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Is $\mathbb R[x]$ an ideal in $\mathbb R$ ? To me, $\mathbb R[x]$ an ideal. If I took any element from the ring $\mathbb R$ and any element in $\mathbb R[x]$ then the product of those two elements belong to $\mathbb R[x]$

Now about the Hilbert Basis Theorem. The Hilbert Basis Theorem says that every ideal in the ring of polynomials in finitely many variables has a basis.

I am trying to use an example so that I could convince myself. Suppose that I'm interested in the following polynomial:

$3x^2 + 4xy + 2y^2 + 7x - 5y + 12 = 2y^2 + (4x - 5)y + (3x^2 + 7x + 12)$

The LHS is a polynomial in $\ R[x,y]$ and the RHS is a polynomial in $\ (R[x]) [y]$. If I knew that $R$ has the property that every ideal in $R$ is finitely generated, would that then mean that every ideal in $\ R[x,y]$ and $\ (R[x]) [y]$ is finitely generated? Is this the proper interpretation of Hilbert Basis Theorem?

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$\mathbb{R}[x]$ isn't even in $\mathbb{R}$, so it can't be an ideal of it. – anon Mar 29 '12 at 3:55
$\mathbb{R}$ is a field, so it's only proper ideal is itself. – akgjh Sep 18 '15 at 0:35
up vote 4 down vote accepted

$\mathbb{R}[x]$ is a $\mathbb{R}$-module, but it is not an ideal of $\mathbb{R}$, because an ideal $I$ of a ring $R$ is a subset of $R$ that has the property of being an $R$-module. If you're not familiar with $R$-modules, they are algebraic structures that have a multiplication by elements of $R$ that satisfies a natural set of axioms. (When $R$ is a field, such as $R = \mathbb{R}$, then an $R$-module is an $R$-vector space.)

As far as the Hilbert Basis Theorem goes, it says that if $R$ is a ring where every ideal is finitely generated, then that same property holds for $R[x]$. If you apply the Hilbert Basis Theorem twice to a ring $R$ where every ideal is finitely generated (such a ring is called Noetherian), you get that $R[x]$ is Noetherian after the first application and then that $(R[x])[y]$ is Noetherian from the second application. Then you observe that $R[x][y]$ is (canonically) isomorphic to $R[x,y]$ (as your polynomial calculation illustrates in a specific example) to deduce that $R[x,y]$ is Noetherian. Similarly, if you apply the Hilbert Basis Theorem $n$ times, you can deduce that $R[x_1,\dots,x_n]$ is Noetherian for any $n \geq 1$.

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No, $\mathbb{R}[x]$ is not an ideal of $\mathbb{R}$ because it isn't a subset of $\mathbb{R}$. Perhaps you should make sure you understand this, as well as what it means for a polynomial to be finitely-generated.

Hilbert's theorem says that if $A$ is Noetherian, so is $A[x]$. Hence, applying the theorem $n$ times, so is $A[x_1, \dots, x_n]$. However, it is crucial that $A$ be Noetherian.

In geometric language, Hilbert's theorem says that a scheme of finite type over a Noetherian scheme is Noetherian.

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