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A <commutative ring> $R$ is a <Noetherian ring> if every <ideal> has a finite <generating set of an ideal>. Equivalently, it satisfies the <ascending chain condition> on <ideals>. Indeed, the union of an ascending chain is an <ideal>, and finitely many generators all belong to one term, after which the chain stabilizes. Conversely, an <ideal> without a finite <generating set of an ideal> would allow us to choose successively $a_{n+1}\notin(a_1,\ldots,a_n)$, producing a strictly ascending chain of <ideals>.
Let $\phi:R\to S$ be a <surjective> <ring homomorphism>, where $R$ is a <Noetherian ring>. For any <ideal> $J$ of $S$, its inverse image is an <ideal> of $R$, say $\phi^{-1}(J)=(a_1,\ldots,a_r)$. Taking images gives
$$
J=(\phi(a_1),\ldots,\phi(a_r)).
$$
Thus <homomorphic images of Noetherian rings are Noetherian>. This also applies to the image of an arbitrary <ring homomorphism>, by taking its image as the codomain.
The <Hilbert basis theorem> states that if $R$ is a <Noetherian ring>, then the <polynomial ring> $R[t]$ is a <Noetherian ring>. Repeating this gives the result for finitely many variables.
To prove it, fix an <ideal> $K\subseteq R[t]$. For $n\geq0$, define the <coefficient ideals of a polynomial ideal>
$$
A_n=\{a\in R:\text{some }f\in K\text{ of degree at most }n\text{ has coefficient }a\text{ at }t^n\}.
$$
Here the zero <polynomial> is allowed. Each $A_n$ is an <ideal>: addition and multiplication by constants preserve the required degree bound. Multiplying a representing <polynomial> by $t$ gives $A_n\subseteq A_{n+1}$. The <ascending chain condition> therefore supplies $N$ such that $A_n=A_N$ for every $n\geq N$.
For each $0\leq n\leq N$, choose a finite <generating set of an ideal> $a_{n1},\ldots,a_{nr_n}$ for $A_n$, omitting zero generators, and choose $f_{nj}\in K$ of <polynomial degree> $n$ with <leading coefficient> $a_{nj}$. If $A_n=0$, choose no such <polynomials>. We claim that the finitely many $f_{nj}$ generate $K$.
Take a nonzero $f\in K$ of <polynomial degree> $d$. Put $n=\min(d,N)$. Its <leading coefficient> belongs to $A_d=A_n$, so write it as $\sum_jb_ja_{nj}$. The <polynomial>
$$
g=f-\sum_jb_jt^{d-n}f_{nj}
$$
belongs to $K$ and is either zero or has <polynomial degree> strictly below $d$. Induction on <polynomial degree> now expresses $f$ as a combination of the chosen $f_{nj}$; when $d=0$, the displayed difference is zero. Hence $K$ is finitely generated. Since $K$ was arbitrary, $R[t]$ is a <Noetherian ring>.
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