x]/I\) for some ideal \(I\) of \(R[x]\). In general, a finitely generated commutative ring with unity and characteristic \(n\) is isomorphic to
\[\mathbb{Z}_n[x_1,\cdots,x_m]/I,\]where \(I\) is an ideal of \(\mathbb{Z}_n[x_1,\cdots,x_m]\).
Let \(F\) be a field, and \(f(x)\) and \(g(x)\) be polynomials in \(F[x]\). Suppose that \(g(x)\) is not the zero polynomial. Then there exist unique polynomials \(q(x)\) and \(r(x)\) such that
- \(f(x) = g(x)q(x) + r(x)\),
- \(r(x)=0\), or \(\deg r(x) < \deg g(x)\).