Theorem. let $f : S \to T$ be a morphism of semigroups. Then $f$ preserves subsemigroups.
Proof. let $A \subseteq S$ then $\forall a_1,a_2 \in A: a1_a2 \in A$. So consider the image $f(A)$ and let $t_1,t_2 \in f(A)$ then there exists $a, b$ such that $f(a) = t_1$ and $f(b) = t_2$. Now consider the product $f(a)f(b)$ then by the definition of semigroup homomorphisms this is equal to $f(ab)$ but we have that $ab \in A$ so $t_1t_2 = f(ab) \in f(A)$ so that $f(A)$ is composition closed. $\square$
Theorem. let $f: S \to T$ be a morphism of semigroups. Then $f$ reflects subsemigroups.
Proof. let $B \subseteq T$ then $f^{-1}(B)$ is a subset of $S$. Suppose that $x,y \in f^{-1}(B)$ then $f(x) \in B$ and $f(y) \in B$. The product $f(xy)$ is equal to $f(x)f(y)$ by the definition of semigroup homomorphisms. Then since $f(x)$ is in $B$, $f(y)$ is in $B$ and $B$ is composition closed $f(xy) = f(x)f(y)$ is in $B$ which implies that $f^{-1}(B)$ is composition closed. $\square$
Theorem. let $f: M \to N$ be a morphism of monoids. Then $f$ preserves and reflect submonoids.
Proof. (1) let $S \subseteq M$ be a submonoid of $M$ then $1 \in S$ and by the definition of monoid homomorphisms $f(1)$ is the identity of $M$ so $1 \in f(S)$. (2) In the other direction, suppose that $S \subseteq M$ is a submonoid so that $1_N \in S$ then by the definition of monoid homomorphisms $f(1_M) = 1_M$ so that $1_M \in f^{-1}(S)$ so that monoid maps reflect identities. (3) then since $f$ preserves and reflects identities as well as subsemigroups it necessarily must also do so for submonoids as well. $\square$
Theorem. let $f: G \to H$ be a group homomorphism. Then $g$ preserves and reflects subgroups.
Proof. (1) let $S \subseteq G$ be a subgroup. Then for all $s \in S$ we have that $-s \in S$. Then consider $f(S)$ if $h$ is in $f(S)$ then we have that there exists $g$ such that $f(g) = h$. Then consider $-h$ by the definition of group homomorphisms $f(-g) = -h$ but then $-g$ is in $S$ because $S$ is a subgroup and therefore inverse closed. It follows that $f(-g) = -h$ is in $f(S)$ so that the image is a subgroup.
(2) in the other direction suppose that $S \subseteq H$ is a subgroup. Then $\forall s \in S : -s \in S$. Suppose that $g \in G$ and $f(g) \in S$. Then $f(-g) = -f(g)$ by the definition of group homomorphisms. By the fact that $S$ is a subgroup and $f(g)$ is in $S$ we have that $-f(g)$ is in $S$ which implies that $f^{-1}(S)$ is inverse closed for all $g \in f^{-1}(S)$. It follows that it is a subgroup. $\square$
We saw that monotone maps reflect ideals over thin categories. A generalization of this which is applicable to semigroups involves the reflection of semigroup ideals, which can be either left, right, or two sided. This produces a functor from the category of semigroups to the category of topological spaces.
Theorem. let $f : S \to T$ be a morphism of semigroups. Then $f$ reflects left, right, or two sided ideals.
Proof. let $I$ be a left ideal of $T$ then we have that for all $i \in I$ and $t \in T$ it holds that $it \in T$. So consider $f^{-1}(I)$ then if we have $x \in f^{-1}(I)$ it follows that $f(x) \in I$. Let $s \ in S$ then $f(sx) = f(s)f(x)$ and since $f(x)$ is in $I$ and $I$ is a left ideal $f(s)f(x)$ is in $I$. So $f$ reflects left ideals and by dualizing it reflects right ideals. By combining this $f$ reflects two sided ideals. $\square$
It is not enough for us to show that semigroup homomorphisms reflect subsemigroups. We would also like to know that they reflect prime ideals. This will then allow us to reproduce the familiar theorems of ring theory dealing with prime ideals.
Theorem. let $f : A \to B$ be a morphism of semigroups then $f$ reflects prime subsemigroups.
Proof. let $P \subseteq B$ be a prime subsemigroup then we have that $Q = B-P$ is a subsemigroup and in total we have $P,Q$ are subsemigroups with $P \cup Q = B$ and $P \cap Q = \emptyset$. Every element of $A$ produces some image, so by the fact that inverse images are lattice homomorphisms we have $f^{-1}(P) \cup f^{-1}(Q) = A$ and $f^{-1}(P) \cap f^{-1}(Q) = \emptyset$. So by the fact that $f$ reflects subsemigroups it follows that $f^{-1}(P)$ is a subsemigroup with complement subsemigroup $f^{-1}(Q)$. $\square$
Corollary. let $f : S \to T$ be a morphism of semigroups. Then $f$ reflects prime ideals.
The one other class of ideals we are really interested in are the radical ideals, especially in commutative algebra because of theri relationship to algebraic varieties under Hilbert'z nullstellensatz. There is a corresponding notion for semigroups, and so by considering that we will get closer to solving some problems related to radical subsemigroups and ideals.
Theorem. let $f : S \to T$ be a morphism of semigroups. Then $f$ reflects radical subsemigroups.
Proof. let $B \subseteq T$ be a radical subsemigroup. Then $f^{-1}(B)$ is a subsemigroup of $S$ because $f$ reflects subsemigroups, so it remains to show that $f^{-1}(B)$ is a radical subsemigroup. Let $x \in f^{-1}(B)$ and suppose that $y^n = x$ Then we can apply $f$ to both sides to get $f(y^n) = f(x)$ and expanding this $f(y)^n = f(x)$. Now since, $f(x)$ is in $B$ and $B$ is radical this implies that $f(y) \in B$. So to sum up this means that $f^{-1}(B)$ is a radical subsemigroup. $\square$
Corollary. let $f : S \to T$ be a morphism of semigroups. Then $f$ reflects semigroup radical ideals.
We can apply these semigroup theorems to our favourite algebraic structures, all of which are constructed out of semigroups. Like lattices (thin categories with all products and coproducts), semirings, and rings. In particular, this is enough to recover that the pre image of a prime ideal of a ring is a prime ideal.
Corollaries.
- Lattice homomorphisms preserve and reflect sublattices
- Semigroup homomorphisms preserve and reflect subsemirings
- Ring homomorphisms preserve and reflect subrings
- Ring homomorphisms reflect left, right, and two sided ideals, prime ideals, and radical ideals
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