Why quotient ring $ k[x_1,...x_n]/I(Y)$ is k - algebra? Let $Y \subseteq \mathbb{A}^n $ and $f,g \in k[x_1,...,x_n]$ have the same restriction to $T$ if and only if $f-g \in I(Y)$. Why the quotient ring $ k[x_1,...x_n]/I(Y)$ is k - algebra?
Definition k-algebra:
Let $K$ be a field, and let $A$ be a vector space over $K$ equipped with an additional binary operation from $A \times A$ to $A$, denoted here by $·$ (that is, if $x $ and $y$ are any two elements of $A$, then $x · y$ is an element of $A$ that is called the product of $x$ and $y$). Then $A$ is an algebra over $K$ if the following identities hold for all elements $x, y, z \in A$ , and all elements (often called scalars) $a$ and $b$ in $K$:
Right distributivity: $(x + y) · z = x · z + y · z$
Left distributivity: $z · (x + y) = z · x + z · y$
Compatibility with scalars: $(ax) · (by) = (ab) (x · y)$.
An algebra over K is sometimes also called a K-algebra.
I try calculate it.
Let $f,g,h \in  k[x_1,...,x_n]$ have the same restriction then we have $(f+ g) · h $  and $(f+g)-h  \in I(Y)$. But I don't know how can I calculate it.
 A: We don't need the special case of $k[x_1,\dots,x_n]/I(Y)$ here. It's a more general fact that if $A$ is a $k$-algebra and $A/I$ a quotient ring of $A$, then $A/I$ is a $k$-algebra as well, where the multiplication by elements $\lambda\in k$ is simply defined as $\lambda\cdot \bar a:=\overline{\lambda\cdot a}$, where $\bar a$ is the equivalence class of $a$ in $A/I$.
Firstly, this multiplication is well-defined, since if $a,b$ are in the same equivalence class in $A/I$ that means there exists $x\in I$ such that $a=b+x$, but then $\lambda a=\lambda(b+x)=\lambda b+\lambda x$, and since $I$ is an ideal, $\lambda x\in I$ and thus $\lambda a$ is in the same equivalence class as $\lambda b$, so this multiplication does not depend on the representative of $\bar a$.
And then we can check the axioms like this (using $\overline{a+b}=\bar a+\bar b,~\overline{ab}=\bar a\bar b$ and the fact that multiplication $\lambda a$ for $\lambda\in k,a\in A$ already fulfills the axioms):
Right distributivity:
$$(\lambda+\mu)\cdot \bar a=\overline{(\lambda+\mu)a}=\overline{\lambda a+\mu a}=\overline{\lambda a}+\overline{\mu a}=\lambda\bar a+\mu\bar a$$
Left distributivity:
$$\lambda(\bar a+\bar b)=\lambda\overline{(a+b)}=\overline{\lambda(a+b)}=\overline{\lambda a+\lambda b}=\overline{\lambda a}+\overline{\lambda b}=\lambda\bar a+\lambda\bar b$$
Scalar compatibility:
$$(\lambda \bar a)(\mu \bar b)=\overline{\lambda a}\cdot\overline{\mu b}=\overline{(\lambda a)(\mu b)}=\overline{(\lambda\mu)(ab)}=\lambda\mu\overline{ab}=\lambda\mu\bar a\bar b$$
(A quicker, but more abstract way to see this is that the definition above boils down to $k\cdot\bar a$ being defined via the ring homomorphism $k\longrightarrow A\longrightarrow A/I$, where the first arrow is the ring homomorphism defining multiplication by $k$ in $A$ and the second arrow is just the quotient map)
