The property of category of Semirings I’m now thinking about the property of category of semirings Rig.
Is it complete or co-complete?
I think that Rig has projective and inductive limits, and finite products and co-products, so it should have (general) products and co-products; and it also has equalizers. And I think it has co-equalizer.
I just think like below:
Projective and inductive limits is equal to projective and inductive limits as in the category of Set.
And finite product is product, and finite co-product is tensor-product over $\mathcal{N}_0$.
(General) (co)products can be represented by projective limits and finite products so Rig has (co)products.
Equalizer is equalizer as in the Set, so Rig is complete.
Coequalizer is may be constructed like below:
Take the most smallest equivalence relation compatible with additions and multiplications, and take the equivalent class (like taking ideal and taking quotient class).
 A: For most "working mathematicians", it's probably enough to be aware of the following result. 
Theorem 0: For any Lawvere theory $T$, let $C^T$ be the category of $T$-algebras in $C$, where $C$ is a complete, cocomplete cartesian closed category (e.g., $C = \mathbf{Set}$). Then $C^T$ is again complete and cocomplete. 
I'll go through this from soup to nuts (with the final proof of Theorem 0 "below the fold"); I believe many such results are scattered throughout the literature, but it could be tricky to find everything collected in one place under these hypotheses. I've tried to keep the arguments simple and brief. 
Recall that "Lawvere theories" are Lawvere's slick way of doing finitary universal algebra. A Lawvere theory is a category $T$ with finite (cartesian) products, equipped with an object $x$ such that every object of $T$ is isomorphic to a finite power $x^n$. A different way of saying it is that, letting $\text{Fin}$ be the category of finite sets, the (unique-up-to-isomorphism) product-preserving functor $i: \text{Fin}^{op} \to T$ that sends the one-element set $1$ to $x$ is essentially surjective. 
For a variety of algebras (such as the variety of rigs or semirings), one should think of its corresponding Lawvere theory $T$ as the category opposite to that of finitely generated free algebras. Then elements of the free algebra on $n$ generators are in natural bijection with morphisms $x^n \to x$ in $T$, which should be thought of as encoding definable operations of the theory modulo provable equality. For $C$ a category with finite cartesian products, we define a $T$-algebra to be a functor $A: T \to C$ that preserves finite products: this is given by an object $A(x)$ of $C$ plus a bunch of operations $A(x)^n \to A(x)$, one for each morphism $x^n \to x$ of $T$. Morphisms between $T$-algebras are just natural transformations; this amounts to preservation of operations. If $C^T = \mathbf{Prod}(T, C)$ is the category of $T$-algebras, then the forgetful functor $U: C^T \to C$ takes $A: T \to C$ to $A(x) = A(i(1))$. 
Theorem 1: The forgetful functor $U: C^T \to C$ is monadic. 
The proof we offer consists in checking that the hypotheses of a crude monadicity theorem hold: 


*

*$U$ has a left adjoint (Proposition 0); 

*$U$ reflects isomorphisms (Proposition 1); 

*$C^T$ has reflexive coequalizers, and $U: C^T \to C$ preserves them (Proposition 2). 
Proposition 0: $U: C^T \to C$ has a left adjoint. 
Proof: (Sketch) The left adjoint $F: C \to C^T$ (the free $T$-algebra functor) takes an object $c$ to a functor $Fc: T \to C$ defined by a coend over the category $\text{Fin}$ of finite sets: 
$$\int^{n \in \text{Fin}} T(i(n), -) \cdot c^n$$ 
as calculated pointwise in $C$ (here $T(i(n), -)$ denotes a representable functor $\hom_T(i(n), -)$). In other words, we have families of canonical maps 
$$T(i(n), -) \cdot \text{Fin}(n, m) \to T(i(m), -), \qquad \text{Fin}(n, m)  \cdot c^m \to c^n$$
which we use to construct a pair of parallel maps whose coequalizer is $F c$:  
$$\sum_{m, n \in \text{Fin}} T(i(n), -) \cdot \text{Fin}(n, m) \cdot c^m \rightrightarrows \sum_{n \in \text{Fin}} T(i(n), -) \cdot c^n \stackrel{\text{coeq}}{\to} F c$$ 
There are a number of details to check that this works, but the arguments can be made soft and conceptual. See for example here for details. QED 
Proposition 1: The functor $U: C^T \to C$ reflects isomorphisms.
Proof: $U$ is a composite 
$$\mathbf{Prod}(T, C) \stackrel{\mathbf{Prod}(i, C)}{\to} \mathbf{Prod}(\text{Fin}^{op}, C) \simeq C$$ 
where $\mathbf{Prod}(i, C)$ reflects isomorphisms because $i: \text{Fin}^{op} \to T$ is essentially surjective. QED 
Proposition 2: $C^T$ has reflexive coequalizers, and they are preserved by $U: C^T \to C$. 
The proof follows from Lemmas 1 and 2 below. 
Lemma 1: Let $C$ be a category with reflexive coequalizers, and with products that distribute over reflexive coequalizers (as in the case of cartesian closed categories). Then the product functor $C \times C \to C: (c, d) \mapsto c \times d$ preserves reflexive coequalizers. 
Proof: The key observation is that the generic reflexive fork $D = \left(a \stackrel{i}{\to} b \stackrel{p}{\underset{q}{\rightrightarrows}} a\right)$ (with $pi = 1_a = qi$ and no other imposed equations) is a sifted category, meaning that the diagonal functor $D \to D \times D$ is final. 
Now suppose we have a reflexive fork diagram $D \to C \times C$ given by two reflexive fork diagrams $F, G: D \to C$ in our category $C$. We have 
$$\array{
(\text{colim}_{d \in D} F(d)) \times (\text{colim}_{d' \in D} G(d')) & \cong & \text{colim}_{d \in D} [F(d) \times \text{colim}_{d' \in D} G(d')] \\ 
 & \cong & \text{colim}_{d \in D} \text{colim}_{d' \in D} [F(d) \times G(d')] \\ 
 & \cong & \text{colim}_{(d, d') \in D \times D} F(d) \times G(d') \\ 
 & \cong & \text{colim}_{d \in D} F(d) \times G(d)
}$$ 
where the first two isomorphisms are come from products distributing over reflexive coequalizers, the third comes from a "Fubini theorem", and the last from the finality of $\Delta: D \to D \times D$. This shows the product applied to a reflexive coequalizer of $\langle F, G\rangle: D \to C \times C$ is canonically isomorphic to the reflexive coequalizer of the product $F \times G: D \to C$, as was to be shown. QED
Lemma 2: If products distribute over reflexive coequalizers in $C$, then $C^T = \mathbf{Prod}(T, C)$ admits reflexive coequalizers and $U: \mathbf{Prod}(T, C) \to C$ preserves them. 
Proof:: Let $[T, C]$ denote the category of all functors $T \to C$; this certainly has reflexive coequalizers if $C$ has them. For any category $A$, let $A^D$ be the category of reflexive fork diagrams in $A$. Let 
$$j: \mathbf{Prod}(T, C) \hookrightarrow [T, C]$$ 
be the full inclusion. We show $\mathbf{Prod}(T, C)$ is closed under reflexive coequalizers as computed in $[T, C]$, i.e., that the composite (where "colim" means coequalizer) 
$$\mathbf{Prod}(T, C)^D \underset{j^D}{\hookrightarrow} [T, C]^D \underset{\text{colim}}{\to} [T, C]$$ 
factors through the full inclusion $j: \mathbf{Prod}(T, C) \hookrightarrow [T, C]$. This will mean both that $\mathbf{Prod}(T, C)$ has reflexive coequalizers and that $j$ preserves them, whence $U$ which is the composite 
$$\mathbf{Prod}(T, C) \underset{j}{\hookrightarrow} [T, C] \underset{[x, 1_C]}{\to} [1, C] = C$$ 
also preserves them. 
But if $\theta_1, \theta_2$ are two objects of $T$, and if $F: D \to \mathbf{Prod}(T, C)$ is a reflexive fork, then $\text{colim}\; j F$ preserves the product $\theta_1 \times \theta_2$ since 
$$\array{
[\text{colim}_{d \in D} F(d)](\theta_1) \times [\text{colim}_{d' \in D} F(d')](\theta_2) & \cong & \text{colim}_d [F(d)(\theta_1)] \times \text{colim}_{d'} [F(d')(\theta_2)] \\ 
 & \cong & \text{colim}_d [F(d)(\theta_1) \times F(d)(\theta_2)] \\ 
 & \cong & \text{colim}_d F(d)(\theta_1 \times \theta_2) \\ 
 & \cong & [\text{colim}_d F(d)](\theta_1 \times \theta_2)
}$$ 
where the first and last isomorphisms hold since colimits in $[T, C]$ are computed pointwise, the second isomorphism holds by Lemma 1, and the third holds since each $F(d): T \to C$ is product-preserving. QED 

The proof of Theorem 0 is immediate from Theorem 1, Proposition 2, and the following result, originally due to Linton (Coequalizers in categories of algebras, Seminar on triples and categorical homology theory, Lecture   Notes in Mathematics 80, Springer-Verlag, Berlin, Heidelberg, New  York (1969), 75-90). 
Theorem 2: For any monadic functor $U: A \to C$, if $C$ is complete and cocomplete and $A$ has reflexive coequalizers, then $A$ is complete and cocomplete. 
Proof: It is well-known that monadic functors reflect limits, so if $C$ is complete, then $A$ must also be complete. We now show that $A$ has coproducts and coequalizers. 
Denote the monad by $M = UF$ where $F$ is left adjoint to $U$; let $\varepsilon: FU \to 1_A$ be the counit of the adjunction, and $\eta: 1_C \to UF$ the unit. To show $A$ has coproducts, let $(c_i, \xi_i)$ be a collection of algebras. Then $F(\sum_i U c_i)$ is the coproduct $\sum_i F U c_i$ in $A$ (since $F$ preserves coproducts and $C$ has them). We have a reflexive fork 
$$\array{
\sum_i F U c_i & \stackrel{\sum_i F \eta U c_i}{\to} & \sum_i F U F U c_i & \stackrel{\overset{\sum_i \varepsilon F U c_i}{\longrightarrow}}{\underset{\sum_i F \xi_i}{\longrightarrow}} & \sum_i F U c_i
}$$ 
and it is not difficult to show that the coequalizer in $A$ of this diagram is the coproduct $\sum_i c_i$. 
Finally, general coequalizers in $A$ are constructed from coproducts and reflexive coequalizers: given a parallel pair $f, g: c \rightrightarrows d$ in $C^T$, the coequalizer of $f$ and $g$ is the colimit of the reflexive fork 
$$\array{
d & \to & c + d & \stackrel{(f, 1_d)}{\underset{(g, 1_d)}{\rightrightarrows}} & d}$$ 
where the first arrow is the coproduct coprojection. QED 
