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I have found a few resources such as 2.0.3 here. There however are a series of "identifications" such as saying $k[x,y,z] = k[y,z][x]$

  1. Why can I make these identifications? can I ALWAYS do so? I see that in the polynomial in their example $x^2 + y^2 + z^2$ that we have the same degree polynomial and there is no term with mixed variables. What if I wanted to check reducibility on something such as $xy + z^2$?

  2. The example goes on to say that it suffices to show that $y^2 + z^2$ is divisible by some prime $p$ in $k(z)[y]$, and not by $p^2$. Thus It suffices to show that $y^2 + z^2$ is not a unit, and has no repeated factor, in $k(z)[y]$. So it seems as if we are just taking the polynomials in $k(z)$ and appending the variable y to it, is this correct i.e. $k(z) + a \cdot y$?

Now I understand Eisenstein's criteria but all of the resources I find only apply it in a single variable - otherwise they mention that it "suffices to show" and I fail to see the connection. I could really use a walkthrough of how these work, more concretely, a walkthrough on why $xy + z^2$ is irrreducible.

Thanks.

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  • $\begingroup$ Not a complete answer, just a couple of quick points (I didn't read your reference, so this may be in there): $z^2+xy \in k[x,y][z]$ is irreducible by Eisenstein with $x$, for example. And $x^2+y^2+z^2$ can be shown to be irreducible as long as $k$ has characteristic $\neq 2$; otherwise $x^2+y^2+z^2 = (x+y+z)^2$. If $k$ doesn't have characteristic $2$, then $y^2+z^2$ factors iff $k$ has a square root of $-1$, but in this case the factors are not associates, so as remarked in your question 2, Eisenstein still applies. $\endgroup$ – Dustan Levenstein Jan 23 '15 at 17:45
  • $\begingroup$ Thanks Dustan, your comment is helpful - I see that if charecteristic is 2 then we have that, for example $(x+y)^2 mod 2 = (x^2 + 2xy + y^2) mod 2 \equiv x^2 + y^2$ as the $(2xy) mod \, 2 \equiv 0$. I don't see why I can treat $z^2 + xy$ as breaking it up between $k[x,y][z]$ - is it just because there is no overlap between the terms i.e. no $(z)(x)$ as ideals for example $z^nx^m, n,m > 0$. I think I am having a bit of a problem with the notation. $\endgroup$ – Relative0 Jan 23 '15 at 18:51
  • $\begingroup$ There are canonical isomorphisms $k[x,y] \simeq (k[x])[y] \simeq (k[y])[x]$: a polynomial in $y$ with coefficients in $k[x]$ can be distributed out into a polynomial in $x$ and $y$ with coefficients in $k$, and similarly for a polynomial in $x$ with coefficients in $k[y]$. $\endgroup$ – Dustan Levenstein Jan 23 '15 at 20:45
  • $\begingroup$ ... and similarly for more than $2$ variables. $\endgroup$ – Dustan Levenstein Jan 23 '15 at 20:46
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(1). Yes, you can always make an identification like $k[x,y,z]=k[x,y][z].$ It can be done more generally: $k[x_1, x_2, \dots ,x_n]=k[x1, \dots , x_{i-1}, x_{i+1}, \dots ,x_n][x_i],$ for any $1 \leq i \leq n.$ To see this note that any polynomial is of the form $\sum \alpha_{r_1, \cdots , r_n}x_1^{r_1}\cdots x_n^{r_n}.$ But this can also be written as $\sum \alpha_{r_1, \cdots , r_n}(x_1^{r_1}\cdots x_{i-1}^{r_{i-1}}x_{i+1}^{r_{i+1}} \cdots x_n^{r_n})x_i^{r_i}.$

(2). Consider the polynomial $y^2 + z^2 \in k[y,z].$ We want to show that $y^2+z^2$ is divisible by a prime $p \in k[x,y],$ but not by $p^2.$ (I think there is a little mistake at this point. Because it's clear why $y^2+z^2$ should be reducible. May be more correct statement would be: "if it is not irreducible then we want to show that $y^2+z^2$ is divisible by a prime $p \in k[x,y],$ but not by $p^2.$") Using Gauss Lemma, we can check it in the polynomial ring $k(z)[y].$ (Here one passes from $k[y,z]$ to $k(z)[y]$ to make sure that we can use division algorithm.)

(3). As for your example $xy+z^2 \in k[x,y,z],$ the first thing one occurs to mind is to consider it a polynomial in $z$ over the ring $k[x,y].$ In general we don't have division algorithm in the polynomial ring over a field with more than one variable. But in this case, if we consider it a polynomial in $z$ over the ring $k[x,y],$ then we have an advantage that then it will be a monic polynomial in $z$ and so we can use division algorithm again as a variable of $z.$ Where as if we consider it as a polynomial in $x$ or $y$ over $k[y,z]$ or over $k[x,z],$ then we will loose this advantage. But if a polynomial has the property that all of it's monomial contains at least two variables, then we have no general choice.

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