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Could someone please clarify the definition of polynomial ring in two variables $\mathbb{R}[x,y]$? Does it consist of all objects in form $a_0 + a_1 x^{k_{1x}}y^{k_{1y}} + a_2 x^{k_{2x}}y^{k_{2y}} + \cdots$? In addition, What would be the ideal $\langle x,y \rangle$ in this case? Is it the subring with polynomials in form $f(x,y)x + g(x,y)y$?

3x89g2
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3 Answers3

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Yes, you are right for both. Another way of seeing $\langle x,y\rangle$ is to say that this is the kernel of the evaluation map at the point $(0,0)$ ; that is the set of polynomials that vanish at $(0,0)$.

C. Falcon
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$\mathbf R[x,y]$ consists of all finite sums of monomials in two variables $cx^i y^j$, $c\in\mathbf R$, $i,j\in \mathbf N$ (there is a more abstract and rigourous definition, which is valid for any commutative ring).

The ideal $\langle x,y\rangle$ consists of all polynomials with constant term $0$. It is not a subring with the usual conventions since it does not have an identity element.

Bernard
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  • @OP Be aware that is not a rigorous definition without defining "variable", what it means to juxtapose variables,, etc. – Bill Dubuque Nov 08 '16 at 20:43
  • @Misakov: it is the set of all linear combinations of $f_1(x,y), \dots, f_n(x,y)$ with polynomial coefficients, i.e. poynomials of the form $;g_1(x,y)f_1(x,y)+\dots+g_n(x,y)f_n(x,y)$. – Bernard Nov 08 '16 at 20:51
  • I am confused. I see now that $\langle x,y\rangle$ is not a subring.. But if it's not a subring, how could it be an ideal? I thought an ideal has to be a subring in the first place? – 3x89g2 Nov 08 '16 at 20:51
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    No. First an ideal does not contain a unit element (this is a common convention for rings in commutative algebra). More importantly, a subring is stable under multiplication of any two of its elements, and an ideal is stable under multiplication of any one of its elements by any element of the larger ring – somewhat like a vector subspace. – Bernard Nov 08 '16 at 20:57
  • On my textbook, Abstract Algebra, an Introduction by Hungerford, it says "A subring $I$ of a ring $R$ is an ideal, provided that whenever $r\in R$ and $a\in I$, we have $ra \in I$ and $ar\in I$". So the book is wrong? – 3x89g2 Nov 08 '16 at 21:16
  • This is not different from what I say, except the book considers subrings without a unit element (which is uncommon in commutative algebra). In the case of an ideal, if it contains the unit element, it is the whole ring. – Bernard Nov 08 '16 at 21:27
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If you know already the definition of the polynomial ring $R[x]$, then you can just take $R[x,y]=R[x][y]$ as a definition. If you take the general definition from, say, wikipedia, then you can prove the following result:

Proposition: There exists a unique isomorphism between the polynomial ring $R[x,y]$ and the ring $R[x][y]$ of polynomials in $y$ with coefficients from $R[x]$, which is identity on $R$, and which sends $x$ to $x$ and $y$ to $y$.

Concerning the ideal $\langle x,y\rangle$, we can say different things. For example, $I= \langle x,y \rangle\subset k[x,y]$ is not principal, see here.

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Dietrich Burde
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