Products of ideals of K[x1, x2, x3, x4]

[Math Amateur]

Products of ideals of K[x1, x2, x3, x3]

I am reading R.Y. Sharpe: Steps in Commutative Algebra. In chapter 2 on Ideals, on page 28 we find Exercise 2.27 which reads as follows: (see attachment)

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2.27 Exercise: Let \(\displaystyle K\) be a field, and let \(\displaystyle R = K[x_1, x_2, x_3, x_4] \), the ring of polynomials over K in indeterminates x_1, x_2, x_3, x_4.

Set \(\displaystyle I = Rx_1 + Rx_2 \) and \(\displaystyle J = Rx_3 + Rx_4 \)

Show that \(\displaystyle IJ \ne \{fg: \ f \in I, g \in J \} \)
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My problem is I seemed to have ended up showing that \(\displaystyle IJ = \{fg: \ f \in I, g \in J \} \) ... so obviously something is wrong with my working ...

Can someone please explain my error(s)?

My working is as follows:

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\(\displaystyle Rx_1 = \{ fx_1 \ | \ f \in R \} \)

and Rx_2, Rx_3, Rx_4 are defined similarly.\(\displaystyle I = Rx_1 + Rx_2 \)

\(\displaystyle = \{ h+k \ | \ h \in Rx_1 , k \in Rx_2 \} \)

\(\displaystyle = \{ fx_1 + gx_2 \ | \ f, g \in R \} \)

and similarly

\(\displaystyle J = \{ hx_1 + kx_2 \ | \ h, k \in R \} \)

Then \(\displaystyle IJ = \) set of all finite sums of elements of the form \(\displaystyle lm \) with \(\displaystyle l \in I, m \in J \)

\(\displaystyle = \{ {\sum}_{i=1}^{n} l_im_i \ | \ n \in \mathbb{N}, l_i \in I, m_i \in J \} \)

\(\displaystyle = \{ {\sum}_{i=1}^{n} (f_ix_1 + g_ix_2)(h_ix_3 + k_ix_4) \ | \ f_i, g_i, h_i, k_i \in R \} \)

\(\displaystyle = \{ {\sum}_{i=1}^{n} f_ih_ix_1x_3 + f_ik_ix_1x_4 + g_ih_ix_2x_3 + g_ik_ix_2x_4 \ | \ f_i, g_i, h_i, k_i \in R \} \)

\(\displaystyle = \{ {\sum}_{i=1}^{n} l_ix_1x_3 + m_ix_1x_4 + p _ix_2x_3 + q_ix_2x_4 \ | \ l_i, m_i, p_i, q_i \in R \} \)

\(\displaystyle = lx_1x_3 + mx_1x_4 + px_2x_3 + x_2x_4 \ | \ l, m, p, q \in R \}\)

since we can put \(\displaystyle l_1 + l_2 + ... \ ... l_n = l \) and similarly with \(\displaystyle m, p, q \)Now consider the set \(\displaystyle \{ fg: \ f\in I, g \in J \} \)

\(\displaystyle \{ fg: \ f\in I, g \in J \} \)

\(\displaystyle = \{ (l_1x_1 + m_1x_2)(p_1x_3 + q_1x_4) \ | \ l_1, m_1, p_1, q_1 \in R \} \)

\(\displaystyle = \{ l_1p_1x_1x_3 + l_1q_1x_1x_4 + m_1p_1x_2x_3 + m_1q_1x_2x_4 \ | \ l_1, m_1, p_1, q_1 \in R \} \)

\(\displaystyle = \{ lx_1x_3 + mx_1x_4 + px_2x_3 + qx_2x_4 \ | \ l, m, p, q \in R \} \)BUT then \(\displaystyle IJ = \{fg: \ f \in I, g \in J \} \) ?

Can someone please explain my error(s)

Peter

 

[Deveno]

Well clearly we have:

$\{fg: f \in I, g \in J\} \subseteq IJ$, so there must be something in $IJ$ that is NOT in $\{fg: f \in I, g \in J\}$.

Consider $x_1x_3 + x_2x_4$.

Clearly, we have:

$x_1 = x_1(1) + x_2(0) \in I$
$x_3 = x_3(1) + x_4(0) \in J$

$x_2 = x_1(0) + x_2(1) \in I$
$x_4 = x_3(1) + x_4(0) \in J$

so $x_1x_3,x_2x_4 \in IJ$, and thus so is their sum.

Now if:

$x_1x_3 + x_2x_4 = (x_1f + x_2g)(x_3h + x_4k),\ f,g,h,k \in K[x_1,x_2,x_3,x_4]$

$x_1x_3 + x_2x_4 = x_1x_3(fh) + x_1x_4(fk) + x_2x_3(gh) + x_2x_4(gk)$

so:

$fh = gk = 1 \implies f,g,h,k \neq 0$
$fk = gh = 0 \implies f = 0$ or $k = 0$, and $g = 0$ or $h = 0$.

In general, we only get "some" $R$-linear combinations of $x_1x_3,x_1x_4,x_2x_3,x_2x_4$ in the set $\{fg: f\in I,g \in J\}$ whereas in $IJ$ we get ALL of them. In other words showing that every element of $\{fg: f\in I,g \in J\}$ is of the same form as a "typical" element of $IJ$ only shows containment one way, you have to show that ANY such element of $IJ$ can be obtained that way, and it suffices to exhibit just ONE that cannot.

 

[Math Amateur]

Well clearly we have:

$\{fg: f \in I, g \in J\} \subseteq IJ$, so there must be something in $IJ$ that is NOT in $\{fg: f \in I, g \in J\}$.

Consider $x_1x_3 + x_2x_4$.

Clearly, we have:

$x_1 = x_1(1) + x_2(0) \in I$
$x_3 = x_3(1) + x_4(0) \in J$

$x_2 = x_1(0) + x_2(1) \in I$
$x_4 = x_3(1) + x_4(0) \in J$

so $x_1x_3,x_2x_4 \in IJ$, and thus so is their sum.

Now if:

$x_1x_3 + x_2x_4 = (x_1f + x_2g)(x_3h + x_4k),\ f,g,h,k \in K[x_1,x_2,x_3,x_4]$

$x_1x_3 + x_2x_4 = x_1x_3(fh) + x_1x_4(fk) + x_2x_3(gh) + x_2x_4(gk)$

so:

$fh = gk = 1 \implies f,g,h,k \neq 0$
$fk = gh = 0 \implies f = 0$ or $k = 0$, and $g = 0$ or $h = 0$.

In general, we only get "some" $R$-linear combinations of $x_1x_3,x_1x_4,x_2x_3,x_2x_4$ in the set $\{fg: f\in I,g \in J\}$ whereas in $IJ$ we get ALL of them. In other words showing that every element of $\{fg: f\in I,g \in J\}$ is of the same form as a "typical" element of $IJ$ only shows containment one way, you have to show that ANY such element of $IJ$ can be obtained that way, and it suffices to exhibit just ONE that cannot.

Thanks so much for the help, Deveno ... ... but just a clarification question ...

It is clear from the first part of your analysis that

\(\displaystyle x_1x_3 + x_2x_4 \in IJ \)

Now presumably (?), what we want to do next is to show that

\(\displaystyle x_1x_3 + x_2x_4 \notin \{ fg \ | \ f \in I, g \in J \} \) ... ... (1)

Now, presumably, the steps after you write "Now if:" do exactly that - but I cannot see how what you have done shows (1) above.

Can you please explain the logic ...

Thanks again,

Peter

 

[Deveno]

If a set $A$ has an element the set $B$ does not, then the two sets cannot possibly be the SAME set.

For example, the set $A = \{a\}$ and the set $B = \{a,b\}$ are not the same (provided $a \neq b$), since $b \in B$, but $b \not \in A$.

In general, the problem with the set:

$K = \{xy: x \in I, y\in J\}$ for two ideals $I,J$ of a ring $R$ is that it is usually not closed under addition, so $(K,+)$ is not a subgroup of $(R,+)$.

The exercise you have been given to prove is typically the "standard example" for demonstrating this, compare exercise 3 (b) here:

http://math.berkeley.edu/~daffyd/113s11/hw6sol.pdf

 

[Math Amateur]

If a set $A$ has an element the set $B$ does not, then the two sets cannot possibly be the SAME set.

For example, the set $A = \{a\}$ and the set $B = \{a,b\}$ are not the same (provided $a \neq b$), since $b \in B$, but $b \not \in A$.

In general, the problem with the set:

$K = \{xy: x \in I, y\in J\}$ for two ideals $I,J$ of a ring $R$ is that it is usually not closed under addition, so $(K,+)$ is not a subgroup of $(R,+)$.

The exercise you have been given to prove is typically the "standard example" for demonstrating this, compare exercise 3 (b) here:

http://math.berkeley.edu/~daffyd/113s11/hw6sol.pdf

Thanks Deveno ... I believe I understand what you have said in this post ... but the mechanics in the previous post still escape me ... I think that I did not explain my difficulties clearly enough ... so I will try to explain myself more explicitly and carefully ...So ... I can see that

\(\displaystyle x_1x_3 + x_2x_4 \in IJ \)

Now we want to show that

\(\displaystyle x_1x_3 + x_2x_4 \notin \{ fg \ | \ f \in I, g \in J \} \)So in doing this you begin:

"Now if:

$x_1x_3 + x_2x_4 = (x_1f + x_2g)(x_3h + x_4k),\ f,g,h,k \in K[x_1,x_2,x_3,x_4]$"

Here you are just expanding out the fact that

\(\displaystyle x_1x_3 + x_2x_4 \in IJ \) given that

\(\displaystyle I = Rx_1 + Rx_2 \) and \(\displaystyle J = Rx_3 + Rx_4 \)
Then you write:

"$x_1x_3 + x_2x_4 = x_1x_3(fh) + x_1x_4(fk) + x_2x_3(gh) + x_2x_4(gk)$" ... ... (1)

This is just multiplying out ... ...Then you write:

"so:

$fh = gk = 1 \implies f,g,h,k \neq 0$
$fk = gh = 0 \implies f = 0$ or $k = 0$, and $g = 0$ or $h = 0$."

These are just conditions on \(\displaystyle f, g, h, k \) that follow from equation (1) above ...
BUT ... how/why exactly do the above steps show that

\(\displaystyle IJ \ne \{fg: \ f \in I, g \in J \} \)

Sorry to be so pedestrian with this ... but hoping you can help ...

Peter

 

[Deveno]

$f$ (for example) cannot be 0 AND non-zero at the same time. Therefore, such an $f$ DOES NOT EXIST.

If we accept that $f \neq 0$, then $k$ HAS to be 0, which is again a contradiction. We cannot escape this...we have 4 non-zero polynomials, yet 2 of them (it doesn't really matter WHICH two) have to be 0.

So the existence of the $f,g,h,k$ we have to have if $x_1x_3 + x_2x_4$ is to be in the product set is impossible, which means that there is NO such factorization of $x_1x_3 + x_2x_4$, which means it is not in the set: $\{fg: f \in I,g \in J\}$.

 

[Math Amateur]

$f$ (for example) cannot be 0 AND non-zero at the same time. Therefore, such an $f$ DOES NOT EXIST.

If we accept that $f \neq 0$, then $k$ HAS to be 0, which is again a contradiction. We cannot escape this...we have 4 non-zero polynomials, yet 2 of them (it doesn't really matter WHICH two) have to be 0.

So the existence of the $f,g,h,k$ we have to have if $x_1x_3 + x_2x_4$ is to be in the product set is impossible, which means that there is NO such factorization of $x_1x_3 + x_2x_4$, which means it is not in the set: $\{fg: f \in I,g \in J\}$.

My apologies Deveno .. I do see the point regarding f, g, h, and k but that leaves me more perplexed ...

I will try to explain clearly ...

Well ... the fact that \(\displaystyle x_1x_3 + x_2x_4 \in IJ \) and that \(\displaystyle I = Rx_1 + Rx_2 \) and \(\displaystyle J = Rx_3 + Rx_4 \)

\(\displaystyle \Longrightarrow \)

$x_1x_3 + x_2x_4 = (x_1f + x_2g)(x_3h + x_4k),\ f,g,h,k \in K[x_1,x_2,x_3,x_4]$ ... ... (1)

(Is this my error ... I am assuming that (1) is simply coming from \(\displaystyle x_1x_3 + x_2x_4 \in IJ \) and the definitions of I and J.)

So then we show ... ? ... that no such f, g, h , k exist ?

So on my assumption about (1) this (ridiculously) seems to show that \(\displaystyle x_1x_3 + x_2x_4 \notin IJ \) ? and not that
\(\displaystyle x_1x_3 + x_2x_4 \notin \{fg \ | \ f \in I, g \in J \} \)

If the equation (1) was a condition for \(\displaystyle x_1x_3 + x_2x_4 \) to belong to \(\displaystyle \{fg \ | \ f \in I, g \in J \} \) then I could understand the proof.

Can you help further?

Many apologies for being slow here.

Peter

 

[Deveno]

No...condition (1) is indeed what we need for $x_1x_3 + x_2x_4$ to belong to the "product set" (not the "product ideal").

It should be self-evident that $x_1x_3 + x_2x_4 \in IJ$, since each term of the sum is. Since I later show it is not in the product set, the two sets cannot be equal.

 

[Math Amateur]

No...condition (1) is indeed what we need for $x_1x_3 + x_2x_4$ to belong to the "product set" (not the "product ideal").

It should be self-evident that $x_1x_3 + x_2x_4 \in IJ$, since each term of the sum is. Since I later show it is not in the product set, the two sets cannot be equal.

Yes, just realized this a moment ago ... thanks for all the posts on this exercise ...

Peter