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This question is self-contained. In the book "Monomial Ideals", by Herzog and Hibi, p. 45, we have the following definition:

Definition: Let $K$ be a field. A one-parameter flat family of $K$-algebras is a family of $K$-algebras $R_a, \, a \in K$, for which there exists a $K$-algebra $R$ and a flat $K$-algebra homomorphism $K[t] \rightarrow R$ whose fibres $R/(t-a)R$ are isomorphic to $R_a$ for all $a \in K$ ($t$ is an indeterminate over $K$).

Question: i don't understand two things. First what does the quantity $R/(t-a)R$ mean, since $t$ is not in general an element of $R$. Is it the image of $t$ under the $K$-algebra morphism $K[t] \rightarrow R$? Second, why are the quantities $R/(t-a)R$ called fibres? I know that in topology fibre is the inverse image of a closed set under a continuous morphism and in commutative algebra i have seen the word fibre been used in the context of inverse images of prime ideals.

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Yes, $t \in R$ also denotes the image of $t \in K[t]$. This is somewhat confusing, especially when the homomorphism $K[t] \to R$ is something like below.

You can image $K[t] \to R$ as a morphism of (affine) schemes $\mathrm{Spec}(R) \to \mathbb{A}^1_K$. The scheme-theoretic fiber over a $K$-point $a$ corresponding to the maximal ideal $(t-a)$ is $\mathrm{Spec}(R \otimes_{K[t]} K[t]/(t-a)) = \mathrm{Spec}(R/(t-a))$. This explains the terminology. It is well-known that the underlying topological space of the scheme-theoretic fiber is the usual topological fiber.

In this language, the definition becomes really geometric (as already indicated by the terminology): A flat family is given (affine) schemes $X_a$ parametrized by the $K$-points $a$ of $\mathbb{A}^1_K$, such that there is a flat (affine) scheme $X$ and a morphism $X \to \mathbb{A}^1_K$ whose fiber at $a$ identifies with $X_a$.

You can also understand this without the language of schemes. If $R = K[t]$, then $K[t] \to R$ corresponds to a polynomial $p \in R[t]$ (the image of $t$) (which you may imagine as the morphism $p : \mathbb{A}^1_K \to \mathbb{A}^1_K$), and the fiber of $R$ at $t-a$ is $R/(p-a) = K[t]/(p-a)$. If $K$ is algebraically closed, $p$ is monic, and not constant $a$, we may write $p-a = (t-\lambda_1)^{k_1} \cdot \dotsc \cdot (t-\lambda_n)^{k_n}$, where $\{\lambda_1,\dotsc,\lambda_n\}$ equals the usual fiber $\{\lambda \in K : p(\lambda)=a\}$. The Chinese Remainder Theorem gives $R/(p-a) \cong \prod_i K[t]/(t-\lambda_i)^{k_i}$. As you can see, the fiber of $p$ at $a$ is encoded in this algebra, including multiplicities.

Specific example: Take $K=\mathbb{C}$, $p = x^3$ (i.e. we look at $\mathbb{A}^1 \to \mathbb{A}^1,~ t \mapsto t^3$) and say $a=1$. If $\zeta_3$ is a third primitive root of unity, then the fiber is $\mathbb{C}[t]/(p-a) = \mathbb{C}[t]/(t-1) \times \mathbb{C}[t]/(t-\zeta_3) \times \mathbb{C}[t]/(t-\zeta_3^2) \cong \mathbb{C}^3$, a product of three fields. However, for $a=0$, the fiber is $\mathbb{C}[t]/(t^3)$, not reduced.

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