Affine Varieties - the x-axis in R^2

[Math Amateur]

In Dummit and Foote, Chapter 15, Section 15.2 Radicals and Affine Varieties, Example 2, page 681 begins as follows:

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"The x-axis in \(\displaystyle \mathbb{R}^2 \) is irreducible since it has coordinate ring

\(\displaystyle \mathbb{R}[x,y]/(y) \cong \mathbb{R}[x] \)

which is an integral domain."

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Can someone please help me to show formally and rigorously how the isomorphism

\(\displaystyle \mathbb{R}[x,y]/(y) \cong \mathbb{R}[x] \) is established.I suspect it comes from applying the First (or Fundamental) Isomorphism Theorem for rings ... but I am unsure of the mappings involved and how they are established

Would appreciate some help>

Peter

 

[jvicens]

Hello Peter,

You are correct that the isomorphism between the coordinate ring of the x-axis and the polynomial ring in one variable is established using the First Isomorphism Theorem for rings.

Let's break down the steps involved in this proof:

1. First, we need to define the coordinate ring of the x-axis in \mathbb{R}^2. This is done by taking the quotient of the polynomial ring \mathbb{R}[x,y] by the ideal (y), which represents the set of all polynomials in \mathbb{R}[x,y] that have y as a factor. So the coordinate ring is given by \mathbb{R}[x,y]/(y).

2. Next, we need to define a map from \mathbb{R}[x,y] to \mathbb{R}[x]. This map is given by the natural projection map \pi: \mathbb{R}[x,y] \rightarrow \mathbb{R}[x], which sends each polynomial in \mathbb{R}[x,y] to its corresponding polynomial in \mathbb{R}[x] by simply setting y=0.

3. Now, we need to show that this map \pi is a ring homomorphism. This means that it preserves addition and multiplication. It is easy to see that \pi preserves addition since adding two polynomials in \mathbb{R}[x,y] and then setting y=0 gives the same result as setting y=0 for each polynomial first and then adding the results. For multiplication, we need to use the distributive property to show that \pi(fg) = \pi(f)\pi(g) for any f,g \in \mathbb{R}[x,y]. This can be done by expanding the product fg and then setting y=0, which gives the same result as setting y=0 for each polynomial first and then multiplying the results.

4. Now, we need to show that the kernel of \pi is equal to the ideal (y). This means that any polynomial in \mathbb{R}[x,y] that is sent to 0 by \pi must be in the ideal (y). This is true since setting y=0 in any polynomial in \mathbb{R}[x,y] will eliminate the y-term, leaving only polynomials in \mathbb{R}[x] which are in the ideal (y).

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