In this lecture, we talked about rings, ideals, and ring homomorphisms.
1Rings and ideals¶
Last time, we covered rings and ideals. A given ring $R$ always has at least two ideals: $(0)$, and $R$ (trivial). If $R$ is a field, those are the only two ideals.
1.1Principal ideals¶
A principal ideal is an ideal generated by a single element ($(a) = aR$ for some $a \in R$). For $R = \mathbb{Z}$ or $R = F[X]$ (where $F$ is a field), every ideal is principal.
Examples of nonprincipal ideals:
 $(2, x)$ in $\mathbb{Z}[x]$, meaning, all the $f(x)$ such that $f(0)$ is even (recall that $\mathbb{Z}$ is not a field)
 In $F[x, y]$ (multivariable polynomials with coefficients from a field), we have the set of polynomials with constant coefficients, which we can write as $\{\sum a_{nm} x^ny^m$ such that $a_{00} = 0\}$. (To show that this is not a principal ideal, we note that the gcd of $x^n$ and $y^m$ is 1, but 1 is not part of this ideal or something.)
1.2Ring homomorphisms¶
A ring homomorphism is a function from one ring ($R_1$) to another ($R_2$) that agrees with the structure of the rings, namely, that:
 $f(1_{R_1}) = 1_{R_2}$^{1}
 $f(a + b) = f(a) + f(b)$ (with $a, b \in R_1$)
 $f(a \cdot b) = f(a) \cdot f(b)$ (with $a, b \in R_1$)
Note that the zero element is also preserved (i.e., we have $f(0_{R_1}) = 0_{R_2}$). This does not need to be its own property because it follows directly from the second axiom (we can call it "property 0" if we like). The proof is left as an exercise for the reader (it's simple  it relies on the fact that rings have additive inverses). Incidentally, property 1 does not generally follow from property 3, since elements in a ring don't necessarily have multiplicative inverses, and so it's necessary to make it explicit.
Another property that follows from the axiomatic ones (including "property 0") is that $f(a) = f(a)$. Proof: $f(a) + f(a) = f(a a) = f(0) = 0$, then we just add $f(a)$ to both sides.
1.2.1Examples of ring homomorphisms¶
 $f: R \to R \times R$ such that $f(a) = (a, 0)$ is NOT a ring homomorphism, since it doesn't satisfy property 1.
 The identity function (trivial though): $f:R\to R$, $f(a) = a$
 Inclusion of one ring into another. Since $\mathbb{Z} \subset \mathbb{Q} \subset \mathbb R \subset \mathbb C$, then we can have a function from, e.g., $\mathbb Z \to \mathbb Q$ that sends 1 to 1/1, 2 to 2/1, etc.
 $f: \mathbb Z \to \mathbb Z / n \mathbb Z$. $f(a) = [a] = a + n\mathbb Z$ (the congruence class of $a$). Incidentally, this is not injective (clearly not onetoone)
 $f: F[X] \to F$, maps $p(x)$ to $p(0)$ (so the value of the constant coefficient). To prove that this is a ring homomorphism, we first prove that $f(1) = 1$, then that it satisfies properties 2 and 3, etc. This was question 3 on assignment 3 (link to solutions).
 Let $F[\epsilon] = \{a + b \epsilon \}$ where $\epsilon^2 = 0$ be the ring of dual elements. Then $ev'_a$ is a function $F[X] \to F[\epsilon]$, that maps $p(x) \to p(a) = p'(a)\epsilon$ for some given $a$ (where $p'(a)$ is the derivative). Proof that this is a ring homomorphism: 1. $ev'_a(1) = 1(a) + 0 = 1+0\epsilon$. 2. Leaving it as an exercise for the reader. Involves associativity/commutativity of addition, etc. Same for 3.
1.2.2Related terminology¶
The image of a ring homomorphism $f: R_1 \to R_2$ is given by $\{f(a), \,\forall a \in R_1\}$. This is a subset of $R_2$, or equal to $R_2$. It is also a subring of $R_2$, since it contains 1 (by property 1 of a ring homomorphism), and is closed under addition and multiplication. Note that the image is equal to $R_2$ if and only if the function is surjective.
The kernel of a ring homomorphism is given by $\{a \in R_1$ such that $f(a) = 0\}$ (so the things in the domain that map to 0). This is a subset of $R_1$, and also an ideal.

Some definitions of ring homomorphisms that I've seen on the Internet don't include this property. not sure if it matters. ↩