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I am studying Algebraic Geometry, where we introduced the notion of polynomial ring, denoted by $K[x_1, ... x_n]$, that is, a commutative ring over a field $K$, whose elements are polynomials (functions with coefficients from $K$ in the form $p_0 + p_1 x^1_1 + ...$.). I also provide the definition of Noetherian ring - it is such ring that every ideal is finitely generated.

Then there is a claim implied by Hilbert Basis Theorem, saying that

"For any field K, and any natural number $n$, the ring $K[x_1, . . . , x_n]$ is a Noetherian ring."

My question is regarding this proof of the claim above:

"The field $K$ is a noetherian ring, because the only ideal in it is the zero ideal which is finitely generated. By induction, we deduce the result."

Why is the only ideal of $K$ a zero ideal? I thought that ideal $I$ of a ring $K$, in general, is defined as an additive subgroup and $\forall f \in I, k \in K: k f \in I$. I just dont see why this can only consist of ${0}$? (My question is about $K$, not $K[x_1, . . . , x_n]$.)

Thank you very much for explaining.

(Source: I am reading mostly Fulton´s Algebraic Curves.)

EDIT: I am not sure about how to best title this question, recommendation is welcome.

Tereza Tizkova
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    Once cannot over-emphasize the comment below: it is essential to master basic ring theory and commutative algebra before beginning to study algebraic geometry. – Bill Dubuque Aug 18 '22 at 19:12

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If $I$ is an ideal in a field $K$ that contains some nonzero element $x$ then (since it's an ideal) $1 = xx^{-1} \in I$ so $I = K$.

Ethan Bolker
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  • He was asking about the polynomial ring in $n$ elements over the field $K$, not the field $K$ itself. – Alan Aug 18 '22 at 18:00
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    @Alan The question is about the field $K$. The ring $K[x_1,...,x_n]$ clearly has nontrivial ideals. (just as any commutative ring which is not a field) – Mark Aug 18 '22 at 18:05
  • Doh, I'm half awake and it's been too long since I've done algebra. Woops! – Alan Aug 18 '22 at 18:06
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    Have to say, it's a surprising question. Usually one starts learning algebraic geometry only after having good knowledge in commutative algebra. But that is a very basic question about commutative rings. – Mark Aug 18 '22 at 18:08
  • @Mark My thought too. – Ethan Bolker Aug 18 '22 at 18:11
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here – Bill Dubuque Aug 18 '22 at 19:01
  • @Mark You are right. My story of studying graduate math is quite complicated, thats why I sometimes have to learn concepts in unusual order. Thank you all for the advice though. – Tereza Tizkova Aug 18 '22 at 19:57
  • @EthanBolker So, to understand your question also intuitively, if $K$ wasnt a field, it wouldnt have the multiplicative inverse? (For the multiplicative identity, I am not sure either, as I sometimes see rings defined with it, sometimes without.) – Tereza Tizkova Aug 18 '22 at 20:06
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    @TerezaTizkova Well, you should at least be familiar with the objects you work with. In algebraic geometry rings are usually assumed to be commutative with an identity element. (specifically the rings you asked about are such rings) If $I$ is an ideal of such a ring $R$ and contains an invertible element then $I=R$. Try to prove it as an easy exercise to make sure you understand this. Now, if $R$ is a field then any nonzero ideal trivially contains an invertible element. – Mark Aug 18 '22 at 20:13
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    @Mark Thank you very much, that makes sense to me. – Tereza Tizkova Aug 18 '22 at 20:19