*Algebra*

## 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.