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Consider the ring $R[x][y]$. I am not sure that I know exactly what is going on here. A typical element of the ring is a polynomial in $y$ with coefficients that are themselves polynomials in $x$.

For example: $p=(x^2+ax+b)y^3+(x+c)y$ where $a,b,c \in R$.

So, the polynomial above is an element of $R[x][y]$, but am I allowed to distribute the $y^3$ and $y$ over each of the coefficients? Or is this not allowed by the ring $R[x][y]$?

Now, $(x^2+ax+b) \in R[x][y]$ and $y^3 \in R[x][y]$ and so $(x^2+ax+b) \cdot y^3=x^2y^3+axy^3+by^3 \in R[x][y]$ by distributivity in $R[x][y]$.

But the then does $x^2y^3+axy^3+by^3=(x^2+ax+b)y^3$?

In other words, is multiplication in $R[x][y]$ the operation that holds the coefficients in $R[x]$ and the unknowns in $y$ together? Or are the polynomials in $R[x][y]$ merely syntactical and can't be manipulated as above?

violet
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    Note that $R[x][y] = R[x, y]$. – Jonas Hardt Apr 15 '23 at 21:56
  • @JonasHardt are they equal or isomorphic? because the post below claims isomorphic – violet Apr 15 '23 at 22:07
  • You can do this even if the coefficients aren't polynomials. For example, you can write $x^2+x+1$ as $2x^2-x^2+3x-x-x-1+2$ if you want. – anon Apr 15 '23 at 22:08
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    They are canonically isomorphic which is pretty equal. –  Apr 15 '23 at 22:13
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    Some people define $R[x, y]$ to mean $R[x][y]$, while others don't. – Jonas Hardt Apr 15 '23 at 22:15
  • Such (re)distribution equalities hold in any commutive ring (or more generally for $x,y\in S\supset R$ such that $x$, $y$ and all $,r\in R$ all commute with each other - which is necessary (& sufficient) to rewrite to form $,\sum r_i x^{i_1} y^{i_2}$) – Bill Dubuque Apr 15 '23 at 22:21
  • I am not sure to understand your doubts. You write $$"(x^2+ax+b) \cdot y^3=x^2y^3+axy^3+by^3 $$ by distributivity " and then ask $$"x^2y^3+axy^3+by^3=(x^2+ax+b)y^3?".$$ You seem to believe there are two multiplications in $R[x][y].$ – Anne Bauval Apr 15 '23 at 22:25
  • This is what I am trying to ask: Let $$ be the multiplication operation in $R[x]$. Write an element out in $R[x]$: $r_nx^n+r_{n-1}x^{n-1}...+r_0$. By this, do we actually mean: $r_nx^n+r_{n-1}*x^{n-1}...+r_0$? Or is it the case that each monomial in the sum is merely syntactical: that is, there is no operation holding $x^i$ and a coefficient in $R$ together? – violet Apr 15 '23 at 22:45
  • @AnneBauval is this last question more clear? – violet Apr 15 '23 at 22:46
  • Not really. Now you are talking of $R[x],$ not of $R[x][y].$ But there is as well only one multiplication in that ring. You can use the $$ sign to denote it, or as usual no sign at all. So: you can write $rx*x,$ or as usual $rx^2.$ – Anne Bauval Apr 15 '23 at 22:54
  • I don't think anyone should define $R[x][y]$ to mean $R[x,y]$, as $S[X]$ should have a meaning for any commutative ring $S$ and and any set $X$. –  Apr 21 '23 at 13:02
  • Re: your elaborated question in the prior comment: you may find helpful this answer, which touches on syntax vs,. semantics of polynomial rings, and related matters. – Bill Dubuque Aug 07 '24 at 18:29

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It is a theorem that for any commutative ring $R$ there is an isomorphism $R[x][y]\cong R[x,y]\cong R[y][x]$ given by switching your perspective (polynomial in $y$ with coefficients polynomials $x$ is the same thing as a polynomial in $x$ with coefficients polynomials in $y$). This is a good supplementary point to your question, as not only is it valid to distribute and rearrange everything (which is permissible just by ring axioms) but you can formalise the idea that these rearrangements can turn any polynomial in $y$ into one in $x$, that there is essentially no difference.

The ring structure of $R[x][y]$ does permit something like $(x^2+1)y^2+y=x^2y^2+(y^2+y)$. All you have done there is use associativity and the distributivity of multiplication. $1\in R[x],x^2\in R[x]$, … there is no problem making that equality an equality in $R[x][y]$.

All of your equalities are valid equalities in $R[x][y]$. In asking: “is $x^2y^3+axy^3=(x^2+ax)y^3$ in $R[x][y]$” (where $a\in R$) you are asking: “is $cy^3+dy^3=(c+d)y^3$ in $K[y]$ where $c,d\in K$” (where $K=R[x]$ and $c=x^2,d=ax$) and yes, that is true by definition of $K[y]$.

FShrike
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  • The distributive property is in fact a ring axiom. – anon Apr 15 '23 at 22:09
  • The more precise version of what I meant is the ring isomorphisms, which are a theorem. @runway44 But I’ll edit to make what I meant clearer (distribute + rearrange + shift your perspective) – FShrike Apr 15 '23 at 22:10
  • @FShrike so when I write $p(x)y^n+... \in R[x][y]$ the "thing" holding $p(x)$ and $y^n$ together is the multiplication operation in $R[x][y]$ correct? – violet Apr 15 '23 at 22:21
  • @violet I guess..? I’m not sure what precisely you’re getting at by the “thing holding it together” but the ring structure of $R[x][y]$ is just that of $K[y]$ where $K$ is the ring $R[x]$. $p(x)y^n+q(x)y^m$ is just $ky^n+k’y^m$ for some coefficients $k,k’\in K$ and it makes sense in $K[y]$ by very definition of $K[y]$ – FShrike Apr 15 '23 at 22:24
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Every polynomial ring $S[y]$, over some commutative ring $S$, is itself a (commutative) ring and as such, satisfies the distributivity property $$(P(y)+Q(y))T(y)=P(y)T(y)+ Q(y)T(y),$$ e.g. $$(s_1+s_2)y^3=s_1y^3+s_2y^3$$ for $s_i\in S,$ and same for sums of three terms.

Apply this to $S=R[x]$ and $$(s_1,s_2,s_3)=(x^2,ax,b).$$

Anne Bauval
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