Ideal Generated by a Finite Number of Polynomials?

You were close.

The ideal $(f_1,...,f_s)$ of $k[x_1,..., x_n]$ is the set of all polynomials which can be expressed as $h_1f_1+\cdots +h_sf_s$, for some $h_1,...,h_s \in k[x_1,..., x_n]$.

In other words, "linear combinations" of $f_1,...,f_s$ where the "linear coefficients" are arbitrary elements $h_1,...,h_s$ of $k[x_1,..., x_n]$.

Regarding the author's point about common zeros . . .

If $\bar{k}$ is an algebraically closed extension of $k$, and if there is some $a = (a_1,...,a_n) \in \left(\bar{k}\right)^n$ such that $f_1(a) = \cdots = f_s(a) = 0$, then $a$ is automatically a zero of any polynomial of the form $h_1f_1+\cdots +h_sf_s$, hence $a$ is a common zero for all members of the ideal $(f_1,...,f_s)$.

In particular, if you've identified a common zero $a \in \left(\bar{k}\right)^n$ for $f_1,...,f_s$, then if $g \in k[x_1,..., x_n]$ is such that $g(a) \ne 0$, it follows that $g$ is not in the ideal $(f_1,...,f_s)$.

An important, nontrivial result is a partial converse: If $g \in k[x_1,...,x_n]$ is such that $g,g^2,g^3,...$ are not in the ideal $(f_1,...,f_s)$, there is some $a \in \left(\bar{k}\right)^n$ such that $a$ is a common zero of $f_1,...,f_n$, but $a$ is not a zero of $g$.

Perhaps a good way to understand this (or to understand anything, for that matter) is to look at examples.

The definition of ideal makes sense in any ring, so let’s look at $\Bbb Z$ first. Every ideal, you soon persuade yourself, is a set $d\Bbb Z\subset\Bbb Z$, that is, just the set of all multiples of a given number $d$. What if you try to take two numbers and look at $\langle d_1,d_2\rangle\subset\Bbb Z$? Please do this with specific numbers. Like what about $d_1=8$ and $d_2=6$? You want to describe the totality of all numbers writable in the form $8m+6n$. You rapidly see that this is an ideal in the sense of the definition, and you see that the set is equal to $2\Bbb Z$. This is basic number theory.

Now do the same thing with a ring of polynomials, but in just one variable, $R=\Bbb Q[x]$. One proves (you prove it, with Euclidean division of polynomials) that every ideal of $R$ is of form $fR$, where $f$ is a well-chosen polynomial. In fact, for a nonzero ideal $I$, you take $f$ to be a nonzero element of $I$ of least degree. So, what is $\langle f_1,f_2\rangle$, when $f_1$ and $f_2$ are two polynomials in $x$ that are given? Just as with numbers, it’s the ideal $gR$ where $g$ is the greatest common divisor of $f$ and $g$. Convince yourself that $\langle x^3-1,x^2-2x+1\rangle$ is the set of all multiples of $x-1$.

Things are no longer so simple when you have polynomials in more than one indeterminate. But at least you get some insight by looking at the simplest cases, and I hope you see that the set of all possible $h_1f_1+\cdots+h_nf_n$ is an ideal.

Would it help to see the see the set with explicit polynomials in place of the $f_1, f_2, \dots, f_n$?

For example, lets look at an explicit example when $n = 2$, so an ideal generated by 2 polynomials. Also, we'll work in a polynomial ring in two variables over $k$, i.e. $k[x,y]$.

Here is the ideal generated by the polynomials $x^2 - 1$, $yx+x$.$$\langle \, x^2 - 1, \, yx+x \, \rangle = \{ \, f\cdot(x^2 - 1) + g\cdot(yx + x) \, | \, f,g \in k[x,y] \, \}.$$

So the elements of the set are any polynomial that can be written in the form $f \cdot (x^2 - 1) + g \cdot (yx+x)$. But we can choose $f,g$ to be $\textit{any}$ polynomial we want. For example, we know $x^2 - 1$ itself is in that set because we can choose $f = 1, g = 0$. We also know that the polynomial $x^3-x+yx ^2 + x^2$ is in the set, because we can choose $f = x, g = x$.

If you are familiar with linear algebra you can maybe, in a way, draw a connection between an ideal generated by polynomials and the span of a set of vectors. You can think of it as, the set of all things that can be made from the objects defining it.