Brian Bi
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Section 11.2. Polynomial Rings

Exercise 11.2.1 Suppose $$x^2 + x + 1$$ times some polynomial equals $$x^4 + 3x^3 + x^2 + 7x + 5$$. Clearly the other factor must be monic and quadratic, and its constant term must be 5. So we can write it in the form $$x^2 + ax + 5$$. Expanding out the product $$(x^2 + x + 1) (x^2 + ax + 5)$$ yields $$x^4 + (a + 1)x^3 + (a + 6)x^2 + (a + 5)x + 5 = x^4 + 3x^3 + x^2 + 7x + 5$$. Equating terms of like powers yields \begin{align*} a + 1 &= 3 \\ a + 6 &= 1 \\ a + 5 &= 7 \end{align*} The second and third equations taken together imply that $$1 = -6$$, which is only true in $$\mathbb{Z}/(7)$$ and the degenerate case $$\mathbb{Z}/(1)$$ (the zero ring). In the former case the solution $$a = 2$$ exists, while in the latter case, both polynomials are simply zero, so one certainly divides the other. Therefore, the only solutions are $$n = 1$$ and $$n = 7$$.

Exercise 11.2.2 The set $$F[[t]]$$ can be given the structure of a ring by defining:

• $$\left[\sum_{i=0}^\infty a_i t^i\right] + \left[\sum_{i=0}^\infty b_i t^i\right] = \sum_{i=0}^\infty (a_i + b_i)t^i$$
• $$\left[\sum_{i=0}^\infty a_i t^i\right] \left[\sum_{i=0}^\infty b_i t^i\right] = \sum_{i=0}^\infty \sum_{j + k = i} a_j b_k t^i$$

It is clear that $$(F[[t]], +)$$ is isomorphic to a direct product of countably many copies of $$F$$ considered as a group, and therefore forms an abelian group with the zero power series $$0 + 0t + 0t^2 + \ldots$$ as the identity. It is also clear that multiplication is commutative, from the way we have defined it.

Suppose $$a_0 = 1$$ and $$a_i = 0$$ for all $$i \ge 1$$. Then $\left[\sum_{i=0}^\infty a_i t^i\right] \left[\sum_{i=0}^\infty b_i t^i\right] = \sum_{i=0}^\infty \sum_{j + k = i} a_j b_k t^i = \sum_{i=0}^\infty a_0 b_i t^i = \sum_{i=0}^\infty b_i t^i$ so the power series $$1 + 0t + 0t^2 + \ldots$$ is the multiplicative identity.

Let $$a = a_0 + a_1 t + \ldots, b = b_0 + b_1 t + \ldots, c = c_0 + c_1 t + \ldots$$ be given. Then \begin{align*} ac &= \sum_{i=0}^\infty \sum_{j + k = i} a_j c_k t^i \\ bc &= \sum_{i=0}^\infty \sum_{j + k = i} b_j c_k t^i \\ ac + bc &= \sum_{i=0}^\infty \left[\sum_{j + k = i} a_j c_k + \sum_{j+k = i} b_j c_k \right] t^i = \sum_{i=0}^\infty \sum_{j + k = i} (a_j c_k + b_j c_k) t^i = \sum_{i=0}^\infty \sum_{j+k = i} (a_j + b_j)c_k t^i \\ a + b &= \sum_{i=0}^\infty (a_i + b_i) t^i \\ (a + b)c &= \sum_{i=0}^\infty \sum_{j+k = i} (a_j + b_j)c_k t^i \end{align*} so we always have $$(a + b)c = ac + bc$$; $$(F[[t]], +, \cdot)$$ satisfies the distributive property.

Finally, let $$a, b, c$$ be given. Then \begin{align*} (ab)c &= \left[\sum_{i=0}^\infty \sum_{j+k} a_j b_k t^i\right] \left[\sum_{i=0}^\infty c_i t^i \right] \\ &= \sum_{i=0}^\infty \sum_{m + n = i} \left[\sum_{j + k = m} a_j b_k\right] c_n t^i \\ &= \sum_{i=0}^\infty \sum_{m + n = i} \sum_{j + k = m} a_j b_k c_n t^i \\ &= \sum_{i=0}^\infty \sum_{j + k + n = i} a_j b_k c_n t^i \\ a(bc) &= (bc)a \\ &= \sum_{i=0}^\infty \sum_{j + k + n = i} b_j c_k a_n t^i \\ &= \sum_{i=0}^\infty \sum_{k + n + j = i} b_k c_n a_j t^i \\ &= (ab)c \end{align*} so multiplication is also associative. The above results establish that $$(F[[t]], +, \cdot)$$ is a ring.

We claim that $$a_0 + a_1 t + \ldots$$ is a unit in $$F[[t]]$$ iff $$a_0 \ne 0$$. One direction is easy: if $$a_0 = 0$$ then the constant term will be 0 in any product, so $$a_0$$ is not a unit. For the other direction, define $$b_0 = a_0^{-1}$$ and for $$i \ge 1$$, define $b_i = -a_0^{-1} \sum_{j=0}^{i-1} b_j a_{i-j}$ where $$b_i$$ depends on $$b_0, b_1, \ldots, b_{i-1}$$. If $$c_0 + c_1 t + \ldots = (a_0 + a_1 t + \ldots)(b_0 + b_1 t + \ldots)$$, then $$c_0 = a_0 b_0 = 1$$ and for $$i \ge 1$$, \begin{align*} c_i &= \sum_{j + k = i} a_j b_k \\ &= \sum_{j=0}^i a_{i-j} b_j \\ &= a_0 b_i + \sum_{j=0}^{i-1} a_{i-j} b_i \\ &= -\sum_{j=0}^{i-1} b_j a_{i-j} + \sum_{j=0}^{i-1} a_{i-j} b_i \\ &= 0 \end{align*} therefore $$c_0 + c_1 t + \ldots = 1$$, and $$b_0 + b_1 t + \ldots$$ is the inverse of $$a_0 + a_1 t + \ldots$$. So the latter is a unit.