MHB How can you simplify generated ideals in a commutative unital ring?

[Fermat1]

Let $x,y$ be members of a commutative unital ring. By using various 'rules' show that
$<y^4+3x^3-2x^2,7y^4+5(xy+yx^2),x^3+2y^3>+<x^3,xy^2,xy^3,yx^2,xy^2,y^4$>
$=<x^2,xy,y^3>$, where $<.>$ denotes the ideal generated by$.$

Can you tell me the rules for simplyifing these generated ideals (and I will complete the question)? My teacher went through them a while back but I lost my notes. Thanks

 

[Fermat1]

To start myself off, I have proved that $<xy>=<x><y>$ but $<x>+<y>$ contains $<x+y>$. Can anyone help with simplifying the expression?

 

[Opalg]

Let $x,y$ be members of a commutative unital ring. By using various 'rules' show that
$<y^4+3x^3-2x^2,7y^4+5(xy+yx^2),x^3+2y^3>+<x^3,xy^2,xy^3,yx^2,xy^2,y^4$>
$=<x^2,xy,y^3>$, where $<.>$ denotes the ideal generated by$.$

Can you tell me the rules for simplyifing these generated ideals (and I will complete the question)? My teacher went through them a while back but I lost my notes. Thanks

I don't know a list of rules for dealing with this, but there is one property that seems fairly obvious. Namely, if two ideals are given in terms of generators then their sum is given by the union of those generators: $\langle a_1, a_2, \ldots, a_m \rangle + \langle b_1, b_2, \ldots, b_n \rangle = \langle a_1, a_2, \ldots, a_m, b_1, b_2, \ldots, b_n \rangle.$ In this case, $$\begin{aligned}\langle y^4+3x^3-2x^2,&7y^4+5(xy+yx^2), x^3+2y^3 \rangle +\langle x^3,xy^2,xy^3,yx^2,xy^2,y^4 \rangle \\ &= \langle y^4+3x^3-2x^2, 7y^4+5(xy+yx^2),x^3+2y^3, x^3,xy^2,xy^3,yx^2,xy^2,y^4 \rangle .\end{aligned}$$ After that, you can reduce the number of generators by common sense methods. For example, the set of linear combinations of $x^3+2y^3$ and $x^3$ is the same as the set of linear combinations of $2y^3$ and $x^3.$ So you can simplify the list of generators by replacing $x^3+2y^3$ by $2y^3.$ Also, if $xy^2$ is a generator then $xy^3 = (xy^2)y$ is automatically in the ideal, so is not needed as a generator. In that way, you should be able to simplify the list of generators to $\langle -2x^2, 5xy, 2y^3, x^3,xy^2,yx^2,xy^2,y^4 \rangle .$

To get any further than that, I had to make the assumption that the ring has characteristic $0$ (or at least does not have characteristic $2$ or $5$). If so, then you can multiply the first three of the generators in that last set by scalars and replace them by $x^2$, $xy$ and $y^3$. the remaining generators in the list are then multiples of those three, which therefore form a complete list of generators for the ideal.

 

[Fermat1]

I don't know a list of rules for dealing with this, but there is one property that seems fairly obvious. Namely, if two ideals are given in terms of generators then their sum is given by the union of those generators: $\langle a_1, a_2, \ldots, a_m \rangle + \langle b_1, b_2, \ldots, b_n \rangle = \langle a_1, a_2, \ldots, a_m, b_1, b_2, \ldots, b_n \rangle.$ In this case, $$\begin{aligned}\langle y^4+3x^3-2x^2,&7y^4+5(xy+yx^2), x^3+2y^3 \rangle +\langle x^3,xy^2,xy^3,yx^2,xy^2,y^4 \rangle \\ &= \langle y^4+3x^3-2x^2, 7y^4+5(xy+yx^2),x^3+2y^3, x^3,xy^2,xy^3,yx^2,xy^2,y^4 \rangle .\end{aligned}$$ After that, you can reduce the number of generators by common sense methods. For example, the set of linear combinations of $x^3+2y^3$ and $x^3$ is the same as the set of linear combinations of $2y^3$ and $x^3.$ So you can simplify the list of generators by replacing $x^3+2y^3$ by $2y^3.$ Also, if $xy^2$ is a generator then $xy^3 = (xy^2)y$ is automatically in the ideal, so is not needed as a generator. In that way, you should be able to simplify the list of generators to $\langle -2x^2, 5xy, 2y^3, x^3,xy^2,yx^2,xy^2,y^4 \rangle .$

To get any further than that, I had to make the assumption that the ring has characteristic $0$ (or at least does not have characteristic $2$ or $5$). If so, then you can multiply the first three of the generators in that last set by scalars and replace them by $x^2$, $xy$ and $y^3$. the remaining generators in the list are then multiples of those three, which therefore form a complete list of generators for the ideal.

Thanks. I don't understand the last part though. For example, $x^2$ can be generated by $5x^2$ simply by multiplying by $1/5$. What has it got to do with the characteristic?

 

[Opalg]

Thanks. I don't understand the last part though. For example, $x^2$ can be generated by $5x^2$ simply by multiplying by $1/5$. What has it got to do with the characteristic?

The problem is that $5$ may not have a reciprocal in the ring. For example, the ring of integers contains only whole numbers, not fractions. Come to think of it, that has nothing to do with the ring's characteristic, it's more basic than that.