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The quaternions are a division ring but not a field.

Let \( R \) be a ring and let \( x, y \in R \).

• \( 0_R x = 0_R = x 0_R \)
• \( (-x)y = -(xy) = x(-y) \)
• \( (-x)(-y) = xy \)

Let \( R \) be a commutative ring with unity. The following are equivalent.

• \( R \) is an integral domain.
• \( R \) has no non-zero zero-divisors.
• For all \( x, y, z \in R \) with \( x \neq 0 \) we have \( xy = xz \) implies \( y = z \).

Let \( R \) be a ring with unity. The characteristic of \( R \) is the additive order of \( 1_R \).

A subring \( S \subseteq R \) is an ideal if and only if

1. \( S \neq \emptyset \),
2. for all \( x, y \in S \) we have \( x - y \in S \), and
3. for all \( x \in S \) and all \( r \in R \) we have \( rx, xr \in S \).

Let \( R \) be a ring with ideal \( I \). Then \( R / I \) is a ring under the operations \[ (x + I) + (y + I) = (x + y) + I , \] and \[ (x + I) \cdot (y + I) = (xy) + I . \] Moreover, the function \( \pi_I \colon R \to R / I \) defined by \( \pi_I(x) = x + I \) is a ring homomorphism.

Let \( f \colon R \to S \) be a ring homomorphism and \( I \) an ideal. Then \( f \) factors through \( \pi_I \) (i.e., there is an \( \bar{f} \colon R / I \to S \) with \( f = \bar{f} \circ \pi_I \)) if and only if \( I \subseteq \ker(f) \). Moreover, if it does factor through \( \pi_I \), then this factorization is unique.

Let \( f \colon R \to S \) be a ring homomorphism. Then \( R / \ker(f) \cong \mathrm{im}(f) \).

Note: Our proof in class uses the homomorphism theorem to do all of the heavy lifting. It may be useful for students to work through a more element-wise proof on your own. You can use the proof of the first isomorphism theorem from group theory as inspiration.

Let \( S \subseteq R \) and \( I \leq R \).

1. \( S + I = \{ s + x : s \in S, x \in I\} \) is a subring of \( R \).
2. If \( S \leq R \), then \( S + I \leq R \).

Let \( S \) be a subring of \( R \) and \( I \) an ideal.

1. \( I \cap S \) is an ideal of \( S \).
2. \( I \) is an ideal of \( I + S \).
3. \( S / (I \cap S) \cong (I + S) / I \).

Let \( R \) be a ring with ideals \( I \) and \( J \) such that \( J \subseteq I \).

1. \( J \) is an ideal of \( I \).
2. \( I / J \) is an ideal of \( R / J \).
3. \( (R / J) / (I / J) \cong R / I \).

A polynomial \( p \in \mathbb{F}[x] \) has a zero at \( \alpha \in \mathbb{F} \) if and only if \( x - \alpha \mid p(x) \).

Let \( p \) be a polynomial over a field \( \mathbb{F} \). If \( p(x) = (x - \alpha) q(x) + r(x) \) as in the Poly QRT, then \( r(x) = p(\alpha) \) is a constant polynomial.

Note: we discussed how we really proved this when we proved the prior proposition.

Let \( p \) be a polynomial over field \( \mathbb{F} \) with degree \( n \). Then \( p \) has at most \( n \) distinct zeroes.

Let \( \mathbb{F} \) be a field and let \( a \) and \( b \) be polynomials over \( \mathbb{F} \). If not both \( a(x) = 0 \) and \( b(x) = 0 \), then \( \gcd(a(x), b(x)) \) exists. Moreover, there are \( s(x), t(x) \in \mathbb{F}[x] \) satisfying \( \gcd(a(x), b(x)) = a(x)s(x) + b(x)t(x) \). Finally, the greatest common divisor is unique.